864

How can I load a Python module given its full path? Note that the file can be anywhere in the filesystem, as it is a configuration option.

edited Feb 16 '14 at 15:36

ferkulat

1,98421630

asked Sep 15 '08 at 22:30

derfred

5,43731821

add a comment |

864

How can I load a Python module given its full path? Note that the file can be anywhere in the filesystem, as it is a configuration option.

edited Feb 16 '14 at 15:36

ferkulat

1,98421630

asked Sep 15 '08 at 22:30

derfred

5,43731821

add a comment |

864

299

How can I load a Python module given its full path? Note that the file can be anywhere in the filesystem, as it is a configuration option.

edited Feb 16 '14 at 15:36

ferkulat

1,98421630

asked Sep 15 '08 at 22:30

derfred

5,43731821

How can I load a Python module given its full path? Note that the file can be anywhere in the filesystem, as it is a configuration option.

python configuration python-import python-module

edited Feb 16 '14 at 15:36

ferkulat

1,98421630

asked Sep 15 '08 at 22:30

derfred

5,43731821

edited Feb 16 '14 at 15:36

ferkulat

1,98421630

asked Sep 15 '08 at 22:30

derfred

5,43731821

edited Feb 16 '14 at 15:36

ferkulat

1,98421630

edited Feb 16 '14 at 15:36

ferkulat

1,98421630

edited Feb 16 '14 at 15:36

ferkulat

1,98421630

asked Sep 15 '08 at 22:30

derfred

5,43731821

asked Sep 15 '08 at 22:30

derfred

5,43731821

asked Sep 15 '08 at 22:30

derfred

5,43731821

add a comment |

26 Answers
26

active

oldest

votes

990

For Python 3.5+ use:

import importlib.util

spec = importlib.util.spec_from_file_location("module.name", "/path/to/file.py")

foo = importlib.util.module_from_spec(spec)

spec.loader.exec_module(foo)

foo.MyClass()

For Python 3.3 and 3.4 use:

from importlib.machinery import SourceFileLoader



foo = SourceFileLoader("module.name", "/path/to/file.py").load_module()

foo.MyClass()

(Although this has been deprecated in Python 3.4.)

Python 2 use:

import imp



foo = imp.load_source('module.name', '/path/to/file.py')

foo.MyClass()

There are equivalent convenience functions for compiled Python files and DLLs.

See also. http://bugs.python.org/issue21436.

edited Dec 7 '15 at 14:31

answered Sep 15 '08 at 22:41

Sebastian Rittau

12k32021

33

If I knew the namespace - 'module.name' - I would already use __import__.

– Sridhar Ratnakumar
Aug 10 '09 at 21:54

49

@SridharRatnakumar the value of the first argument of imp.load_source only sets the .__name__ of the returned module. it doesn't effect loading.

– Dan D.
Dec 14 '11 at 4:51

15

@DanD. — the first argument of imp.load_source() determines the key of the new entry created in the sys.modules dictionary, so the first argument does indeed affect loading.

– Brandon Rhodes
Apr 21 '13 at 16:32

8

The imp module is deprecated since version 3.4: The imp package is pending deprecation in favor of importlib.

– Chiel ten Brinke
Dec 8 '13 at 11:20

115

One might think that python imports are getting more and more complicated with each new version.

– AXO
Mar 8 '16 at 13:32

|
show 27 more comments

339

The advantage of adding a path to sys.path (over using imp) is that it simplifies things when importing more than one module from a single package. For example:

import sys

# the mock-0.3.1 dir contains testcase.py, testutils.py & mock.py

sys.path.append('/foo/bar/mock-0.3.1')



from testcase import TestCase

from testutils import RunTests

from mock import Mock, sentinel, patch

answered Sep 24 '08 at 19:36

Daryl Spitzer

51.9k59143162

7

How do we use sys.path.append to point to a single python file instead of a directory?

– Phani
Jan 13 '14 at 17:46

20

:-) Perhaps your question would be better suited as a StackOverflow question, not a comment on an answer.

– Daryl Spitzer
Mar 6 '15 at 0:12

2

To all people who were trying to include a file to their path... by definition "the shell path is a colon delimited list of directories". I'm relatively new to python, but the python path also follows the unix design principle from what I have seen. Please correct me if I am wrong.

– Michael Baptist
Apr 15 '15 at 5:37

2

The python path can contain zip archives, "eggs" (a complex kind of zip archives), etc. Modules can be imported out of them. So the path elements are indeed containers of files, but they are not necessarily directories.

– alexis
Apr 30 '15 at 21:21

7

Beware of the fact that Python caches import statements. In the rare case that you have two different folders sharing a single class name (classX), the approach of adding a path to sys.path, importing classX, removing the path and repeating for the reamaining paths won't work. Python will always load the class from the first path from its cache. In my case I aimed at creating a plugin system where all plugins implement a specific classX. I ended up using SourceFileLoader, note that its deprecation is controversial.

– ComFreek
Jul 3 '15 at 17:18

|
show 3 more comments

You can also do something like this and add the directory that the configuration file is sitting in to the Python load path, and then just do a normal import, assuming you know the name of the file in advance, in this case "config".

Messy, but it works.

configfile = '~/config.py'



import os

import sys



sys.path.append(os.path.dirname(os.path.expanduser(configfile)))



import config

edited Jan 17 '14 at 22:38

Peter Mortensen

13.6k1984111

answered Sep 15 '08 at 22:44

ctcherry

23.9k45165

add a comment |

It sounds like you don't want to specifically import the configuration file (which has a whole lot of side effects and additional complications involved), you just want to run it, and be able to access the resulting namespace. The standard library provides an API specifically for that in the form of runpy.run_path:

from runpy import run_path

settings = run_path("/path/to/file.py")

That interface is available in Python 2.7 and Python 3.2+

answered May 20 '16 at 6:52

ncoghlan

26.4k85466

I like this method but when I get the result of run_path its a dictionary which I cannot seem to access?

– Stephen Ellwood
Sep 11 '18 at 9:00

What do you mean by "cannot access"? You can't import from it (that's why this is only a good option when import-style access isn't actually required), but the contents should be available via the regular dict API (result[name], result.get('name', default_value), etc)

– ncoghlan
Sep 13 '18 at 3:16

add a comment |

You can use the

load_source(module_name, path_to_file)

method from imp module.

edited Nov 11 '14 at 7:07

twasbrillig

6,41242452

answered Sep 15 '08 at 22:41

zuber

2,52321917

... and imp.load_dynamic(module_name, path_to_file) for DLLs

– HEKTO
Dec 16 '15 at 19:03

26

heads up that imp is deprecated now.

– t1m0
Apr 6 '16 at 18:11

add a comment |

def import_file(full_path_to_module):

    try:

        import os

        module_dir, module_file = os.path.split(full_path_to_module)

        module_name, module_ext = os.path.splitext(module_file)

        save_cwd = os.getcwd()

        os.chdir(module_dir)

        module_obj = __import__(module_name)

        module_obj.__file__ = full_path_to_module

        globals()[module_name] = module_obj

        os.chdir(save_cwd)

    except:

        raise ImportError



import_file('/home/somebody/somemodule.py')

answered Sep 16 '08 at 1:43

Chris Calloway

1,0681712

36

Why write 14 lines of buggy code when this is already addressed by the standard library? You haven't done error checking on format or content of full_path_to_module or the os.whatever operations; and using a catch-all except: clause is rarely a good idea.

– Chris Johnson
Jun 7 '13 at 19:17

You should use more "try-finally"s in here. E.g. save_cwd = os.getcwd() try: … finally: os.chdir(save_cwd)

– kay
Sep 21 '14 at 1:33

9

@ChrisJohnson this is already addressed by the standard library yeah, but python has nasty habit of not being backward-compatible... as the checked answer says there're 2 different ways before and after 3.3. In that case I'd rather like to write my own universal function than check version on the fly. And yes, maybe this code isn't too well error-protected, but it shows an idea (which is os.chdir(), I haven't though about it), basing on which I can write a better code. Hence +1.

– Sushi271
May 15 '15 at 10:27

add a comment |

Here is some code that works in all Python versions, from 2.7-3.5 and probably even others.

config_file = "/tmp/config.py"

with open(config_file) as f:

    code = compile(f.read(), config_file, 'exec')

    exec(code, globals(), locals())

I tested it. It may be ugly but so far is the only one that works in all versions.

answered Jun 3 '16 at 10:04

sorin

74.7k116370579

This answer worked for me where load_source did not because it imports the script and provides the script access to the modules and globals at the time of importing.

– Klik
Nov 22 '17 at 19:13

add a comment |

I have come up with a slightly modified version of @SebastianRittau's wonderful answer (for Python > 3.4 I think), which will allow you to load a file with any extension as a module using spec_from_loader instead of spec_from_file_location:

from importlib.util import spec_from_loader, module_from_spec

from importlib.machinery import SourceFileLoader 



spec = spec_from_loader("module.name", SourceFileLoader("module.name", "/path/to/file.py"))

mod = module_from_spec(spec)

spec.loader.exec_module(mod)

The advantage of encoding the path in an explicit SourceFileLoader is that the machinery will not try to figure out the type of the file from the extension. This means that you can load something like a .txt file using this method, but you could not do it with spec_from_file_location without specifying the loader because .txt is not in importlib.machinery.SOURCE_SUFFIXES.

edited Sep 19 '18 at 18:13

answered Apr 25 '17 at 5:45

Mad Physicist

36.4k1671101

add a comment |

Do you mean load or import?

You can manipulate the sys.path list specify the path to your module, then import your module. For example, given a module at:

/foo/bar.py

You could do:

import sys

sys.path[0:0] = ['/foo'] # puts the /foo directory at the start of your path

import bar

edited Dec 2 '18 at 1:05

Jason Orendorff

30.8k34884

answered Sep 15 '08 at 22:46

Wheat

63239

@Wheat Why sys.path[0:0] instead of sys.path[0]?

– user618677
Jan 9 '12 at 6:56

4

B/c sys.path[0] = xy overwrites the first path item while path[0:0] =xy is equivalent to path.insert(0, xy)

– dom0
Nov 15 '12 at 14:16

2

hm the path.insert worked for me but the [0:0] trick did not.

– jsh
Sep 30 '13 at 3:18

10

sys.path[0:0] = ['/foo']

– Kevin Edwards
Apr 1 '15 at 17:00

add a comment |

I believe you can use imp.find_module() and imp.load_module() to load the specified module. You'll need to split the module name off of the path, i.e. if you wanted to load /home/mypath/mymodule.py you'd need to do:

imp.find_module('mymodule', '/home/mypath/')

...but that should get the job done.

edited Jan 2 '15 at 11:08

Mathieu Rodic

4,72022940

answered Sep 15 '08 at 22:37

Matt

1313

add a comment |

If your top-level module is not a file but is packaged as a directory with __init__.py, then the accepted solution almost works, but not quite. In Python 3.5+ the following code is needed (note the added line that begins with 'sys.modules'):

MODULE_PATH = "/path/to/your/module/__init__.py"

MODULE_NAME = "mymodule"

spec = importlib.util.spec_from_file_location(MODULE_NAME, MODULE_PATH)

module = importlib.util.module_from_spec(spec)

sys.modules[spec.name] = module 

spec.loader.exec_module(module)

Without this line, when exec_module is executed, it tries to bind relative imports in your top level __init__.py to the top level module name -- in this case "mymodule". But "mymodule" isn't loaded yet so you'll get the error "SystemError: Parent module 'mymodule' not loaded, cannot perform relative import". So you need to bind the name before you load it. The reason for this is the fundamental invariant of the relative import system: "The invariant holding is that if you have sys.modules['spam'] and sys.modules['spam.foo'] (as you would after the above import), the latter must appear as the foo attribute of the former" as discussed here.

edited Oct 17 '18 at 16:49

Hadrien TOMA

8151221

answered May 17 '18 at 15:23

Sam Grondahl

1,01611223

1

This is exactly what I was missing! Thanks! (This also correctly answers Proper way to dynamically import a module with relative imports?)

– mforbes
Jul 14 '18 at 7:30

Any idea how to do this with python 2?

– mforbes
Jul 16 '18 at 0:57

add a comment |

To import your module, you need to add its directory to the environment variable, either temporarily or permanently.

Temporarily

import sys

sys.path.append("/path/to/my/modules/")

import my_module

Permanently

Adding the following line to your .bashrc file (in linux) and excecute source ~/.bashrc in the terminal:

export PYTHONPATH="${PYTHONPATH}:/path/to/my/modules/"

Credit/Source: saarrrr, another stackexchange question

answered Nov 15 '18 at 2:30

Miladiouss

527614

This "temp" solution is a great answer if you want to prod a project around in a jupyter notebook elsewhere.

– fordy
Nov 16 '18 at 17:20

add a comment |

This should work

path = os.path.join('./path/to/folder/with/py/files', '*.py')

for infile in glob.glob(path):

    basename = os.path.basename(infile)

    basename_without_extension = basename[:-3]



    # http://docs.python.org/library/imp.html?highlight=imp#module-imp

    imp.load_source(basename_without_extension, infile)

edited Jan 4 '12 at 4:28

joran

134k19320380

answered Jan 4 '12 at 2:17

Hengjie

3,5512433

4

A more general way to cut the extension out is: name, ext = os.path.splitext(os.path.basename(infile)). Your method works because the previous restriction to .py extension. Also, you should probably import the module to some variable/dictionary entry.

– ReneSac
Dec 6 '12 at 13:16

add a comment |

This area of Python 3.4 seems to be extremely tortuous to understand! However with a bit of hacking using the code from Chris Calloway as a start I managed to get something working. Here's the basic function.

def import_module_from_file(full_path_to_module):

    """

    Import a module given the full path/filename of the .py file



    Python 3.4



    """



    module = None



    try:



        # Get module name and path from full path

        module_dir, module_file = os.path.split(full_path_to_module)

        module_name, module_ext = os.path.splitext(module_file)



        # Get module "spec" from filename

        spec = importlib.util.spec_from_file_location(module_name,full_path_to_module)



        module = spec.loader.load_module()



    except Exception as ec:

        # Simple error printing

        # Insert "sophisticated" stuff here

        print(ec)



    finally:

        return module

This appears to use non-deprecated modules from Python 3.4. I don't pretend to understand why, but it seems to work from within a program. I found Chris' solution worked on the command line but not from inside a program.

answered Apr 12 '15 at 12:22

Redlegjed

311

add a comment |

I'm not saying that it is better, but for the sake of completeness, I wanted to suggest the exec function, available in both python 2 and 3.
exec allows you to execute arbitrary code in either the global scope, or in an internal scope, provided as a dictionary.

For example, if you have a module stored in "/path/to/module" with the function foo(), you could run it by doing the following:

module = dict()

with open("/path/to/module") as f:

    exec(f.read(), module)

module['foo']()

This makes it a bit more explicit that you're loading code dynamically, and grants you some additional power, such as the ability to provide custom builtins.

And if having access through attributes, instead of keys is important to you, you can design a custom dict class for the globals, that provides such access, e.g.:

class MyModuleClass(dict):

    def __getattr__(self, name):

        return self.__getitem__(name)

answered Jun 2 '15 at 19:57

yoniLavi

1,3561622

2

execfile(), also

– crowder
Jun 20 '15 at 4:13

add a comment |

To import a module from a given filename, you can temporarily extend the path, and restore the system path in the finally block reference:

filename = "directory/module.py"



directory, module_name = os.path.split(filename)

module_name = os.path.splitext(module_name)[0]



path = list(sys.path)

sys.path.insert(0, directory)

try:

    module = __import__(module_name)

finally:

    sys.path[:] = path # restore

answered Oct 2 '15 at 11:14

Peter Zhu

704921

add a comment |

I made a package that uses imp for you. I call it import_file and this is how it's used:

>>>from import_file import import_file

>>>mylib = import_file('c:\mylib.py')

>>>another = import_file('relative_subdir/another.py')

You can get it at:

http://pypi.python.org/pypi/import_file

or at

http://code.google.com/p/import-file/

answered Jun 8 '11 at 19:41

ubershmekel

5,45824665

1

os.chdir ? (minimal characters to approve comment).

– ychaouche
Oct 14 '12 at 10:46

I've spent all day troubleshooting an import bug in a pyinstaller generated exe. In the end this is the only thing that worked for me. Thank you so much for making this!

– frakman1
Nov 29 '18 at 22:12

add a comment |

Import package modules at runtime (Python recipe)

http://code.activestate.com/recipes/223972/

###################

##                #

## classloader.py #

##                #

###################



import sys, types



def _get_mod(modulePath):

    try:

        aMod = sys.modules[modulePath]

        if not isinstance(aMod, types.ModuleType):

            raise KeyError

    except KeyError:

        # The last [''] is very important!

        aMod = __import__(modulePath, globals(), locals(), [''])

        sys.modules[modulePath] = aMod

    return aMod



def _get_func(fullFuncName):

    """Retrieve a function object from a full dotted-package name."""



    # Parse out the path, module, and function

    lastDot = fullFuncName.rfind(u".")

    funcName = fullFuncName[lastDot + 1:]

    modPath = fullFuncName[:lastDot]



    aMod = _get_mod(modPath)

    aFunc = getattr(aMod, funcName)



    # Assert that the function is a *callable* attribute.

    assert callable(aFunc), u"%s is not callable." % fullFuncName



    # Return a reference to the function itself,

    # not the results of the function.

    return aFunc



def _get_class(fullClassName, parentClass=None):

    """Load a module and retrieve a class (NOT an instance).



    If the parentClass is supplied, className must be of parentClass

    or a subclass of parentClass (or None is returned).

    """

    aClass = _get_func(fullClassName)



    # Assert that the class is a subclass of parentClass.

    if parentClass is not None:

        if not issubclass(aClass, parentClass):

            raise TypeError(u"%s is not a subclass of %s" %

                            (fullClassName, parentClass))



    # Return a reference to the class itself, not an instantiated object.

    return aClass





######################

##       Usage      ##

######################



class StorageManager: pass

class StorageManagerMySQL(StorageManager): pass



def storage_object(aFullClassName, allOptions={}):

    aStoreClass = _get_class(aFullClassName, StorageManager)

    return aStoreClass(allOptions)

edited Dec 22 '13 at 2:17

Eric Leschinski

87.1k38321274

answered Sep 15 '08 at 22:43

user10370

371

add a comment |

You can use the pkgutil module (specifically the walk_packages method) to get a list of the packages in the current directory. From there it's trivial to use the importlib machinery to import the modules you want:

import pkgutil

import importlib



packages = pkgutil.walk_packages(path='.')

for importer, name, is_package in packages:

    mod = importlib.import_module(name)

    # do whatever you want with module now, it's been imported!

edited Jan 2 '15 at 11:04

Mathieu Rodic

4,72022940

answered Sep 13 '14 at 19:57

bob_twinkles

1869

add a comment |

In Linux, adding a symbolic link in the directory your python script is located works.

ie:

ln -s /absolute/path/to/module/module.py /absolute/path/to/script/module.py

python will create /absolute/path/to/script/module.pyc and will update it if you change the contents of /absolute/path/to/module/module.py

then include the following in mypythonscript.py

from module import *

edited Apr 11 '18 at 17:21

Micah Smith

2,6211422

answered Nov 18 '14 at 13:06

user2760152

234

1

This is the hack I used, and it has caused me some problems. One of the more painful ones was that IDEA has an issue where it doesn't pickup altered code from within the link, but yet attempts to save what it thinks is there. A race condition where the last to save is what sticks... I lost a decent amount of work because of this.

– Gripp
Jun 16 '15 at 23:23

@Gripp not sure if I am understanding your issue, but I frequently (almost exclusively) edit my scripts on a remote server from my desktop via SFTP with a client like CyberDuck, and in that case as well it is a bad idea to try and edit the symlinked file, instead its much safer to edit the original file. You can catch some of these issues by using git and checking your git status to verify that your changes to the script are actually making it back to the source document and not getting lost in the ether.

– user5359531
Aug 1 '17 at 19:39

add a comment |

quite simple way: suppose you want import file with relative path ../../MyLibs/pyfunc.py



libPath = '../../MyLibs'

import sys

if not libPath in sys.path: sys.path.append(libPath)

import pyfunc as pf

But if you make it without a guard you can finally get a very long path

answered Jan 26 '18 at 4:52

Andrei Keino

612

add a comment |

A simple solution using importlib instead of the imp package (tested for Python 2.7, although it should work for Python 3 too):

import importlib



dirname, basename = os.path.split(pyfilepath) # pyfilepath: '/my/path/mymodule.py'

sys.path.append(dirname) # only directories should be added to PYTHONPATH

module_name = os.path.splitext(basename)[0] # '/my/path/mymodule.py' --> 'mymodule'

module = importlib.import_module(module_name) # name space of defined module (otherwise we would literally look for "module_name")

Now you can directly use the namespace of the imported module, like this:

a = module.myvar

b = module.myfunc(a)

The advantage of this solution is that we don't even need to know the actual name of the module we would like to import, in order to use it in our code. This is useful, e.g. in case the path of the module is a configurable argument.

edited Jul 28 '18 at 2:24

answered May 24 '18 at 12:07

Ataxias

175110

This way you are modifying the sys.path, which does not fit every use case.

– bgusach
Jul 19 '18 at 14:26

@bgusach This may be true, but it is also desirable in some cases (adding a path to sys.path simplifies things when importing more than one module from a single package). At any rate, if this not desirable, one can immediately afterwards do sys.path.pop()

– Ataxias
Jul 20 '18 at 16:38

add a comment |

Create python module test.py

import sys

sys.path.append("<project-path>/lib/")

from tes1 import Client1

from tes2 import Client2

import tes3

Create python module test_check.py

from test import Client1

from test import Client2

from test import test3

We can import the imported module from module.

answered Dec 6 '18 at 12:41

abhimanyu

285

add a comment |

Adding this to the list of answers as I couldn't find anything that worked. This will allow imports of compiled (pyd) python modules in 3.4:

import sys

import importlib.machinery



def load_module(name, filename):

    # If the Loader finds the module name in this list it will use

    # module_name.__file__ instead so we need to delete it here

    if name in sys.modules:

        del sys.modules[name]

    loader = importlib.machinery.ExtensionFileLoader(name, filename)

    module = loader.load_module()

    locals()[name] = module

    globals()[name] = module



load_module('something', r'C:PathTosomething.pyd')

something.do_something()

answered Jan 10 '18 at 15:58

David

314115

add a comment |

This answer is a supplement to Sebastian Rittau's answer responding to the comment: "but what if you don't have the module name?" This is a quick and dirty way of getting the likely python module name given a filename -- it just goes up the tree until it finds a directory without an __init__.py file and then turns it back into a filename. For Python 3.4+ (uses pathlib), which makes sense since Py2 people can use "imp" or other ways of doing relative imports:

import pathlib



def likely_python_module(filename):

    '''

    Given a filename or Path, return the "likely" python module name.  That is, iterate

    the parent directories until it doesn't contain an __init__.py file.



    :rtype: str

    '''

    p = pathlib.Path(filename).resolve()

    paths = 

    if p.name != '__init__.py':

        paths.append(p.stem)

    while True:

        p = p.parent

        if not p:

            break

        if not p.is_dir():

            break



        inits = [f for f in p.iterdir() if f.name == '__init__.py']

        if not inits:

            break



        paths.append(p.stem)



    return '.'.join(reversed(paths))

There are certainly possibilities for improvement, and the optional __init__.py files might necessitate other changes, but if you have __init__.py in general, this does the trick.

answered Sep 8 '18 at 15:29

Michael Scott Cuthbert

1,83121333

add a comment |

-1

The best way, I think, is from the official documentation (29.1. imp — Access the import internals):

import imp

import sys



def __import__(name, globals=None, locals=None, fromlist=None):

    # Fast path: see if the module has already been imported.

    try:

        return sys.modules[name]

    except KeyError:

        pass



    # If any of the following calls raises an exception,

    # there's a problem we can't handle -- let the caller handle it.



    fp, pathname, description = imp.find_module(name)



    try:

        return imp.load_module(name, fp, pathname, description)

    finally:

        # Since we may exit via an exception, close fp explicitly.

        if fp:

            fp.close()

answered Nov 25 '14 at 12:58

Zompa

111

1

This solution does not allow you to provide the path, which is what the question asks for.

– Micah Smith
Apr 11 '18 at 17:20

add a comment |

protected by tripleee Jul 11 '15 at 12:01

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26 Answers
26

active

oldest

votes

26 Answers
26

active

oldest

votes

990

For Python 3.5+ use:

import importlib.util

spec = importlib.util.spec_from_file_location("module.name", "/path/to/file.py")

foo = importlib.util.module_from_spec(spec)

spec.loader.exec_module(foo)

foo.MyClass()

For Python 3.3 and 3.4 use:

from importlib.machinery import SourceFileLoader



foo = SourceFileLoader("module.name", "/path/to/file.py").load_module()

foo.MyClass()

(Although this has been deprecated in Python 3.4.)

Python 2 use:

import imp



foo = imp.load_source('module.name', '/path/to/file.py')

foo.MyClass()

There are equivalent convenience functions for compiled Python files and DLLs.

See also. http://bugs.python.org/issue21436.

edited Dec 7 '15 at 14:31

answered Sep 15 '08 at 22:41

Sebastian Rittau

12k32021

33

If I knew the namespace - 'module.name' - I would already use __import__.

– Sridhar Ratnakumar
Aug 10 '09 at 21:54

49

@SridharRatnakumar the value of the first argument of imp.load_source only sets the .__name__ of the returned module. it doesn't effect loading.

– Dan D.
Dec 14 '11 at 4:51

15

@DanD. — the first argument of imp.load_source() determines the key of the new entry created in the sys.modules dictionary, so the first argument does indeed affect loading.

– Brandon Rhodes
Apr 21 '13 at 16:32

8

The imp module is deprecated since version 3.4: The imp package is pending deprecation in favor of importlib.

– Chiel ten Brinke
Dec 8 '13 at 11:20

115

One might think that python imports are getting more and more complicated with each new version.

– AXO
Mar 8 '16 at 13:32

|
show 27 more comments

990

For Python 3.5+ use:

import importlib.util

spec = importlib.util.spec_from_file_location("module.name", "/path/to/file.py")

foo = importlib.util.module_from_spec(spec)

spec.loader.exec_module(foo)

foo.MyClass()

For Python 3.3 and 3.4 use:

from importlib.machinery import SourceFileLoader



foo = SourceFileLoader("module.name", "/path/to/file.py").load_module()

foo.MyClass()

(Although this has been deprecated in Python 3.4.)

Python 2 use:

import imp



foo = imp.load_source('module.name', '/path/to/file.py')

foo.MyClass()

There are equivalent convenience functions for compiled Python files and DLLs.

See also. http://bugs.python.org/issue21436.

edited Dec 7 '15 at 14:31

answered Sep 15 '08 at 22:41

Sebastian Rittau

12k32021

33

If I knew the namespace - 'module.name' - I would already use __import__.

– Sridhar Ratnakumar
Aug 10 '09 at 21:54

49

@SridharRatnakumar the value of the first argument of imp.load_source only sets the .__name__ of the returned module. it doesn't effect loading.

– Dan D.
Dec 14 '11 at 4:51

15

@DanD. — the first argument of imp.load_source() determines the key of the new entry created in the sys.modules dictionary, so the first argument does indeed affect loading.

– Brandon Rhodes
Apr 21 '13 at 16:32

8

The imp module is deprecated since version 3.4: The imp package is pending deprecation in favor of importlib.

– Chiel ten Brinke
Dec 8 '13 at 11:20

115

One might think that python imports are getting more and more complicated with each new version.

– AXO
Mar 8 '16 at 13:32

|
show 27 more comments

990

For Python 3.5+ use:

import importlib.util

spec = importlib.util.spec_from_file_location("module.name", "/path/to/file.py")

foo = importlib.util.module_from_spec(spec)

spec.loader.exec_module(foo)

foo.MyClass()

For Python 3.3 and 3.4 use:

from importlib.machinery import SourceFileLoader



foo = SourceFileLoader("module.name", "/path/to/file.py").load_module()

foo.MyClass()

(Although this has been deprecated in Python 3.4.)

Python 2 use:

import imp



foo = imp.load_source('module.name', '/path/to/file.py')

foo.MyClass()

There are equivalent convenience functions for compiled Python files and DLLs.

See also. http://bugs.python.org/issue21436.

edited Dec 7 '15 at 14:31

answered Sep 15 '08 at 22:41

Sebastian Rittau

12k32021

For Python 3.5+ use:

import importlib.util

spec = importlib.util.spec_from_file_location("module.name", "/path/to/file.py")

foo = importlib.util.module_from_spec(spec)

spec.loader.exec_module(foo)

foo.MyClass()

For Python 3.3 and 3.4 use:

from importlib.machinery import SourceFileLoader



foo = SourceFileLoader("module.name", "/path/to/file.py").load_module()

foo.MyClass()

(Although this has been deprecated in Python 3.4.)

Python 2 use:

import imp



foo = imp.load_source('module.name', '/path/to/file.py')

foo.MyClass()

There are equivalent convenience functions for compiled Python files and DLLs.

See also. http://bugs.python.org/issue21436.

edited Dec 7 '15 at 14:31

answered Sep 15 '08 at 22:41

Sebastian Rittau

12k32021

edited Dec 7 '15 at 14:31

answered Sep 15 '08 at 22:41

Sebastian Rittau

12k32021

answered Sep 15 '08 at 22:41

Sebastian Rittau

12k32021

answered Sep 15 '08 at 22:41

Sebastian Rittau

12k32021

33

If I knew the namespace - 'module.name' - I would already use __import__.

– Sridhar Ratnakumar
Aug 10 '09 at 21:54

49

@SridharRatnakumar the value of the first argument of imp.load_source only sets the .__name__ of the returned module. it doesn't effect loading.

– Dan D.
Dec 14 '11 at 4:51

15

@DanD. — the first argument of imp.load_source() determines the key of the new entry created in the sys.modules dictionary, so the first argument does indeed affect loading.

– Brandon Rhodes
Apr 21 '13 at 16:32

8

The imp module is deprecated since version 3.4: The imp package is pending deprecation in favor of importlib.

– Chiel ten Brinke
Dec 8 '13 at 11:20

115

One might think that python imports are getting more and more complicated with each new version.

– AXO
Mar 8 '16 at 13:32

|
show 27 more comments

33

If I knew the namespace - 'module.name' - I would already use __import__.

– Sridhar Ratnakumar
Aug 10 '09 at 21:54

49

@SridharRatnakumar the value of the first argument of imp.load_source only sets the .__name__ of the returned module. it doesn't effect loading.

– Dan D.
Dec 14 '11 at 4:51

15

@DanD. — the first argument of imp.load_source() determines the key of the new entry created in the sys.modules dictionary, so the first argument does indeed affect loading.

– Brandon Rhodes
Apr 21 '13 at 16:32

8

The imp module is deprecated since version 3.4: The imp package is pending deprecation in favor of importlib.

– Chiel ten Brinke
Dec 8 '13 at 11:20

115

One might think that python imports are getting more and more complicated with each new version.

– AXO
Mar 8 '16 at 13:32

If I knew the namespace - 'module.name' - I would already use __import__.

– Sridhar Ratnakumar
Aug 10 '09 at 21:54

@SridharRatnakumar the value of the first argument of imp.load_source only sets the .__name__ of the returned module. it doesn't effect loading.

– Dan D.
Dec 14 '11 at 4:51

@DanD. — the first argument of imp.load_source() determines the key of the new entry created in the sys.modules dictionary, so the first argument does indeed affect loading.

– Brandon Rhodes
Apr 21 '13 at 16:32

The imp module is deprecated since version 3.4: The imp package is pending deprecation in favor of importlib.

– Chiel ten Brinke
Dec 8 '13 at 11:20

115

One might think that python imports are getting more and more complicated with each new version.

– AXO
Mar 8 '16 at 13:32

|
show 27 more comments

339

The advantage of adding a path to sys.path (over using imp) is that it simplifies things when importing more than one module from a single package. For example:

import sys

# the mock-0.3.1 dir contains testcase.py, testutils.py & mock.py

sys.path.append('/foo/bar/mock-0.3.1')



from testcase import TestCase

from testutils import RunTests

from mock import Mock, sentinel, patch

answered Sep 24 '08 at 19:36

Daryl Spitzer

51.9k59143162

7

How do we use sys.path.append to point to a single python file instead of a directory?

– Phani
Jan 13 '14 at 17:46

20

:-) Perhaps your question would be better suited as a StackOverflow question, not a comment on an answer.

– Daryl Spitzer
Mar 6 '15 at 0:12

2

To all people who were trying to include a file to their path... by definition "the shell path is a colon delimited list of directories". I'm relatively new to python, but the python path also follows the unix design principle from what I have seen. Please correct me if I am wrong.

– Michael Baptist
Apr 15 '15 at 5:37

2

The python path can contain zip archives, "eggs" (a complex kind of zip archives), etc. Modules can be imported out of them. So the path elements are indeed containers of files, but they are not necessarily directories.

– alexis
Apr 30 '15 at 21:21

7

Beware of the fact that Python caches import statements. In the rare case that you have two different folders sharing a single class name (classX), the approach of adding a path to sys.path, importing classX, removing the path and repeating for the reamaining paths won't work. Python will always load the class from the first path from its cache. In my case I aimed at creating a plugin system where all plugins implement a specific classX. I ended up using SourceFileLoader, note that its deprecation is controversial.

– ComFreek
Jul 3 '15 at 17:18

|
show 3 more comments

339

The advantage of adding a path to sys.path (over using imp) is that it simplifies things when importing more than one module from a single package. For example:

import sys

# the mock-0.3.1 dir contains testcase.py, testutils.py & mock.py

sys.path.append('/foo/bar/mock-0.3.1')



from testcase import TestCase

from testutils import RunTests

from mock import Mock, sentinel, patch

answered Sep 24 '08 at 19:36

Daryl Spitzer

51.9k59143162

7

How do we use sys.path.append to point to a single python file instead of a directory?

– Phani
Jan 13 '14 at 17:46

20

:-) Perhaps your question would be better suited as a StackOverflow question, not a comment on an answer.

– Daryl Spitzer
Mar 6 '15 at 0:12

2

To all people who were trying to include a file to their path... by definition "the shell path is a colon delimited list of directories". I'm relatively new to python, but the python path also follows the unix design principle from what I have seen. Please correct me if I am wrong.

– Michael Baptist
Apr 15 '15 at 5:37

2

The python path can contain zip archives, "eggs" (a complex kind of zip archives), etc. Modules can be imported out of them. So the path elements are indeed containers of files, but they are not necessarily directories.

– alexis
Apr 30 '15 at 21:21

7

Beware of the fact that Python caches import statements. In the rare case that you have two different folders sharing a single class name (classX), the approach of adding a path to sys.path, importing classX, removing the path and repeating for the reamaining paths won't work. Python will always load the class from the first path from its cache. In my case I aimed at creating a plugin system where all plugins implement a specific classX. I ended up using SourceFileLoader, note that its deprecation is controversial.

– ComFreek
Jul 3 '15 at 17:18

|
show 3 more comments

339

The advantage of adding a path to sys.path (over using imp) is that it simplifies things when importing more than one module from a single package. For example:

import sys

# the mock-0.3.1 dir contains testcase.py, testutils.py & mock.py

sys.path.append('/foo/bar/mock-0.3.1')



from testcase import TestCase

from testutils import RunTests

from mock import Mock, sentinel, patch

answered Sep 24 '08 at 19:36

Daryl Spitzer

51.9k59143162

The advantage of adding a path to sys.path (over using imp) is that it simplifies things when importing more than one module from a single package. For example:

import sys

# the mock-0.3.1 dir contains testcase.py, testutils.py & mock.py

sys.path.append('/foo/bar/mock-0.3.1')



from testcase import TestCase

from testutils import RunTests

from mock import Mock, sentinel, patch

answered Sep 24 '08 at 19:36

Daryl Spitzer

51.9k59143162

answered Sep 24 '08 at 19:36

Daryl Spitzer

51.9k59143162

answered Sep 24 '08 at 19:36

Daryl Spitzer

51.9k59143162

answered Sep 24 '08 at 19:36

Daryl Spitzer

51.9k59143162

7

How do we use sys.path.append to point to a single python file instead of a directory?

– Phani
Jan 13 '14 at 17:46

20

:-) Perhaps your question would be better suited as a StackOverflow question, not a comment on an answer.

– Daryl Spitzer
Mar 6 '15 at 0:12

2

To all people who were trying to include a file to their path... by definition "the shell path is a colon delimited list of directories". I'm relatively new to python, but the python path also follows the unix design principle from what I have seen. Please correct me if I am wrong.

– Michael Baptist
Apr 15 '15 at 5:37

2

The python path can contain zip archives, "eggs" (a complex kind of zip archives), etc. Modules can be imported out of them. So the path elements are indeed containers of files, but they are not necessarily directories.

– alexis
Apr 30 '15 at 21:21

7

Beware of the fact that Python caches import statements. In the rare case that you have two different folders sharing a single class name (classX), the approach of adding a path to sys.path, importing classX, removing the path and repeating for the reamaining paths won't work. Python will always load the class from the first path from its cache. In my case I aimed at creating a plugin system where all plugins implement a specific classX. I ended up using SourceFileLoader, note that its deprecation is controversial.

– ComFreek
Jul 3 '15 at 17:18

|
show 3 more comments

7

How do we use sys.path.append to point to a single python file instead of a directory?

– Phani
Jan 13 '14 at 17:46

20

:-) Perhaps your question would be better suited as a StackOverflow question, not a comment on an answer.

– Daryl Spitzer
Mar 6 '15 at 0:12

2

To all people who were trying to include a file to their path... by definition "the shell path is a colon delimited list of directories". I'm relatively new to python, but the python path also follows the unix design principle from what I have seen. Please correct me if I am wrong.

– Michael Baptist
Apr 15 '15 at 5:37

2

The python path can contain zip archives, "eggs" (a complex kind of zip archives), etc. Modules can be imported out of them. So the path elements are indeed containers of files, but they are not necessarily directories.

– alexis
Apr 30 '15 at 21:21

7

Beware of the fact that Python caches import statements. In the rare case that you have two different folders sharing a single class name (classX), the approach of adding a path to sys.path, importing classX, removing the path and repeating for the reamaining paths won't work. Python will always load the class from the first path from its cache. In my case I aimed at creating a plugin system where all plugins implement a specific classX. I ended up using SourceFileLoader, note that its deprecation is controversial.

– ComFreek
Jul 3 '15 at 17:18

How do we use sys.path.append to point to a single python file instead of a directory?

– Phani
Jan 13 '14 at 17:46

:-) Perhaps your question would be better suited as a StackOverflow question, not a comment on an answer.

– Daryl Spitzer
Mar 6 '15 at 0:12

To all people who were trying to include a file to their path... by definition "the shell path is a colon delimited list of directories". I'm relatively new to python, but the python path also follows the unix design principle from what I have seen. Please correct me if I am wrong.

– Michael Baptist
Apr 15 '15 at 5:37

The python path can contain zip archives, "eggs" (a complex kind of zip archives), etc. Modules can be imported out of them. So the path elements are indeed containers of files, but they are not necessarily directories.

– alexis
Apr 30 '15 at 21:21

Beware of the fact that Python caches import statements. In the rare case that you have two different folders sharing a single class name (classX), the approach of adding a path to sys.path, importing classX, removing the path and repeating for the reamaining paths won't work. Python will always load the class from the first path from its cache. In my case I aimed at creating a plugin system where all plugins implement a specific classX. I ended up using SourceFileLoader, note that its deprecation is controversial.

– ComFreek
Jul 3 '15 at 17:18

|
show 3 more comments

Messy, but it works.

configfile = '~/config.py'



import os

import sys



sys.path.append(os.path.dirname(os.path.expanduser(configfile)))



import config

edited Jan 17 '14 at 22:38

Peter Mortensen

13.6k1984111

answered Sep 15 '08 at 22:44

ctcherry

23.9k45165

add a comment |

Messy, but it works.

configfile = '~/config.py'



import os

import sys



sys.path.append(os.path.dirname(os.path.expanduser(configfile)))



import config

edited Jan 17 '14 at 22:38

Peter Mortensen

13.6k1984111

answered Sep 15 '08 at 22:44

ctcherry

23.9k45165

add a comment |

Messy, but it works.

configfile = '~/config.py'



import os

import sys



sys.path.append(os.path.dirname(os.path.expanduser(configfile)))



import config

edited Jan 17 '14 at 22:38

Peter Mortensen

13.6k1984111

answered Sep 15 '08 at 22:44

ctcherry

23.9k45165

Messy, but it works.

configfile = '~/config.py'



import os

import sys



sys.path.append(os.path.dirname(os.path.expanduser(configfile)))



import config

edited Jan 17 '14 at 22:38

Peter Mortensen

13.6k1984111

answered Sep 15 '08 at 22:44

ctcherry

23.9k45165

edited Jan 17 '14 at 22:38

Peter Mortensen

13.6k1984111

edited Jan 17 '14 at 22:38

Peter Mortensen

13.6k1984111

edited Jan 17 '14 at 22:38

Peter Mortensen

13.6k1984111

answered Sep 15 '08 at 22:44

ctcherry

23.9k45165

answered Sep 15 '08 at 22:44

ctcherry

23.9k45165

answered Sep 15 '08 at 22:44

ctcherry

23.9k45165

add a comment |

from runpy import run_path

settings = run_path("/path/to/file.py")

That interface is available in Python 2.7 and Python 3.2+

answered May 20 '16 at 6:52

ncoghlan

26.4k85466

I like this method but when I get the result of run_path its a dictionary which I cannot seem to access?

– Stephen Ellwood
Sep 11 '18 at 9:00

What do you mean by "cannot access"? You can't import from it (that's why this is only a good option when import-style access isn't actually required), but the contents should be available via the regular dict API (result[name], result.get('name', default_value), etc)

– ncoghlan
Sep 13 '18 at 3:16

add a comment |

from runpy import run_path

settings = run_path("/path/to/file.py")

That interface is available in Python 2.7 and Python 3.2+

answered May 20 '16 at 6:52

ncoghlan

26.4k85466

I like this method but when I get the result of run_path its a dictionary which I cannot seem to access?

– Stephen Ellwood
Sep 11 '18 at 9:00

What do you mean by "cannot access"? You can't import from it (that's why this is only a good option when import-style access isn't actually required), but the contents should be available via the regular dict API (result[name], result.get('name', default_value), etc)

– ncoghlan
Sep 13 '18 at 3:16

add a comment |

from runpy import run_path

settings = run_path("/path/to/file.py")

That interface is available in Python 2.7 and Python 3.2+

answered May 20 '16 at 6:52

ncoghlan

26.4k85466

from runpy import run_path

settings = run_path("/path/to/file.py")

That interface is available in Python 2.7 and Python 3.2+

answered May 20 '16 at 6:52

ncoghlan

26.4k85466

answered May 20 '16 at 6:52

ncoghlan

26.4k85466

answered May 20 '16 at 6:52

ncoghlan

26.4k85466

answered May 20 '16 at 6:52

ncoghlan

26.4k85466

I like this method but when I get the result of run_path its a dictionary which I cannot seem to access?

– Stephen Ellwood
Sep 11 '18 at 9:00

What do you mean by "cannot access"? You can't import from it (that's why this is only a good option when import-style access isn't actually required), but the contents should be available via the regular dict API (result[name], result.get('name', default_value), etc)

– ncoghlan
Sep 13 '18 at 3:16

add a comment |

I like this method but when I get the result of run_path its a dictionary which I cannot seem to access?

– Stephen Ellwood
Sep 11 '18 at 9:00

What do you mean by "cannot access"? You can't import from it (that's why this is only a good option when import-style access isn't actually required), but the contents should be available via the regular dict API (result[name], result.get('name', default_value), etc)

– ncoghlan
Sep 13 '18 at 3:16

I like this method but when I get the result of run_path its a dictionary which I cannot seem to access?

– Stephen Ellwood
Sep 11 '18 at 9:00

What do you mean by "cannot access"? You can't import from it (that's why this is only a good option when import-style access isn't actually required), but the contents should be available via the regular dict API (result[name], result.get('name', default_value), etc)

– ncoghlan
Sep 13 '18 at 3:16

add a comment |

You can use the

load_source(module_name, path_to_file)

method from imp module.

edited Nov 11 '14 at 7:07

twasbrillig

6,41242452

answered Sep 15 '08 at 22:41

zuber

2,52321917

... and imp.load_dynamic(module_name, path_to_file) for DLLs

– HEKTO
Dec 16 '15 at 19:03

26

heads up that imp is deprecated now.

– t1m0
Apr 6 '16 at 18:11

add a comment |

You can use the

load_source(module_name, path_to_file)

method from imp module.

edited Nov 11 '14 at 7:07

twasbrillig

6,41242452

answered Sep 15 '08 at 22:41

zuber

2,52321917

... and imp.load_dynamic(module_name, path_to_file) for DLLs

– HEKTO
Dec 16 '15 at 19:03

26

heads up that imp is deprecated now.

– t1m0
Apr 6 '16 at 18:11

add a comment |

You can use the

load_source(module_name, path_to_file)

method from imp module.

edited Nov 11 '14 at 7:07

twasbrillig

6,41242452

answered Sep 15 '08 at 22:41

zuber

2,52321917

You can use the

load_source(module_name, path_to_file)

method from imp module.

edited Nov 11 '14 at 7:07

twasbrillig

6,41242452

answered Sep 15 '08 at 22:41

zuber

2,52321917

edited Nov 11 '14 at 7:07

twasbrillig

6,41242452

edited Nov 11 '14 at 7:07

twasbrillig

6,41242452

edited Nov 11 '14 at 7:07

twasbrillig

6,41242452

answered Sep 15 '08 at 22:41

zuber

2,52321917

answered Sep 15 '08 at 22:41

zuber

2,52321917

answered Sep 15 '08 at 22:41

zuber

2,52321917

... and imp.load_dynamic(module_name, path_to_file) for DLLs

– HEKTO
Dec 16 '15 at 19:03

26

heads up that imp is deprecated now.

– t1m0
Apr 6 '16 at 18:11

add a comment |

... and imp.load_dynamic(module_name, path_to_file) for DLLs

– HEKTO
Dec 16 '15 at 19:03

26

heads up that imp is deprecated now.

– t1m0
Apr 6 '16 at 18:11

... and imp.load_dynamic(module_name, path_to_file) for DLLs

– HEKTO
Dec 16 '15 at 19:03

heads up that imp is deprecated now.

– t1m0
Apr 6 '16 at 18:11

add a comment |

def import_file(full_path_to_module):

    try:

        import os

        module_dir, module_file = os.path.split(full_path_to_module)

        module_name, module_ext = os.path.splitext(module_file)

        save_cwd = os.getcwd()

        os.chdir(module_dir)

        module_obj = __import__(module_name)

        module_obj.__file__ = full_path_to_module

        globals()[module_name] = module_obj

        os.chdir(save_cwd)

    except:

        raise ImportError



import_file('/home/somebody/somemodule.py')

answered Sep 16 '08 at 1:43

Chris Calloway

1,0681712

36

Why write 14 lines of buggy code when this is already addressed by the standard library? You haven't done error checking on format or content of full_path_to_module or the os.whatever operations; and using a catch-all except: clause is rarely a good idea.

– Chris Johnson
Jun 7 '13 at 19:17

You should use more "try-finally"s in here. E.g. save_cwd = os.getcwd() try: … finally: os.chdir(save_cwd)

– kay
Sep 21 '14 at 1:33

9

@ChrisJohnson this is already addressed by the standard library yeah, but python has nasty habit of not being backward-compatible... as the checked answer says there're 2 different ways before and after 3.3. In that case I'd rather like to write my own universal function than check version on the fly. And yes, maybe this code isn't too well error-protected, but it shows an idea (which is os.chdir(), I haven't though about it), basing on which I can write a better code. Hence +1.

– Sushi271
May 15 '15 at 10:27

add a comment |

def import_file(full_path_to_module):

    try:

        import os

        module_dir, module_file = os.path.split(full_path_to_module)

        module_name, module_ext = os.path.splitext(module_file)

        save_cwd = os.getcwd()

        os.chdir(module_dir)

        module_obj = __import__(module_name)

        module_obj.__file__ = full_path_to_module

        globals()[module_name] = module_obj

        os.chdir(save_cwd)

    except:

        raise ImportError



import_file('/home/somebody/somemodule.py')

answered Sep 16 '08 at 1:43

Chris Calloway

1,0681712

36

Why write 14 lines of buggy code when this is already addressed by the standard library? You haven't done error checking on format or content of full_path_to_module or the os.whatever operations; and using a catch-all except: clause is rarely a good idea.

– Chris Johnson
Jun 7 '13 at 19:17

You should use more "try-finally"s in here. E.g. save_cwd = os.getcwd() try: … finally: os.chdir(save_cwd)

– kay
Sep 21 '14 at 1:33

9

@ChrisJohnson this is already addressed by the standard library yeah, but python has nasty habit of not being backward-compatible... as the checked answer says there're 2 different ways before and after 3.3. In that case I'd rather like to write my own universal function than check version on the fly. And yes, maybe this code isn't too well error-protected, but it shows an idea (which is os.chdir(), I haven't though about it), basing on which I can write a better code. Hence +1.

– Sushi271
May 15 '15 at 10:27

add a comment |

def import_file(full_path_to_module):

    try:

        import os

        module_dir, module_file = os.path.split(full_path_to_module)

        module_name, module_ext = os.path.splitext(module_file)

        save_cwd = os.getcwd()

        os.chdir(module_dir)

        module_obj = __import__(module_name)

        module_obj.__file__ = full_path_to_module

        globals()[module_name] = module_obj

        os.chdir(save_cwd)

    except:

        raise ImportError



import_file('/home/somebody/somemodule.py')

answered Sep 16 '08 at 1:43

Chris Calloway

1,0681712

def import_file(full_path_to_module):

    try:

        import os

        module_dir, module_file = os.path.split(full_path_to_module)

        module_name, module_ext = os.path.splitext(module_file)

        save_cwd = os.getcwd()

        os.chdir(module_dir)

        module_obj = __import__(module_name)

        module_obj.__file__ = full_path_to_module

        globals()[module_name] = module_obj

        os.chdir(save_cwd)

    except:

        raise ImportError



import_file('/home/somebody/somemodule.py')

answered Sep 16 '08 at 1:43

Chris Calloway

1,0681712

answered Sep 16 '08 at 1:43

Chris Calloway

1,0681712

answered Sep 16 '08 at 1:43

Chris Calloway

1,0681712

answered Sep 16 '08 at 1:43

Chris Calloway

1,0681712

36

Why write 14 lines of buggy code when this is already addressed by the standard library? You haven't done error checking on format or content of full_path_to_module or the os.whatever operations; and using a catch-all except: clause is rarely a good idea.

– Chris Johnson
Jun 7 '13 at 19:17

You should use more "try-finally"s in here. E.g. save_cwd = os.getcwd() try: … finally: os.chdir(save_cwd)

– kay
Sep 21 '14 at 1:33

9

@ChrisJohnson this is already addressed by the standard library yeah, but python has nasty habit of not being backward-compatible... as the checked answer says there're 2 different ways before and after 3.3. In that case I'd rather like to write my own universal function than check version on the fly. And yes, maybe this code isn't too well error-protected, but it shows an idea (which is os.chdir(), I haven't though about it), basing on which I can write a better code. Hence +1.

– Sushi271
May 15 '15 at 10:27

add a comment |

36

Why write 14 lines of buggy code when this is already addressed by the standard library? You haven't done error checking on format or content of full_path_to_module or the os.whatever operations; and using a catch-all except: clause is rarely a good idea.

– Chris Johnson
Jun 7 '13 at 19:17

You should use more "try-finally"s in here. E.g. save_cwd = os.getcwd() try: … finally: os.chdir(save_cwd)

– kay
Sep 21 '14 at 1:33

9

@ChrisJohnson this is already addressed by the standard library yeah, but python has nasty habit of not being backward-compatible... as the checked answer says there're 2 different ways before and after 3.3. In that case I'd rather like to write my own universal function than check version on the fly. And yes, maybe this code isn't too well error-protected, but it shows an idea (which is os.chdir(), I haven't though about it), basing on which I can write a better code. Hence +1.

– Sushi271
May 15 '15 at 10:27

Why write 14 lines of buggy code when this is already addressed by the standard library? You haven't done error checking on format or content of full_path_to_module or the os.whatever operations; and using a catch-all except: clause is rarely a good idea.

– Chris Johnson
Jun 7 '13 at 19:17

You should use more "try-finally"s in here. E.g. save_cwd = os.getcwd() try: … finally: os.chdir(save_cwd)

– kay
Sep 21 '14 at 1:33

@ChrisJohnson this is already addressed by the standard library yeah, but python has nasty habit of not being backward-compatible... as the checked answer says there're 2 different ways before and after 3.3. In that case I'd rather like to write my own universal function than check version on the fly. And yes, maybe this code isn't too well error-protected, but it shows an idea (which is os.chdir(), I haven't though about it), basing on which I can write a better code. Hence +1.

– Sushi271
May 15 '15 at 10:27

add a comment |

Here is some code that works in all Python versions, from 2.7-3.5 and probably even others.

config_file = "/tmp/config.py"

with open(config_file) as f:

    code = compile(f.read(), config_file, 'exec')

    exec(code, globals(), locals())

I tested it. It may be ugly but so far is the only one that works in all versions.

answered Jun 3 '16 at 10:04

sorin

74.7k116370579

This answer worked for me where load_source did not because it imports the script and provides the script access to the modules and globals at the time of importing.

– Klik
Nov 22 '17 at 19:13

add a comment |

Here is some code that works in all Python versions, from 2.7-3.5 and probably even others.

config_file = "/tmp/config.py"

with open(config_file) as f:

    code = compile(f.read(), config_file, 'exec')

    exec(code, globals(), locals())

I tested it. It may be ugly but so far is the only one that works in all versions.

answered Jun 3 '16 at 10:04

sorin

74.7k116370579

This answer worked for me where load_source did not because it imports the script and provides the script access to the modules and globals at the time of importing.

– Klik
Nov 22 '17 at 19:13

add a comment |

Here is some code that works in all Python versions, from 2.7-3.5 and probably even others.

config_file = "/tmp/config.py"

with open(config_file) as f:

    code = compile(f.read(), config_file, 'exec')

    exec(code, globals(), locals())

I tested it. It may be ugly but so far is the only one that works in all versions.

answered Jun 3 '16 at 10:04

sorin

74.7k116370579

Here is some code that works in all Python versions, from 2.7-3.5 and probably even others.

config_file = "/tmp/config.py"

with open(config_file) as f:

    code = compile(f.read(), config_file, 'exec')

    exec(code, globals(), locals())

I tested it. It may be ugly but so far is the only one that works in all versions.

answered Jun 3 '16 at 10:04

sorin

74.7k116370579

answered Jun 3 '16 at 10:04

sorin

74.7k116370579

answered Jun 3 '16 at 10:04

sorin

74.7k116370579

answered Jun 3 '16 at 10:04

sorin

74.7k116370579

This answer worked for me where load_source did not because it imports the script and provides the script access to the modules and globals at the time of importing.

– Klik
Nov 22 '17 at 19:13

add a comment |

This answer worked for me where load_source did not because it imports the script and provides the script access to the modules and globals at the time of importing.

– Klik
Nov 22 '17 at 19:13

This answer worked for me where load_source did not because it imports the script and provides the script access to the modules and globals at the time of importing.

– Klik
Nov 22 '17 at 19:13

add a comment |

from importlib.util import spec_from_loader, module_from_spec

from importlib.machinery import SourceFileLoader 



spec = spec_from_loader("module.name", SourceFileLoader("module.name", "/path/to/file.py"))

mod = module_from_spec(spec)

spec.loader.exec_module(mod)

edited Sep 19 '18 at 18:13

answered Apr 25 '17 at 5:45

Mad Physicist

36.4k1671101

add a comment |

from importlib.util import spec_from_loader, module_from_spec

from importlib.machinery import SourceFileLoader 



spec = spec_from_loader("module.name", SourceFileLoader("module.name", "/path/to/file.py"))

mod = module_from_spec(spec)

spec.loader.exec_module(mod)

edited Sep 19 '18 at 18:13

answered Apr 25 '17 at 5:45

Mad Physicist

36.4k1671101

add a comment |

from importlib.util import spec_from_loader, module_from_spec

from importlib.machinery import SourceFileLoader 



spec = spec_from_loader("module.name", SourceFileLoader("module.name", "/path/to/file.py"))

mod = module_from_spec(spec)

spec.loader.exec_module(mod)

edited Sep 19 '18 at 18:13

answered Apr 25 '17 at 5:45

Mad Physicist

36.4k1671101

from importlib.util import spec_from_loader, module_from_spec

from importlib.machinery import SourceFileLoader 



spec = spec_from_loader("module.name", SourceFileLoader("module.name", "/path/to/file.py"))

mod = module_from_spec(spec)

spec.loader.exec_module(mod)

edited Sep 19 '18 at 18:13

answered Apr 25 '17 at 5:45

Mad Physicist

36.4k1671101

edited Sep 19 '18 at 18:13

answered Apr 25 '17 at 5:45

Mad Physicist

36.4k1671101

answered Apr 25 '17 at 5:45

Mad Physicist

36.4k1671101

answered Apr 25 '17 at 5:45

Mad Physicist

36.4k1671101

add a comment |

Do you mean load or import?

You can manipulate the sys.path list specify the path to your module, then import your module. For example, given a module at:

/foo/bar.py

You could do:

import sys

sys.path[0:0] = ['/foo'] # puts the /foo directory at the start of your path

import bar

edited Dec 2 '18 at 1:05

Jason Orendorff

30.8k34884

answered Sep 15 '08 at 22:46

Wheat

63239

@Wheat Why sys.path[0:0] instead of sys.path[0]?

– user618677
Jan 9 '12 at 6:56

4

B/c sys.path[0] = xy overwrites the first path item while path[0:0] =xy is equivalent to path.insert(0, xy)

– dom0
Nov 15 '12 at 14:16

2

hm the path.insert worked for me but the [0:0] trick did not.

– jsh
Sep 30 '13 at 3:18

10

sys.path[0:0] = ['/foo']

– Kevin Edwards
Apr 1 '15 at 17:00

add a comment |

Do you mean load or import?

You can manipulate the sys.path list specify the path to your module, then import your module. For example, given a module at:

/foo/bar.py

You could do:

import sys

sys.path[0:0] = ['/foo'] # puts the /foo directory at the start of your path

import bar

edited Dec 2 '18 at 1:05

Jason Orendorff

30.8k34884

answered Sep 15 '08 at 22:46

Wheat

63239

@Wheat Why sys.path[0:0] instead of sys.path[0]?

– user618677
Jan 9 '12 at 6:56

4

B/c sys.path[0] = xy overwrites the first path item while path[0:0] =xy is equivalent to path.insert(0, xy)

– dom0
Nov 15 '12 at 14:16

2

hm the path.insert worked for me but the [0:0] trick did not.

– jsh
Sep 30 '13 at 3:18

10

sys.path[0:0] = ['/foo']

– Kevin Edwards
Apr 1 '15 at 17:00

add a comment |

Do you mean load or import?

You can manipulate the sys.path list specify the path to your module, then import your module. For example, given a module at:

/foo/bar.py

You could do:

import sys

sys.path[0:0] = ['/foo'] # puts the /foo directory at the start of your path

import bar

edited Dec 2 '18 at 1:05

Jason Orendorff

30.8k34884

answered Sep 15 '08 at 22:46

Wheat

63239

Do you mean load or import?

You can manipulate the sys.path list specify the path to your module, then import your module. For example, given a module at:

/foo/bar.py

You could do:

import sys

sys.path[0:0] = ['/foo'] # puts the /foo directory at the start of your path

import bar

edited Dec 2 '18 at 1:05

Jason Orendorff

30.8k34884

answered Sep 15 '08 at 22:46

Wheat

63239

edited Dec 2 '18 at 1:05

Jason Orendorff

30.8k34884

edited Dec 2 '18 at 1:05

Jason Orendorff

30.8k34884

edited Dec 2 '18 at 1:05

Jason Orendorff

30.8k34884

answered Sep 15 '08 at 22:46

Wheat

63239

answered Sep 15 '08 at 22:46

Wheat

63239

answered Sep 15 '08 at 22:46

Wheat

63239

@Wheat Why sys.path[0:0] instead of sys.path[0]?

– user618677
Jan 9 '12 at 6:56

4

B/c sys.path[0] = xy overwrites the first path item while path[0:0] =xy is equivalent to path.insert(0, xy)

– dom0
Nov 15 '12 at 14:16

2

hm the path.insert worked for me but the [0:0] trick did not.

– jsh
Sep 30 '13 at 3:18

10

sys.path[0:0] = ['/foo']

– Kevin Edwards
Apr 1 '15 at 17:00

add a comment |

@Wheat Why sys.path[0:0] instead of sys.path[0]?

– user618677
Jan 9 '12 at 6:56

4

B/c sys.path[0] = xy overwrites the first path item while path[0:0] =xy is equivalent to path.insert(0, xy)

– dom0
Nov 15 '12 at 14:16

2

hm the path.insert worked for me but the [0:0] trick did not.

– jsh
Sep 30 '13 at 3:18

10

sys.path[0:0] = ['/foo']

– Kevin Edwards
Apr 1 '15 at 17:00

@Wheat Why sys.path[0:0] instead of sys.path[0]?

– user618677
Jan 9 '12 at 6:56

B/c sys.path[0] = xy overwrites the first path item while path[0:0] =xy is equivalent to path.insert(0, xy)

– dom0
Nov 15 '12 at 14:16

hm the path.insert worked for me but the [0:0] trick did not.

– jsh
Sep 30 '13 at 3:18

sys.path[0:0] = ['/foo']

– Kevin Edwards
Apr 1 '15 at 17:00

add a comment |

imp.find_module('mymodule', '/home/mypath/')

...but that should get the job done.

edited Jan 2 '15 at 11:08

Mathieu Rodic

4,72022940

answered Sep 15 '08 at 22:37

Matt

1313

add a comment |

imp.find_module('mymodule', '/home/mypath/')

...but that should get the job done.

edited Jan 2 '15 at 11:08

Mathieu Rodic

4,72022940

answered Sep 15 '08 at 22:37

Matt

1313

add a comment |

imp.find_module('mymodule', '/home/mypath/')

...but that should get the job done.

edited Jan 2 '15 at 11:08

Mathieu Rodic

4,72022940

answered Sep 15 '08 at 22:37

Matt

1313

imp.find_module('mymodule', '/home/mypath/')

...but that should get the job done.

edited Jan 2 '15 at 11:08

Mathieu Rodic

4,72022940

answered Sep 15 '08 at 22:37

Matt

1313

edited Jan 2 '15 at 11:08

Mathieu Rodic

4,72022940

edited Jan 2 '15 at 11:08

Mathieu Rodic

4,72022940

edited Jan 2 '15 at 11:08

Mathieu Rodic

4,72022940

answered Sep 15 '08 at 22:37

Matt

1313

answered Sep 15 '08 at 22:37

Matt

1313

answered Sep 15 '08 at 22:37

Matt

1313

add a comment |

MODULE_PATH = "/path/to/your/module/__init__.py"

MODULE_NAME = "mymodule"

spec = importlib.util.spec_from_file_location(MODULE_NAME, MODULE_PATH)

module = importlib.util.module_from_spec(spec)

sys.modules[spec.name] = module 

spec.loader.exec_module(module)

edited Oct 17 '18 at 16:49

Hadrien TOMA

8151221

answered May 17 '18 at 15:23

Sam Grondahl

1,01611223

1

This is exactly what I was missing! Thanks! (This also correctly answers Proper way to dynamically import a module with relative imports?)

– mforbes
Jul 14 '18 at 7:30

Any idea how to do this with python 2?

– mforbes
Jul 16 '18 at 0:57

add a comment |

MODULE_PATH = "/path/to/your/module/__init__.py"

MODULE_NAME = "mymodule"

spec = importlib.util.spec_from_file_location(MODULE_NAME, MODULE_PATH)

module = importlib.util.module_from_spec(spec)

sys.modules[spec.name] = module 

spec.loader.exec_module(module)

edited Oct 17 '18 at 16:49

Hadrien TOMA

8151221

answered May 17 '18 at 15:23

Sam Grondahl

1,01611223

1

This is exactly what I was missing! Thanks! (This also correctly answers Proper way to dynamically import a module with relative imports?)

– mforbes
Jul 14 '18 at 7:30

Any idea how to do this with python 2?

– mforbes
Jul 16 '18 at 0:57

add a comment |

MODULE_PATH = "/path/to/your/module/__init__.py"

MODULE_NAME = "mymodule"

spec = importlib.util.spec_from_file_location(MODULE_NAME, MODULE_PATH)

module = importlib.util.module_from_spec(spec)

sys.modules[spec.name] = module 

spec.loader.exec_module(module)

edited Oct 17 '18 at 16:49

Hadrien TOMA

8151221

answered May 17 '18 at 15:23

Sam Grondahl

1,01611223

MODULE_PATH = "/path/to/your/module/__init__.py"

MODULE_NAME = "mymodule"

spec = importlib.util.spec_from_file_location(MODULE_NAME, MODULE_PATH)

module = importlib.util.module_from_spec(spec)

sys.modules[spec.name] = module 

spec.loader.exec_module(module)

edited Oct 17 '18 at 16:49

Hadrien TOMA

8151221

answered May 17 '18 at 15:23

Sam Grondahl

1,01611223

edited Oct 17 '18 at 16:49

Hadrien TOMA

8151221

edited Oct 17 '18 at 16:49

Hadrien TOMA

8151221

edited Oct 17 '18 at 16:49

Hadrien TOMA

8151221

answered May 17 '18 at 15:23

Sam Grondahl

1,01611223

answered May 17 '18 at 15:23

Sam Grondahl

1,01611223

answered May 17 '18 at 15:23

Sam Grondahl

1,01611223

1

This is exactly what I was missing! Thanks! (This also correctly answers Proper way to dynamically import a module with relative imports?)

– mforbes
Jul 14 '18 at 7:30

Any idea how to do this with python 2?

– mforbes
Jul 16 '18 at 0:57

add a comment |

1

This is exactly what I was missing! Thanks! (This also correctly answers Proper way to dynamically import a module with relative imports?)

– mforbes
Jul 14 '18 at 7:30

Any idea how to do this with python 2?

– mforbes
Jul 16 '18 at 0:57

This is exactly what I was missing! Thanks! (This also correctly answers Proper way to dynamically import a module with relative imports?)

– mforbes
Jul 14 '18 at 7:30

Any idea how to do this with python 2?

– mforbes
Jul 16 '18 at 0:57

add a comment |

To import your module, you need to add its directory to the environment variable, either temporarily or permanently.

Temporarily

import sys

sys.path.append("/path/to/my/modules/")

import my_module

Permanently

Adding the following line to your .bashrc file (in linux) and excecute source ~/.bashrc in the terminal:

export PYTHONPATH="${PYTHONPATH}:/path/to/my/modules/"

Credit/Source: saarrrr, another stackexchange question

answered Nov 15 '18 at 2:30

Miladiouss

527614

This "temp" solution is a great answer if you want to prod a project around in a jupyter notebook elsewhere.

– fordy
Nov 16 '18 at 17:20

add a comment |

To import your module, you need to add its directory to the environment variable, either temporarily or permanently.

Temporarily

import sys

sys.path.append("/path/to/my/modules/")

import my_module

Permanently

Adding the following line to your .bashrc file (in linux) and excecute source ~/.bashrc in the terminal:

export PYTHONPATH="${PYTHONPATH}:/path/to/my/modules/"

Credit/Source: saarrrr, another stackexchange question

answered Nov 15 '18 at 2:30

Miladiouss

527614

This "temp" solution is a great answer if you want to prod a project around in a jupyter notebook elsewhere.

– fordy
Nov 16 '18 at 17:20

add a comment |

To import your module, you need to add its directory to the environment variable, either temporarily or permanently.

Temporarily

import sys

sys.path.append("/path/to/my/modules/")

import my_module

Permanently

Adding the following line to your .bashrc file (in linux) and excecute source ~/.bashrc in the terminal:

export PYTHONPATH="${PYTHONPATH}:/path/to/my/modules/"

Credit/Source: saarrrr, another stackexchange question

answered Nov 15 '18 at 2:30

Miladiouss

527614

To import your module, you need to add its directory to the environment variable, either temporarily or permanently.

Temporarily

import sys

sys.path.append("/path/to/my/modules/")

import my_module

Permanently

Adding the following line to your .bashrc file (in linux) and excecute source ~/.bashrc in the terminal:

export PYTHONPATH="${PYTHONPATH}:/path/to/my/modules/"

Credit/Source: saarrrr, another stackexchange question

answered Nov 15 '18 at 2:30

Miladiouss

527614

answered Nov 15 '18 at 2:30

Miladiouss

527614

answered Nov 15 '18 at 2:30

Miladiouss

527614

answered Nov 15 '18 at 2:30

Miladiouss

527614

This "temp" solution is a great answer if you want to prod a project around in a jupyter notebook elsewhere.

– fordy
Nov 16 '18 at 17:20

add a comment |

This "temp" solution is a great answer if you want to prod a project around in a jupyter notebook elsewhere.

– fordy
Nov 16 '18 at 17:20

This "temp" solution is a great answer if you want to prod a project around in a jupyter notebook elsewhere.

– fordy
Nov 16 '18 at 17:20

add a comment |

This should work

path = os.path.join('./path/to/folder/with/py/files', '*.py')

for infile in glob.glob(path):

    basename = os.path.basename(infile)

    basename_without_extension = basename[:-3]



    # http://docs.python.org/library/imp.html?highlight=imp#module-imp

    imp.load_source(basename_without_extension, infile)

edited Jan 4 '12 at 4:28

joran

134k19320380

answered Jan 4 '12 at 2:17

Hengjie

3,5512433

4

A more general way to cut the extension out is: name, ext = os.path.splitext(os.path.basename(infile)). Your method works because the previous restriction to .py extension. Also, you should probably import the module to some variable/dictionary entry.

– ReneSac
Dec 6 '12 at 13:16

add a comment |

This should work

path = os.path.join('./path/to/folder/with/py/files', '*.py')

for infile in glob.glob(path):

    basename = os.path.basename(infile)

    basename_without_extension = basename[:-3]



    # http://docs.python.org/library/imp.html?highlight=imp#module-imp

    imp.load_source(basename_without_extension, infile)

edited Jan 4 '12 at 4:28

joran

134k19320380

answered Jan 4 '12 at 2:17

Hengjie

3,5512433

4

A more general way to cut the extension out is: name, ext = os.path.splitext(os.path.basename(infile)). Your method works because the previous restriction to .py extension. Also, you should probably import the module to some variable/dictionary entry.

– ReneSac
Dec 6 '12 at 13:16

add a comment |

This should work

path = os.path.join('./path/to/folder/with/py/files', '*.py')

for infile in glob.glob(path):

    basename = os.path.basename(infile)

    basename_without_extension = basename[:-3]



    # http://docs.python.org/library/imp.html?highlight=imp#module-imp

    imp.load_source(basename_without_extension, infile)

edited Jan 4 '12 at 4:28

joran

134k19320380

answered Jan 4 '12 at 2:17

Hengjie

3,5512433

This should work

path = os.path.join('./path/to/folder/with/py/files', '*.py')

for infile in glob.glob(path):

    basename = os.path.basename(infile)

    basename_without_extension = basename[:-3]



    # http://docs.python.org/library/imp.html?highlight=imp#module-imp

    imp.load_source(basename_without_extension, infile)

edited Jan 4 '12 at 4:28

joran

134k19320380

answered Jan 4 '12 at 2:17

Hengjie

3,5512433

edited Jan 4 '12 at 4:28

joran

134k19320380

edited Jan 4 '12 at 4:28

joran

134k19320380

edited Jan 4 '12 at 4:28

joran

134k19320380

answered Jan 4 '12 at 2:17

Hengjie

3,5512433

answered Jan 4 '12 at 2:17

Hengjie

3,5512433

answered Jan 4 '12 at 2:17

Hengjie

3,5512433

4

A more general way to cut the extension out is: name, ext = os.path.splitext(os.path.basename(infile)). Your method works because the previous restriction to .py extension. Also, you should probably import the module to some variable/dictionary entry.

– ReneSac
Dec 6 '12 at 13:16

add a comment |

4

A more general way to cut the extension out is: name, ext = os.path.splitext(os.path.basename(infile)). Your method works because the previous restriction to .py extension. Also, you should probably import the module to some variable/dictionary entry.

– ReneSac
Dec 6 '12 at 13:16

A more general way to cut the extension out is: name, ext = os.path.splitext(os.path.basename(infile)). Your method works because the previous restriction to .py extension. Also, you should probably import the module to some variable/dictionary entry.

– ReneSac
Dec 6 '12 at 13:16

add a comment |

def import_module_from_file(full_path_to_module):

    """

    Import a module given the full path/filename of the .py file



    Python 3.4



    """



    module = None



    try:



        # Get module name and path from full path

        module_dir, module_file = os.path.split(full_path_to_module)

        module_name, module_ext = os.path.splitext(module_file)



        # Get module "spec" from filename

        spec = importlib.util.spec_from_file_location(module_name,full_path_to_module)



        module = spec.loader.load_module()



    except Exception as ec:

        # Simple error printing

        # Insert "sophisticated" stuff here

        print(ec)



    finally:

        return module

answered Apr 12 '15 at 12:22

Redlegjed

311

add a comment |

def import_module_from_file(full_path_to_module):

    """

    Import a module given the full path/filename of the .py file



    Python 3.4



    """



    module = None



    try:



        # Get module name and path from full path

        module_dir, module_file = os.path.split(full_path_to_module)

        module_name, module_ext = os.path.splitext(module_file)



        # Get module "spec" from filename

        spec = importlib.util.spec_from_file_location(module_name,full_path_to_module)



        module = spec.loader.load_module()



    except Exception as ec:

        # Simple error printing

        # Insert "sophisticated" stuff here

        print(ec)



    finally:

        return module

answered Apr 12 '15 at 12:22

Redlegjed

311

add a comment |

def import_module_from_file(full_path_to_module):

    """

    Import a module given the full path/filename of the .py file



    Python 3.4



    """



    module = None



    try:



        # Get module name and path from full path

        module_dir, module_file = os.path.split(full_path_to_module)

        module_name, module_ext = os.path.splitext(module_file)



        # Get module "spec" from filename

        spec = importlib.util.spec_from_file_location(module_name,full_path_to_module)



        module = spec.loader.load_module()



    except Exception as ec:

        # Simple error printing

        # Insert "sophisticated" stuff here

        print(ec)



    finally:

        return module

answered Apr 12 '15 at 12:22

Redlegjed

311

def import_module_from_file(full_path_to_module):

    """

    Import a module given the full path/filename of the .py file



    Python 3.4



    """



    module = None



    try:



        # Get module name and path from full path

        module_dir, module_file = os.path.split(full_path_to_module)

        module_name, module_ext = os.path.splitext(module_file)



        # Get module "spec" from filename

        spec = importlib.util.spec_from_file_location(module_name,full_path_to_module)



        module = spec.loader.load_module()



    except Exception as ec:

        # Simple error printing

        # Insert "sophisticated" stuff here

        print(ec)



    finally:

        return module

answered Apr 12 '15 at 12:22

Redlegjed

311

answered Apr 12 '15 at 12:22

Redlegjed

311

answered Apr 12 '15 at 12:22

Redlegjed

311

answered Apr 12 '15 at 12:22

Redlegjed

311

add a comment |

For example, if you have a module stored in "/path/to/module" with the function foo(), you could run it by doing the following:

module = dict()

with open("/path/to/module") as f:

    exec(f.read(), module)

module['foo']()

This makes it a bit more explicit that you're loading code dynamically, and grants you some additional power, such as the ability to provide custom builtins.

And if having access through attributes, instead of keys is important to you, you can design a custom dict class for the globals, that provides such access, e.g.:

class MyModuleClass(dict):

    def __getattr__(self, name):

        return self.__getitem__(name)

answered Jun 2 '15 at 19:57

yoniLavi

1,3561622

2

execfile(), also

– crowder
Jun 20 '15 at 4:13

add a comment |

For example, if you have a module stored in "/path/to/module" with the function foo(), you could run it by doing the following:

module = dict()

with open("/path/to/module") as f:

    exec(f.read(), module)

module['foo']()

This makes it a bit more explicit that you're loading code dynamically, and grants you some additional power, such as the ability to provide custom builtins.

And if having access through attributes, instead of keys is important to you, you can design a custom dict class for the globals, that provides such access, e.g.:

class MyModuleClass(dict):

    def __getattr__(self, name):

        return self.__getitem__(name)

answered Jun 2 '15 at 19:57

yoniLavi

1,3561622

2

execfile(), also

– crowder
Jun 20 '15 at 4:13

add a comment |

For example, if you have a module stored in "/path/to/module" with the function foo(), you could run it by doing the following:

module = dict()

with open("/path/to/module") as f:

    exec(f.read(), module)

module['foo']()

This makes it a bit more explicit that you're loading code dynamically, and grants you some additional power, such as the ability to provide custom builtins.

And if having access through attributes, instead of keys is important to you, you can design a custom dict class for the globals, that provides such access, e.g.:

class MyModuleClass(dict):

    def __getattr__(self, name):

        return self.__getitem__(name)

answered Jun 2 '15 at 19:57

yoniLavi

1,3561622

For example, if you have a module stored in "/path/to/module" with the function foo(), you could run it by doing the following:

module = dict()

with open("/path/to/module") as f:

    exec(f.read(), module)

module['foo']()

This makes it a bit more explicit that you're loading code dynamically, and grants you some additional power, such as the ability to provide custom builtins.

And if having access through attributes, instead of keys is important to you, you can design a custom dict class for the globals, that provides such access, e.g.:

class MyModuleClass(dict):

    def __getattr__(self, name):

        return self.__getitem__(name)

answered Jun 2 '15 at 19:57

yoniLavi

1,3561622

answered Jun 2 '15 at 19:57

yoniLavi

1,3561622

answered Jun 2 '15 at 19:57

yoniLavi

1,3561622

answered Jun 2 '15 at 19:57

yoniLavi

1,3561622

2

execfile(), also

– crowder
Jun 20 '15 at 4:13

add a comment |

2

execfile(), also

– crowder
Jun 20 '15 at 4:13

execfile(), also

– crowder
Jun 20 '15 at 4:13

add a comment |

To import a module from a given filename, you can temporarily extend the path, and restore the system path in the finally block reference:

filename = "directory/module.py"



directory, module_name = os.path.split(filename)

module_name = os.path.splitext(module_name)[0]



path = list(sys.path)

sys.path.insert(0, directory)

try:

    module = __import__(module_name)

finally:

    sys.path[:] = path # restore

answered Oct 2 '15 at 11:14

Peter Zhu

704921

add a comment |

To import a module from a given filename, you can temporarily extend the path, and restore the system path in the finally block reference:

filename = "directory/module.py"



directory, module_name = os.path.split(filename)

module_name = os.path.splitext(module_name)[0]



path = list(sys.path)

sys.path.insert(0, directory)

try:

    module = __import__(module_name)

finally:

    sys.path[:] = path # restore

answered Oct 2 '15 at 11:14

Peter Zhu

704921

add a comment |

To import a module from a given filename, you can temporarily extend the path, and restore the system path in the finally block reference:

filename = "directory/module.py"



directory, module_name = os.path.split(filename)

module_name = os.path.splitext(module_name)[0]



path = list(sys.path)

sys.path.insert(0, directory)

try:

    module = __import__(module_name)

finally:

    sys.path[:] = path # restore

answered Oct 2 '15 at 11:14

Peter Zhu

704921

To import a module from a given filename, you can temporarily extend the path, and restore the system path in the finally block reference:

filename = "directory/module.py"



directory, module_name = os.path.split(filename)

module_name = os.path.splitext(module_name)[0]



path = list(sys.path)

sys.path.insert(0, directory)

try:

    module = __import__(module_name)

finally:

    sys.path[:] = path # restore

answered Oct 2 '15 at 11:14

Peter Zhu

704921

answered Oct 2 '15 at 11:14

Peter Zhu

704921

answered Oct 2 '15 at 11:14

Peter Zhu

704921

answered Oct 2 '15 at 11:14

Peter Zhu

704921

add a comment |

I made a package that uses imp for you. I call it import_file and this is how it's used:

>>>from import_file import import_file

>>>mylib = import_file('c:\mylib.py')

>>>another = import_file('relative_subdir/another.py')

You can get it at:

http://pypi.python.org/pypi/import_file

or at

http://code.google.com/p/import-file/

answered Jun 8 '11 at 19:41

ubershmekel

5,45824665

1

os.chdir ? (minimal characters to approve comment).

– ychaouche
Oct 14 '12 at 10:46

I've spent all day troubleshooting an import bug in a pyinstaller generated exe. In the end this is the only thing that worked for me. Thank you so much for making this!

– frakman1
Nov 29 '18 at 22:12

add a comment |

I made a package that uses imp for you. I call it import_file and this is how it's used:

>>>from import_file import import_file

>>>mylib = import_file('c:\mylib.py')

>>>another = import_file('relative_subdir/another.py')

You can get it at:

http://pypi.python.org/pypi/import_file

or at

http://code.google.com/p/import-file/

answered Jun 8 '11 at 19:41

ubershmekel

5,45824665

1

os.chdir ? (minimal characters to approve comment).

– ychaouche
Oct 14 '12 at 10:46

I've spent all day troubleshooting an import bug in a pyinstaller generated exe. In the end this is the only thing that worked for me. Thank you so much for making this!

– frakman1
Nov 29 '18 at 22:12

add a comment |

I made a package that uses imp for you. I call it import_file and this is how it's used:

>>>from import_file import import_file

>>>mylib = import_file('c:\mylib.py')

>>>another = import_file('relative_subdir/another.py')

You can get it at:

http://pypi.python.org/pypi/import_file

or at

http://code.google.com/p/import-file/

answered Jun 8 '11 at 19:41

ubershmekel

5,45824665

I made a package that uses imp for you. I call it import_file and this is how it's used:

>>>from import_file import import_file

>>>mylib = import_file('c:\mylib.py')

>>>another = import_file('relative_subdir/another.py')

You can get it at:

http://pypi.python.org/pypi/import_file

or at

http://code.google.com/p/import-file/

answered Jun 8 '11 at 19:41

ubershmekel

5,45824665

answered Jun 8 '11 at 19:41

ubershmekel

5,45824665

answered Jun 8 '11 at 19:41

ubershmekel

5,45824665

answered Jun 8 '11 at 19:41

ubershmekel

5,45824665

1

os.chdir ? (minimal characters to approve comment).

– ychaouche
Oct 14 '12 at 10:46

I've spent all day troubleshooting an import bug in a pyinstaller generated exe. In the end this is the only thing that worked for me. Thank you so much for making this!

– frakman1
Nov 29 '18 at 22:12

add a comment |

1

os.chdir ? (minimal characters to approve comment).

– ychaouche
Oct 14 '12 at 10:46

I've spent all day troubleshooting an import bug in a pyinstaller generated exe. In the end this is the only thing that worked for me. Thank you so much for making this!

– frakman1
Nov 29 '18 at 22:12

os.chdir ? (minimal characters to approve comment).

– ychaouche
Oct 14 '12 at 10:46

I've spent all day troubleshooting an import bug in a pyinstaller generated exe. In the end this is the only thing that worked for me. Thank you so much for making this!

– frakman1
Nov 29 '18 at 22:12

add a comment |

Import package modules at runtime (Python recipe)

http://code.activestate.com/recipes/223972/

###################

##                #

## classloader.py #

##                #

###################



import sys, types



def _get_mod(modulePath):

    try:

        aMod = sys.modules[modulePath]

        if not isinstance(aMod, types.ModuleType):

            raise KeyError

    except KeyError:

        # The last [''] is very important!

        aMod = __import__(modulePath, globals(), locals(), [''])

        sys.modules[modulePath] = aMod

    return aMod



def _get_func(fullFuncName):

    """Retrieve a function object from a full dotted-package name."""



    # Parse out the path, module, and function

    lastDot = fullFuncName.rfind(u".")

    funcName = fullFuncName[lastDot + 1:]

    modPath = fullFuncName[:lastDot]



    aMod = _get_mod(modPath)

    aFunc = getattr(aMod, funcName)



    # Assert that the function is a *callable* attribute.

    assert callable(aFunc), u"%s is not callable." % fullFuncName



    # Return a reference to the function itself,

    # not the results of the function.

    return aFunc



def _get_class(fullClassName, parentClass=None):

    """Load a module and retrieve a class (NOT an instance).



    If the parentClass is supplied, className must be of parentClass

    or a subclass of parentClass (or None is returned).

    """

    aClass = _get_func(fullClassName)



    # Assert that the class is a subclass of parentClass.

    if parentClass is not None:

        if not issubclass(aClass, parentClass):

            raise TypeError(u"%s is not a subclass of %s" %

                            (fullClassName, parentClass))



    # Return a reference to the class itself, not an instantiated object.

    return aClass





######################

##       Usage      ##

######################



class StorageManager: pass

class StorageManagerMySQL(StorageManager): pass



def storage_object(aFullClassName, allOptions={}):

    aStoreClass = _get_class(aFullClassName, StorageManager)

    return aStoreClass(allOptions)

edited Dec 22 '13 at 2:17

Eric Leschinski

87.1k38321274

answered Sep 15 '08 at 22:43

user10370

371

add a comment |

Import package modules at runtime (Python recipe)

http://code.activestate.com/recipes/223972/

###################

##                #

## classloader.py #

##                #

###################



import sys, types



def _get_mod(modulePath):

    try:

        aMod = sys.modules[modulePath]

        if not isinstance(aMod, types.ModuleType):

            raise KeyError

    except KeyError:

        # The last [''] is very important!

        aMod = __import__(modulePath, globals(), locals(), [''])

        sys.modules[modulePath] = aMod

    return aMod



def _get_func(fullFuncName):

    """Retrieve a function object from a full dotted-package name."""



    # Parse out the path, module, and function

    lastDot = fullFuncName.rfind(u".")

    funcName = fullFuncName[lastDot + 1:]

    modPath = fullFuncName[:lastDot]



    aMod = _get_mod(modPath)

    aFunc = getattr(aMod, funcName)



    # Assert that the function is a *callable* attribute.

    assert callable(aFunc), u"%s is not callable." % fullFuncName



    # Return a reference to the function itself,

    # not the results of the function.

    return aFunc



def _get_class(fullClassName, parentClass=None):

    """Load a module and retrieve a class (NOT an instance).



    If the parentClass is supplied, className must be of parentClass

    or a subclass of parentClass (or None is returned).

    """

    aClass = _get_func(fullClassName)



    # Assert that the class is a subclass of parentClass.

    if parentClass is not None:

        if not issubclass(aClass, parentClass):

            raise TypeError(u"%s is not a subclass of %s" %

                            (fullClassName, parentClass))



    # Return a reference to the class itself, not an instantiated object.

    return aClass





######################

##       Usage      ##

######################



class StorageManager: pass

class StorageManagerMySQL(StorageManager): pass



def storage_object(aFullClassName, allOptions={}):

    aStoreClass = _get_class(aFullClassName, StorageManager)

    return aStoreClass(allOptions)

edited Dec 22 '13 at 2:17

Eric Leschinski

87.1k38321274

answered Sep 15 '08 at 22:43

user10370

371

add a comment |

Import package modules at runtime (Python recipe)

http://code.activestate.com/recipes/223972/

###################

##                #

## classloader.py #

##                #

###################



import sys, types



def _get_mod(modulePath):

    try:

        aMod = sys.modules[modulePath]

        if not isinstance(aMod, types.ModuleType):

            raise KeyError

    except KeyError:

        # The last [''] is very important!

        aMod = __import__(modulePath, globals(), locals(), [''])

        sys.modules[modulePath] = aMod

    return aMod



def _get_func(fullFuncName):

    """Retrieve a function object from a full dotted-package name."""



    # Parse out the path, module, and function

    lastDot = fullFuncName.rfind(u".")

    funcName = fullFuncName[lastDot + 1:]

    modPath = fullFuncName[:lastDot]



    aMod = _get_mod(modPath)

    aFunc = getattr(aMod, funcName)



    # Assert that the function is a *callable* attribute.

    assert callable(aFunc), u"%s is not callable." % fullFuncName



    # Return a reference to the function itself,

    # not the results of the function.

    return aFunc



def _get_class(fullClassName, parentClass=None):

    """Load a module and retrieve a class (NOT an instance).



    If the parentClass is supplied, className must be of parentClass

    or a subclass of parentClass (or None is returned).

    """

    aClass = _get_func(fullClassName)



    # Assert that the class is a subclass of parentClass.

    if parentClass is not None:

        if not issubclass(aClass, parentClass):

            raise TypeError(u"%s is not a subclass of %s" %

                            (fullClassName, parentClass))



    # Return a reference to the class itself, not an instantiated object.

    return aClass





######################

##       Usage      ##

######################



class StorageManager: pass

class StorageManagerMySQL(StorageManager): pass



def storage_object(aFullClassName, allOptions={}):

    aStoreClass = _get_class(aFullClassName, StorageManager)

    return aStoreClass(allOptions)

edited Dec 22 '13 at 2:17

Eric Leschinski

87.1k38321274

answered Sep 15 '08 at 22:43

user10370

371

Import package modules at runtime (Python recipe)

http://code.activestate.com/recipes/223972/

###################

##                #

## classloader.py #

##                #

###################



import sys, types



def _get_mod(modulePath):

    try:

        aMod = sys.modules[modulePath]

        if not isinstance(aMod, types.ModuleType):

            raise KeyError

    except KeyError:

        # The last [''] is very important!

        aMod = __import__(modulePath, globals(), locals(), [''])

        sys.modules[modulePath] = aMod

    return aMod



def _get_func(fullFuncName):

    """Retrieve a function object from a full dotted-package name."""



    # Parse out the path, module, and function

    lastDot = fullFuncName.rfind(u".")

    funcName = fullFuncName[lastDot + 1:]

    modPath = fullFuncName[:lastDot]



    aMod = _get_mod(modPath)

    aFunc = getattr(aMod, funcName)



    # Assert that the function is a *callable* attribute.

    assert callable(aFunc), u"%s is not callable." % fullFuncName



    # Return a reference to the function itself,

    # not the results of the function.

    return aFunc



def _get_class(fullClassName, parentClass=None):

    """Load a module and retrieve a class (NOT an instance).



    If the parentClass is supplied, className must be of parentClass

    or a subclass of parentClass (or None is returned).

    """

    aClass = _get_func(fullClassName)



    # Assert that the class is a subclass of parentClass.

    if parentClass is not None:

        if not issubclass(aClass, parentClass):

            raise TypeError(u"%s is not a subclass of %s" %

                            (fullClassName, parentClass))



    # Return a reference to the class itself, not an instantiated object.

    return aClass





######################

##       Usage      ##

######################



class StorageManager: pass

class StorageManagerMySQL(StorageManager): pass



def storage_object(aFullClassName, allOptions={}):

    aStoreClass = _get_class(aFullClassName, StorageManager)

    return aStoreClass(allOptions)

edited Dec 22 '13 at 2:17

Eric Leschinski

87.1k38321274

answered Sep 15 '08 at 22:43

user10370

371

edited Dec 22 '13 at 2:17

Eric Leschinski

87.1k38321274

edited Dec 22 '13 at 2:17

Eric Leschinski

87.1k38321274

edited Dec 22 '13 at 2:17

Eric Leschinski

87.1k38321274

answered Sep 15 '08 at 22:43

user10370

371

answered Sep 15 '08 at 22:43

user10370

371

answered Sep 15 '08 at 22:43

user10370

371

add a comment |

import pkgutil

import importlib



packages = pkgutil.walk_packages(path='.')

for importer, name, is_package in packages:

    mod = importlib.import_module(name)

    # do whatever you want with module now, it's been imported!

edited Jan 2 '15 at 11:04

Mathieu Rodic

4,72022940

answered Sep 13 '14 at 19:57

bob_twinkles

1869

add a comment |

import pkgutil

import importlib



packages = pkgutil.walk_packages(path='.')

for importer, name, is_package in packages:

    mod = importlib.import_module(name)

    # do whatever you want with module now, it's been imported!

edited Jan 2 '15 at 11:04

Mathieu Rodic

4,72022940

answered Sep 13 '14 at 19:57

bob_twinkles

1869

add a comment |

import pkgutil

import importlib



packages = pkgutil.walk_packages(path='.')

for importer, name, is_package in packages:

    mod = importlib.import_module(name)

    # do whatever you want with module now, it's been imported!

edited Jan 2 '15 at 11:04

Mathieu Rodic

4,72022940

answered Sep 13 '14 at 19:57

bob_twinkles

1869

import pkgutil

import importlib



packages = pkgutil.walk_packages(path='.')

for importer, name, is_package in packages:

    mod = importlib.import_module(name)

    # do whatever you want with module now, it's been imported!

edited Jan 2 '15 at 11:04

Mathieu Rodic

4,72022940

answered Sep 13 '14 at 19:57

bob_twinkles

1869

edited Jan 2 '15 at 11:04

Mathieu Rodic

4,72022940

edited Jan 2 '15 at 11:04

Mathieu Rodic

4,72022940

edited Jan 2 '15 at 11:04

Mathieu Rodic

4,72022940

answered Sep 13 '14 at 19:57

bob_twinkles

1869

answered Sep 13 '14 at 19:57

bob_twinkles

1869

answered Sep 13 '14 at 19:57

bob_twinkles

1869

add a comment |

In Linux, adding a symbolic link in the directory your python script is located works.

ie:

ln -s /absolute/path/to/module/module.py /absolute/path/to/script/module.py

python will create /absolute/path/to/script/module.pyc and will update it if you change the contents of /absolute/path/to/module/module.py

then include the following in mypythonscript.py

from module import *

edited Apr 11 '18 at 17:21

Micah Smith

2,6211422

answered Nov 18 '14 at 13:06

user2760152

234

1

This is the hack I used, and it has caused me some problems. One of the more painful ones was that IDEA has an issue where it doesn't pickup altered code from within the link, but yet attempts to save what it thinks is there. A race condition where the last to save is what sticks... I lost a decent amount of work because of this.

– Gripp
Jun 16 '15 at 23:23

@Gripp not sure if I am understanding your issue, but I frequently (almost exclusively) edit my scripts on a remote server from my desktop via SFTP with a client like CyberDuck, and in that case as well it is a bad idea to try and edit the symlinked file, instead its much safer to edit the original file. You can catch some of these issues by using git and checking your git status to verify that your changes to the script are actually making it back to the source document and not getting lost in the ether.

– user5359531
Aug 1 '17 at 19:39

add a comment |

In Linux, adding a symbolic link in the directory your python script is located works.

ie:

ln -s /absolute/path/to/module/module.py /absolute/path/to/script/module.py

python will create /absolute/path/to/script/module.pyc and will update it if you change the contents of /absolute/path/to/module/module.py

then include the following in mypythonscript.py

from module import *

edited Apr 11 '18 at 17:21

Micah Smith

2,6211422

answered Nov 18 '14 at 13:06

user2760152

234

1

This is the hack I used, and it has caused me some problems. One of the more painful ones was that IDEA has an issue where it doesn't pickup altered code from within the link, but yet attempts to save what it thinks is there. A race condition where the last to save is what sticks... I lost a decent amount of work because of this.

– Gripp
Jun 16 '15 at 23:23

@Gripp not sure if I am understanding your issue, but I frequently (almost exclusively) edit my scripts on a remote server from my desktop via SFTP with a client like CyberDuck, and in that case as well it is a bad idea to try and edit the symlinked file, instead its much safer to edit the original file. You can catch some of these issues by using git and checking your git status to verify that your changes to the script are actually making it back to the source document and not getting lost in the ether.

– user5359531
Aug 1 '17 at 19:39

add a comment |

In Linux, adding a symbolic link in the directory your python script is located works.

ie:

ln -s /absolute/path/to/module/module.py /absolute/path/to/script/module.py

python will create /absolute/path/to/script/module.pyc and will update it if you change the contents of /absolute/path/to/module/module.py

then include the following in mypythonscript.py

from module import *

edited Apr 11 '18 at 17:21

Micah Smith

2,6211422

answered Nov 18 '14 at 13:06

user2760152

234

In Linux, adding a symbolic link in the directory your python script is located works.

ie:

ln -s /absolute/path/to/module/module.py /absolute/path/to/script/module.py

python will create /absolute/path/to/script/module.pyc and will update it if you change the contents of /absolute/path/to/module/module.py

then include the following in mypythonscript.py

from module import *

edited Apr 11 '18 at 17:21

Micah Smith

2,6211422

answered Nov 18 '14 at 13:06

user2760152

234

edited Apr 11 '18 at 17:21

Micah Smith

2,6211422

edited Apr 11 '18 at 17:21

Micah Smith

2,6211422

edited Apr 11 '18 at 17:21

Micah Smith

2,6211422

answered Nov 18 '14 at 13:06

user2760152

234

answered Nov 18 '14 at 13:06

user2760152

234

answered Nov 18 '14 at 13:06

user2760152

234

1

This is the hack I used, and it has caused me some problems. One of the more painful ones was that IDEA has an issue where it doesn't pickup altered code from within the link, but yet attempts to save what it thinks is there. A race condition where the last to save is what sticks... I lost a decent amount of work because of this.

– Gripp
Jun 16 '15 at 23:23

@Gripp not sure if I am understanding your issue, but I frequently (almost exclusively) edit my scripts on a remote server from my desktop via SFTP with a client like CyberDuck, and in that case as well it is a bad idea to try and edit the symlinked file, instead its much safer to edit the original file. You can catch some of these issues by using git and checking your git status to verify that your changes to the script are actually making it back to the source document and not getting lost in the ether.

– user5359531
Aug 1 '17 at 19:39

add a comment |

1

This is the hack I used, and it has caused me some problems. One of the more painful ones was that IDEA has an issue where it doesn't pickup altered code from within the link, but yet attempts to save what it thinks is there. A race condition where the last to save is what sticks... I lost a decent amount of work because of this.

– Gripp
Jun 16 '15 at 23:23

@Gripp not sure if I am understanding your issue, but I frequently (almost exclusively) edit my scripts on a remote server from my desktop via SFTP with a client like CyberDuck, and in that case as well it is a bad idea to try and edit the symlinked file, instead its much safer to edit the original file. You can catch some of these issues by using git and checking your git status to verify that your changes to the script are actually making it back to the source document and not getting lost in the ether.

– user5359531
Aug 1 '17 at 19:39

This is the hack I used, and it has caused me some problems. One of the more painful ones was that IDEA has an issue where it doesn't pickup altered code from within the link, but yet attempts to save what it thinks is there. A race condition where the last to save is what sticks... I lost a decent amount of work because of this.

– Gripp
Jun 16 '15 at 23:23

@Gripp not sure if I am understanding your issue, but I frequently (almost exclusively) edit my scripts on a remote server from my desktop via SFTP with a client like CyberDuck, and in that case as well it is a bad idea to try and edit the symlinked file, instead its much safer to edit the original file. You can catch some of these issues by using git and checking your git status to verify that your changes to the script are actually making it back to the source document and not getting lost in the ether.

– user5359531
Aug 1 '17 at 19:39

add a comment |

quite simple way: suppose you want import file with relative path ../../MyLibs/pyfunc.py



libPath = '../../MyLibs'

import sys

if not libPath in sys.path: sys.path.append(libPath)

import pyfunc as pf

But if you make it without a guard you can finally get a very long path

answered Jan 26 '18 at 4:52

Andrei Keino

612

add a comment |

quite simple way: suppose you want import file with relative path ../../MyLibs/pyfunc.py



libPath = '../../MyLibs'

import sys

if not libPath in sys.path: sys.path.append(libPath)

import pyfunc as pf

But if you make it without a guard you can finally get a very long path

answered Jan 26 '18 at 4:52

Andrei Keino

612

add a comment |

quite simple way: suppose you want import file with relative path ../../MyLibs/pyfunc.py



libPath = '../../MyLibs'

import sys

if not libPath in sys.path: sys.path.append(libPath)

import pyfunc as pf

But if you make it without a guard you can finally get a very long path

answered Jan 26 '18 at 4:52

Andrei Keino

612

quite simple way: suppose you want import file with relative path ../../MyLibs/pyfunc.py



libPath = '../../MyLibs'

import sys

if not libPath in sys.path: sys.path.append(libPath)

import pyfunc as pf

But if you make it without a guard you can finally get a very long path

answered Jan 26 '18 at 4:52

Andrei Keino

612

answered Jan 26 '18 at 4:52

Andrei Keino

612

answered Jan 26 '18 at 4:52

Andrei Keino

612

answered Jan 26 '18 at 4:52

Andrei Keino

612

add a comment |

A simple solution using importlib instead of the imp package (tested for Python 2.7, although it should work for Python 3 too):

import importlib



dirname, basename = os.path.split(pyfilepath) # pyfilepath: '/my/path/mymodule.py'

sys.path.append(dirname) # only directories should be added to PYTHONPATH

module_name = os.path.splitext(basename)[0] # '/my/path/mymodule.py' --> 'mymodule'

module = importlib.import_module(module_name) # name space of defined module (otherwise we would literally look for "module_name")

Now you can directly use the namespace of the imported module, like this:

a = module.myvar

b = module.myfunc(a)

edited Jul 28 '18 at 2:24

answered May 24 '18 at 12:07

Ataxias

175110

This way you are modifying the sys.path, which does not fit every use case.

– bgusach
Jul 19 '18 at 14:26

@bgusach This may be true, but it is also desirable in some cases (adding a path to sys.path simplifies things when importing more than one module from a single package). At any rate, if this not desirable, one can immediately afterwards do sys.path.pop()

– Ataxias
Jul 20 '18 at 16:38

add a comment |

A simple solution using importlib instead of the imp package (tested for Python 2.7, although it should work for Python 3 too):

import importlib



dirname, basename = os.path.split(pyfilepath) # pyfilepath: '/my/path/mymodule.py'

sys.path.append(dirname) # only directories should be added to PYTHONPATH

module_name = os.path.splitext(basename)[0] # '/my/path/mymodule.py' --> 'mymodule'

module = importlib.import_module(module_name) # name space of defined module (otherwise we would literally look for "module_name")

Now you can directly use the namespace of the imported module, like this:

a = module.myvar

b = module.myfunc(a)

edited Jul 28 '18 at 2:24

answered May 24 '18 at 12:07

Ataxias

175110

This way you are modifying the sys.path, which does not fit every use case.

– bgusach
Jul 19 '18 at 14:26

@bgusach This may be true, but it is also desirable in some cases (adding a path to sys.path simplifies things when importing more than one module from a single package). At any rate, if this not desirable, one can immediately afterwards do sys.path.pop()

– Ataxias
Jul 20 '18 at 16:38

add a comment |

A simple solution using importlib instead of the imp package (tested for Python 2.7, although it should work for Python 3 too):

import importlib



dirname, basename = os.path.split(pyfilepath) # pyfilepath: '/my/path/mymodule.py'

sys.path.append(dirname) # only directories should be added to PYTHONPATH

module_name = os.path.splitext(basename)[0] # '/my/path/mymodule.py' --> 'mymodule'

module = importlib.import_module(module_name) # name space of defined module (otherwise we would literally look for "module_name")

Now you can directly use the namespace of the imported module, like this:

a = module.myvar

b = module.myfunc(a)

edited Jul 28 '18 at 2:24

answered May 24 '18 at 12:07

Ataxias

175110

A simple solution using importlib instead of the imp package (tested for Python 2.7, although it should work for Python 3 too):

import importlib



dirname, basename = os.path.split(pyfilepath) # pyfilepath: '/my/path/mymodule.py'

sys.path.append(dirname) # only directories should be added to PYTHONPATH

module_name = os.path.splitext(basename)[0] # '/my/path/mymodule.py' --> 'mymodule'

module = importlib.import_module(module_name) # name space of defined module (otherwise we would literally look for "module_name")

Now you can directly use the namespace of the imported module, like this:

a = module.myvar

b = module.myfunc(a)

edited Jul 28 '18 at 2:24

answered May 24 '18 at 12:07

Ataxias

175110

edited Jul 28 '18 at 2:24

answered May 24 '18 at 12:07

Ataxias

175110

answered May 24 '18 at 12:07

Ataxias

175110

answered May 24 '18 at 12:07

Ataxias

175110

This way you are modifying the sys.path, which does not fit every use case.

– bgusach
Jul 19 '18 at 14:26

@bgusach This may be true, but it is also desirable in some cases (adding a path to sys.path simplifies things when importing more than one module from a single package). At any rate, if this not desirable, one can immediately afterwards do sys.path.pop()

– Ataxias
Jul 20 '18 at 16:38

add a comment |

This way you are modifying the sys.path, which does not fit every use case.

– bgusach
Jul 19 '18 at 14:26

@bgusach This may be true, but it is also desirable in some cases (adding a path to sys.path simplifies things when importing more than one module from a single package). At any rate, if this not desirable, one can immediately afterwards do sys.path.pop()

– Ataxias
Jul 20 '18 at 16:38

This way you are modifying the sys.path, which does not fit every use case.

– bgusach
Jul 19 '18 at 14:26

@bgusach This may be true, but it is also desirable in some cases (adding a path to sys.path simplifies things when importing more than one module from a single package). At any rate, if this not desirable, one can immediately afterwards do sys.path.pop()

– Ataxias
Jul 20 '18 at 16:38

add a comment |

Create python module test.py

import sys

sys.path.append("<project-path>/lib/")

from tes1 import Client1

from tes2 import Client2

import tes3

Create python module test_check.py

from test import Client1

from test import Client2

from test import test3

We can import the imported module from module.

answered Dec 6 '18 at 12:41

abhimanyu

285

add a comment |

Create python module test.py

import sys

sys.path.append("<project-path>/lib/")

from tes1 import Client1

from tes2 import Client2

import tes3

Create python module test_check.py

from test import Client1

from test import Client2

from test import test3

We can import the imported module from module.

answered Dec 6 '18 at 12:41

abhimanyu

285

add a comment |

Create python module test.py

import sys

sys.path.append("<project-path>/lib/")

from tes1 import Client1

from tes2 import Client2

import tes3

Create python module test_check.py

from test import Client1

from test import Client2

from test import test3

We can import the imported module from module.

answered Dec 6 '18 at 12:41

abhimanyu

285

Create python module test.py

import sys

sys.path.append("<project-path>/lib/")

from tes1 import Client1

from tes2 import Client2

import tes3

Create python module test_check.py

from test import Client1

from test import Client2

from test import test3

We can import the imported module from module.

answered Dec 6 '18 at 12:41

abhimanyu

285

answered Dec 6 '18 at 12:41

abhimanyu

285

answered Dec 6 '18 at 12:41

abhimanyu

285

answered Dec 6 '18 at 12:41

abhimanyu

285

add a comment |

Adding this to the list of answers as I couldn't find anything that worked. This will allow imports of compiled (pyd) python modules in 3.4:

import sys

import importlib.machinery



def load_module(name, filename):

    # If the Loader finds the module name in this list it will use

    # module_name.__file__ instead so we need to delete it here

    if name in sys.modules:

        del sys.modules[name]

    loader = importlib.machinery.ExtensionFileLoader(name, filename)

    module = loader.load_module()

    locals()[name] = module

    globals()[name] = module



load_module('something', r'C:PathTosomething.pyd')

something.do_something()

answered Jan 10 '18 at 15:58

David

314115

add a comment |

Adding this to the list of answers as I couldn't find anything that worked. This will allow imports of compiled (pyd) python modules in 3.4:

import sys

import importlib.machinery



def load_module(name, filename):

    # If the Loader finds the module name in this list it will use

    # module_name.__file__ instead so we need to delete it here

    if name in sys.modules:

        del sys.modules[name]

    loader = importlib.machinery.ExtensionFileLoader(name, filename)

    module = loader.load_module()

    locals()[name] = module

    globals()[name] = module



load_module('something', r'C:PathTosomething.pyd')

something.do_something()

answered Jan 10 '18 at 15:58

David

314115

add a comment |

Adding this to the list of answers as I couldn't find anything that worked. This will allow imports of compiled (pyd) python modules in 3.4:

import sys

import importlib.machinery



def load_module(name, filename):

    # If the Loader finds the module name in this list it will use

    # module_name.__file__ instead so we need to delete it here

    if name in sys.modules:

        del sys.modules[name]

    loader = importlib.machinery.ExtensionFileLoader(name, filename)

    module = loader.load_module()

    locals()[name] = module

    globals()[name] = module



load_module('something', r'C:PathTosomething.pyd')

something.do_something()

answered Jan 10 '18 at 15:58

David

314115

Adding this to the list of answers as I couldn't find anything that worked. This will allow imports of compiled (pyd) python modules in 3.4:

import sys

import importlib.machinery



def load_module(name, filename):

    # If the Loader finds the module name in this list it will use

    # module_name.__file__ instead so we need to delete it here

    if name in sys.modules:

        del sys.modules[name]

    loader = importlib.machinery.ExtensionFileLoader(name, filename)

    module = loader.load_module()

    locals()[name] = module

    globals()[name] = module



load_module('something', r'C:PathTosomething.pyd')

something.do_something()

answered Jan 10 '18 at 15:58

David

314115

answered Jan 10 '18 at 15:58

David

314115

answered Jan 10 '18 at 15:58

David

314115

answered Jan 10 '18 at 15:58

David

314115

add a comment |

import pathlib



def likely_python_module(filename):

    '''

    Given a filename or Path, return the "likely" python module name.  That is, iterate

    the parent directories until it doesn't contain an __init__.py file.



    :rtype: str

    '''

    p = pathlib.Path(filename).resolve()

    paths = 

    if p.name != '__init__.py':

        paths.append(p.stem)

    while True:

        p = p.parent

        if not p:

            break

        if not p.is_dir():

            break



        inits = [f for f in p.iterdir() if f.name == '__init__.py']

        if not inits:

            break



        paths.append(p.stem)



    return '.'.join(reversed(paths))

There are certainly possibilities for improvement, and the optional __init__.py files might necessitate other changes, but if you have __init__.py in general, this does the trick.

answered Sep 8 '18 at 15:29

Michael Scott Cuthbert

1,83121333

add a comment |

import pathlib



def likely_python_module(filename):

    '''

    Given a filename or Path, return the "likely" python module name.  That is, iterate

    the parent directories until it doesn't contain an __init__.py file.



    :rtype: str

    '''

    p = pathlib.Path(filename).resolve()

    paths = 

    if p.name != '__init__.py':

        paths.append(p.stem)

    while True:

        p = p.parent

        if not p:

            break

        if not p.is_dir():

            break



        inits = [f for f in p.iterdir() if f.name == '__init__.py']

        if not inits:

            break



        paths.append(p.stem)



    return '.'.join(reversed(paths))

There are certainly possibilities for improvement, and the optional __init__.py files might necessitate other changes, but if you have __init__.py in general, this does the trick.

answered Sep 8 '18 at 15:29

Michael Scott Cuthbert

1,83121333

add a comment |

import pathlib



def likely_python_module(filename):

    '''

    Given a filename or Path, return the "likely" python module name.  That is, iterate

    the parent directories until it doesn't contain an __init__.py file.



    :rtype: str

    '''

    p = pathlib.Path(filename).resolve()

    paths = 

    if p.name != '__init__.py':

        paths.append(p.stem)

    while True:

        p = p.parent

        if not p:

            break

        if not p.is_dir():

            break



        inits = [f for f in p.iterdir() if f.name == '__init__.py']

        if not inits:

            break



        paths.append(p.stem)



    return '.'.join(reversed(paths))

There are certainly possibilities for improvement, and the optional __init__.py files might necessitate other changes, but if you have __init__.py in general, this does the trick.

answered Sep 8 '18 at 15:29

Michael Scott Cuthbert

1,83121333

import pathlib



def likely_python_module(filename):

    '''

    Given a filename or Path, return the "likely" python module name.  That is, iterate

    the parent directories until it doesn't contain an __init__.py file.



    :rtype: str

    '''

    p = pathlib.Path(filename).resolve()

    paths = 

    if p.name != '__init__.py':

        paths.append(p.stem)

    while True:

        p = p.parent

        if not p:

            break

        if not p.is_dir():

            break



        inits = [f for f in p.iterdir() if f.name == '__init__.py']

        if not inits:

            break



        paths.append(p.stem)



    return '.'.join(reversed(paths))

There are certainly possibilities for improvement, and the optional __init__.py files might necessitate other changes, but if you have __init__.py in general, this does the trick.

answered Sep 8 '18 at 15:29

Michael Scott Cuthbert

1,83121333

answered Sep 8 '18 at 15:29

Michael Scott Cuthbert

1,83121333

answered Sep 8 '18 at 15:29

Michael Scott Cuthbert

1,83121333

answered Sep 8 '18 at 15:29

Michael Scott Cuthbert

1,83121333

add a comment |

-1

The best way, I think, is from the official documentation (29.1. imp — Access the import internals):

import imp

import sys



def __import__(name, globals=None, locals=None, fromlist=None):

    # Fast path: see if the module has already been imported.

    try:

        return sys.modules[name]

    except KeyError:

        pass



    # If any of the following calls raises an exception,

    # there's a problem we can't handle -- let the caller handle it.



    fp, pathname, description = imp.find_module(name)



    try:

        return imp.load_module(name, fp, pathname, description)

    finally:

        # Since we may exit via an exception, close fp explicitly.

        if fp:

            fp.close()

answered Nov 25 '14 at 12:58

Zompa

111

1

This solution does not allow you to provide the path, which is what the question asks for.

– Micah Smith
Apr 11 '18 at 17:20

add a comment |

-1

The best way, I think, is from the official documentation (29.1. imp — Access the import internals):

import imp

import sys



def __import__(name, globals=None, locals=None, fromlist=None):

    # Fast path: see if the module has already been imported.

    try:

        return sys.modules[name]

    except KeyError:

        pass



    # If any of the following calls raises an exception,

    # there's a problem we can't handle -- let the caller handle it.



    fp, pathname, description = imp.find_module(name)



    try:

        return imp.load_module(name, fp, pathname, description)

    finally:

        # Since we may exit via an exception, close fp explicitly.

        if fp:

            fp.close()

answered Nov 25 '14 at 12:58

Zompa

111

1

This solution does not allow you to provide the path, which is what the question asks for.

– Micah Smith
Apr 11 '18 at 17:20

add a comment |

-1

The best way, I think, is from the official documentation (29.1. imp — Access the import internals):

import imp

import sys



def __import__(name, globals=None, locals=None, fromlist=None):

    # Fast path: see if the module has already been imported.

    try:

        return sys.modules[name]

    except KeyError:

        pass



    # If any of the following calls raises an exception,

    # there's a problem we can't handle -- let the caller handle it.



    fp, pathname, description = imp.find_module(name)



    try:

        return imp.load_module(name, fp, pathname, description)

    finally:

        # Since we may exit via an exception, close fp explicitly.

        if fp:

            fp.close()

answered Nov 25 '14 at 12:58

Zompa

111

The best way, I think, is from the official documentation (29.1. imp — Access the import internals):

import imp

import sys



def __import__(name, globals=None, locals=None, fromlist=None):

    # Fast path: see if the module has already been imported.

    try:

        return sys.modules[name]

    except KeyError:

        pass



    # If any of the following calls raises an exception,

    # there's a problem we can't handle -- let the caller handle it.



    fp, pathname, description = imp.find_module(name)



    try:

        return imp.load_module(name, fp, pathname, description)

    finally:

        # Since we may exit via an exception, close fp explicitly.

        if fp:

            fp.close()

answered Nov 25 '14 at 12:58

Zompa

111

answered Nov 25 '14 at 12:58

Zompa

111

answered Nov 25 '14 at 12:58

Zompa

111

answered Nov 25 '14 at 12:58

Zompa

111

1

This solution does not allow you to provide the path, which is what the question asks for.

– Micah Smith
Apr 11 '18 at 17:20

add a comment |

1

This solution does not allow you to provide the path, which is what the question asks for.

– Micah Smith
Apr 11 '18 at 17:20

This solution does not allow you to provide the path, which is what the question asks for.

– Micah Smith
Apr 11 '18 at 17:20

add a comment |

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