How to get full path of current file's directory in Python?

555

I want to get the current file's directory path.

I tried:

>>> os.path.abspath(__file__)

'C:\python27\test.py'

But how can I retrieve the directory's path?
For example:

'C:\python27\'

edited Dec 18 '14 at 21:58

Ciro Santilli 新疆改造中心六四事件法轮功

147k34558473

asked Aug 7 '10 at 12:17

Shubham

6,313154678

3

possible duplicate of Find current directory and file's directory

– user2284570
May 23 '14 at 15:01

4

__file__ is not defined when you run python as an interactive shell. The first piece of code in your question looks like it's from an interactive shell, but would actually produce a NameError, at least on python 2.7.3, but others too I guess.

– drevicko
May 31 '15 at 1:04

add a comment |

555

I want to get the current file's directory path.

I tried:

>>> os.path.abspath(__file__)

'C:\python27\test.py'

But how can I retrieve the directory's path?
For example:

'C:\python27\'

edited Dec 18 '14 at 21:58

Ciro Santilli 新疆改造中心六四事件法轮功

147k34558473

asked Aug 7 '10 at 12:17

Shubham

6,313154678

3

possible duplicate of Find current directory and file's directory

– user2284570
May 23 '14 at 15:01

4

__file__ is not defined when you run python as an interactive shell. The first piece of code in your question looks like it's from an interactive shell, but would actually produce a NameError, at least on python 2.7.3, but others too I guess.

– drevicko
May 31 '15 at 1:04

add a comment |

555

115

I want to get the current file's directory path.

I tried:

>>> os.path.abspath(__file__)

'C:\python27\test.py'

But how can I retrieve the directory's path?
For example:

'C:\python27\'

edited Dec 18 '14 at 21:58

Ciro Santilli 新疆改造中心六四事件法轮功

147k34558473

asked Aug 7 '10 at 12:17

Shubham

6,313154678

I want to get the current file's directory path.

I tried:

>>> os.path.abspath(__file__)

'C:\python27\test.py'

But how can I retrieve the directory's path?
For example:

'C:\python27\'

python filesystems

edited Dec 18 '14 at 21:58

Ciro Santilli 新疆改造中心六四事件法轮功

147k34558473

asked Aug 7 '10 at 12:17

Shubham

6,313154678

edited Dec 18 '14 at 21:58

Ciro Santilli 新疆改造中心六四事件法轮功

147k34558473

asked Aug 7 '10 at 12:17

Shubham

6,313154678

edited Dec 18 '14 at 21:58

Ciro Santilli 新疆改造中心六四事件法轮功

147k34558473

edited Dec 18 '14 at 21:58

Ciro Santilli 新疆改造中心六四事件法轮功

147k34558473

edited Dec 18 '14 at 21:58

Ciro Santilli 新疆改造中心六四事件法轮功

147k34558473

asked Aug 7 '10 at 12:17

Shubham

6,313154678

asked Aug 7 '10 at 12:17

Shubham

6,313154678

asked Aug 7 '10 at 12:17

Shubham

6,313154678

3

possible duplicate of Find current directory and file's directory

– user2284570
May 23 '14 at 15:01

4

__file__ is not defined when you run python as an interactive shell. The first piece of code in your question looks like it's from an interactive shell, but would actually produce a NameError, at least on python 2.7.3, but others too I guess.

– drevicko
May 31 '15 at 1:04

add a comment |

3

possible duplicate of Find current directory and file's directory

– user2284570
May 23 '14 at 15:01

4

__file__ is not defined when you run python as an interactive shell. The first piece of code in your question looks like it's from an interactive shell, but would actually produce a NameError, at least on python 2.7.3, but others too I guess.

– drevicko
May 31 '15 at 1:04

possible duplicate of Find current directory and file's directory

– user2284570
May 23 '14 at 15:01

__file__ is not defined when you run python as an interactive shell. The first piece of code in your question looks like it's from an interactive shell, but would actually produce a NameError, at least on python 2.7.3, but others too I guess.

– drevicko
May 31 '15 at 1:04

add a comment |

11 Answers
11

active

oldest

votes

1163

If you mean the directory of the script being run:

import os

os.path.dirname(os.path.abspath(__file__))

If you mean the current working directory:

import os

os.getcwd()

Note that before and after file is two underscores, not just one.

Also note that if you are running interactively or have loaded code from something other than a file (eg: a database or online resource), __file__ may not be set since there is no notion of "current file". The above answer assumes the most common scenario of running a python script that is in a file.

edited Jan 3 at 22:42

answered Aug 7 '10 at 12:24

Bryan Oakley

221k22274434

34

abspath() is mandatory if you do not want to discover weird behaviours on windows, where dirname(file) may return an empty string!

– sorin
Oct 25 '11 at 10:10

4

should be os.path.dirname(os.path.abspath(os.__file__))?

– DrBailey
Mar 27 '14 at 12:28

2

@DrBailey: no, there's nothing special about ActivePython. __file__ (note that it's two underscores on either side of the word) is a standard part of python. It's not available in C-based modules, for example, but it should always be available in a python script.

– Bryan Oakley
Apr 17 '14 at 21:32

7

I would recommend using realpath instead of abspath to resolve a possible symbolic link.

– TTimo
Jan 9 '15 at 21:37

7

@cph2117: this will only work if you run it in a script. There is no __file__ if running from an interactive prompt.

– Bryan Oakley
Aug 11 '16 at 21:57

|
show 12 more comments

In Python 3:

from pathlib import Path



mypath = Path().absolute()

print(mypath)

edited May 15 '18 at 20:56

A-B-B

24.2k66370

answered Apr 30 '18 at 10:51

Ron Kalian

806515

7

I had to do Path(__file__).parent to get the folder that is containing the file

– YellowPillow
Jun 6 '18 at 4:09

That is correct @YellowPillow, Path(__file__) gets you the file. .parent gets you one level above ie the containing directory. You can add more .parent to that to go up as many directories as you require.

– Ron Kalian
Jun 6 '18 at 8:18

Sorry I should've have made this clearer, but if Path().absolute() exists in some module located at path/to/module and you're calling the module from some script located at path/to/script then would return path/to/script instead of path/to/module

– YellowPillow
Jun 6 '18 at 12:22

@YellowPillow Path(__file__).cwd() is more explicit

– Charles
Feb 8 at 16:00

Path(__file__) doesn't always work, for example, it doesn't work in Jupyter Notebook. Path().absolute() solves that problem.

– Ron Kalian
Feb 8 at 16:09

add a comment |

import os

print os.path.dirname(__file__)

answered Aug 7 '10 at 12:24

chefsmart

3,84283446

19

Sorry but this answer is incorrect, the correct one is the one made by Bryan `dirname(abspath(file)). See comments for details.

– sorin
Oct 25 '11 at 10:11

1

It will give / as output

– Tripathi29
Sep 24 '15 at 6:31

1

@sorin Actually on Python 3.6 they are both the same

– Akshay L Aradhya
Apr 2 '18 at 12:25

add a comment |

You can use os and os.path library easily as follows

import os

os.chdir(os.path.dirname(os.getcwd()))

os.path.dirname returns upper directory from current one.
It lets us change to an upper level without passing any file argument and without knowing absolute path.

edited Apr 2 '17 at 18:38

Paolo

10.3k1467103

answered Oct 16 '14 at 13:05

mulg0r

1,51311222

6

This does not give the directory of the current file. It returns the directory of the current working directory which could be completely different. What you suggest only works if the file is in the current working directory.

– Bryan Oakley
Jul 1 '16 at 19:34

add a comment |

In Python 3.x I do:

from pathlib import Path



path = Path(__file__).parent.absolute()

Explanation:

Path(__file__) is the path to the current file.

.parent gives you the directory the file is in.

.absolute() gives you the full absolute path to it.

Using pathlib is the modern way to work with paths. If you need it as a string later for some reason, just do str(path).

answered Feb 26 at 18:36

Arminius

452514

add a comment |

IPython has a magic command %pwd to get the present working directory. It can be used in following way:

from IPython.terminal.embed import InteractiveShellEmbed



ip_shell = InteractiveShellEmbed()



present_working_directory = ip_shell.magic("%pwd")

On IPython Jupyter Notebook %pwd can be used directly as following:

present_working_directory = %pwd

answered Mar 7 '18 at 5:50

Nafeez Quraishi

595717

2

The question isn't about IPython

– Kiro
Apr 10 '18 at 8:07

1

@Kiro, my solution answers the question using IPython. For example If some one answers a question with a solution using a new library then also imho it remains a pertinent answer to the question.

– Nafeez Quraishi
Apr 11 '18 at 9:36

add a comment |

To keep the migration consistency across platforms (macOS/Windows/Linux), try:

path = r'%s' % os.getcwd().replace('\','/')

edited Apr 12 '18 at 7:33

answered Apr 12 '18 at 5:16

Qiao Zhang

11716

add a comment |

I have made a function to use when running python under IIS in CGI in order to get the current folder:

import os 

   def getLocalFolder():

        path=str(os.path.dirname(os.path.abspath(__file__))).split('\')

        return path[len(path)-1]

answered Dec 27 '18 at 10:35

Gil Allen

882920

add a comment |

System: MacOS

Version: Python 3.6 w/ Anaconda

import os rootpath = os.getcwd() os.chdir(rootpath)

answered Jan 30 at 1:42

Suyang Xu

112

add a comment |

USEFUL PATH PROPERTIES IN PYTHON:

 from pathlib import Path



    #Returns the path of the directory, where your script file is placed

    mypath = Path().absolute()

    print('Absolute path : {}'.format(mypath))



    #if you want to go to any other file inside the subdirectories of the directory path got from above method

    filePath = mypath/'data'/'fuel_econ.csv'

    print('File path : {}'.format(filePath))



    #To check if file present in that directory or Not

    isfileExist = filePath.exists()

    print('isfileExist : {}'.format(isfileExist))



    #To check if the path is a directory or a File

    isadirectory = filePath.is_dir()

    print('isadirectory : {}'.format(isadirectory))



    #To get the extension of the file

    fileExtension = mypath/'data'/'fuel_econ.csv'

    print('File extension : {}'.format(filePath.suffix))

OUTPUT:
ABSOLUTE PATH IS THE PATH WHERE YOUR PYTHON FILE IS PLACED

Absolute path : D:StudyMachine LearningJupitor NotebookJupytorNotebookTest2Udacity_ScriptsMatplotlib and seaborn Part2

File path : D:StudyMachine LearningJupitor NotebookJupytorNotebookTest2Udacity_ScriptsMatplotlib and seaborn Part2datafuel_econ.csv

isfileExist : True

isadirectory : False

File extension : .csv

edited Mar 10 at 17:21

answered Mar 10 at 4:06

Arpan Saini

428519

add a comment |

-1

## IMPORT MODULES

import os



## CALCULATE FILEPATH VARIABLE

filepath = os.path.abspath('') ## ~ os.getcwd()

## TEST TO MAKE SURE os.getcwd() is EQUIVALENT ALWAYS..

## ..OR DIFFERENT IN SOME CIRCUMSTANCES

edited Jan 6 at 8:47

answered Jan 2 at 13:13

Jerusalem Programmer

162212

2

The current working directory might not be the same as the directory of the current file. Your solution will only work if the current working directory is the same as the directory that holds the file. That can be fairly common during development, but is usually pretty rare for a deployed script.

– Bryan Oakley
Jan 3 at 22:45

Thanks for the info. I updated my answer to reflect the possibility that these may not be equivalent returning the same filepath. Such is reason why I adhere to philosophy of coding called "incremental programming", testing, and lots of print() function calls to test output and/or values return at each step of the way. I hope my code now better reflects the reality of os.path.abspath() vs. os.getcwd() since I do not claim to know everything. However you can see that I put os.getcwd() in comments as not the main solution, but rather as an alt piece of code worth remembering and checking.

– Jerusalem Programmer
Jan 6 at 8:45

P.S. I think someone edited and erased the majority of my answer that I posted here... :( Such discourages people from participating if their answers are going to be censored and chopped.

– Jerusalem Programmer
Jan 6 at 8:49

You're the only person to have edited this answer. Also, your edits still don't properly answer the question. The question is specifically asking about the path to the file's directory, and your answer only mentions getting the current directory. That is not a correct answer, since the current directory is very often not the same as the directory the script is in. This has nothing to do with abspath.

– Bryan Oakley
Jan 6 at 14:49

Most certainly the code I shared gives path to the file's directory if run from the directory and/or file from where you want to know the absolute file path. Most certainly it is a correct answer. Please don't be a troll.

– Jerusalem Programmer
Jan 7 at 4:48

|
show 2 more comments

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11 Answers
11

active

oldest

votes

11 Answers
11

active

oldest

votes

1163

If you mean the directory of the script being run:

import os

os.path.dirname(os.path.abspath(__file__))

If you mean the current working directory:

import os

os.getcwd()

Note that before and after file is two underscores, not just one.

edited Jan 3 at 22:42

answered Aug 7 '10 at 12:24

Bryan Oakley

221k22274434

34

abspath() is mandatory if you do not want to discover weird behaviours on windows, where dirname(file) may return an empty string!

– sorin
Oct 25 '11 at 10:10

4

should be os.path.dirname(os.path.abspath(os.__file__))?

– DrBailey
Mar 27 '14 at 12:28

2

@DrBailey: no, there's nothing special about ActivePython. __file__ (note that it's two underscores on either side of the word) is a standard part of python. It's not available in C-based modules, for example, but it should always be available in a python script.

– Bryan Oakley
Apr 17 '14 at 21:32

7

I would recommend using realpath instead of abspath to resolve a possible symbolic link.

– TTimo
Jan 9 '15 at 21:37

7

@cph2117: this will only work if you run it in a script. There is no __file__ if running from an interactive prompt.

– Bryan Oakley
Aug 11 '16 at 21:57

|
show 12 more comments

1163

If you mean the directory of the script being run:

import os

os.path.dirname(os.path.abspath(__file__))

If you mean the current working directory:

import os

os.getcwd()

Note that before and after file is two underscores, not just one.

edited Jan 3 at 22:42

answered Aug 7 '10 at 12:24

Bryan Oakley

221k22274434

34

abspath() is mandatory if you do not want to discover weird behaviours on windows, where dirname(file) may return an empty string!

– sorin
Oct 25 '11 at 10:10

4

should be os.path.dirname(os.path.abspath(os.__file__))?

– DrBailey
Mar 27 '14 at 12:28

2

@DrBailey: no, there's nothing special about ActivePython. __file__ (note that it's two underscores on either side of the word) is a standard part of python. It's not available in C-based modules, for example, but it should always be available in a python script.

– Bryan Oakley
Apr 17 '14 at 21:32

7

I would recommend using realpath instead of abspath to resolve a possible symbolic link.

– TTimo
Jan 9 '15 at 21:37

7

@cph2117: this will only work if you run it in a script. There is no __file__ if running from an interactive prompt.

– Bryan Oakley
Aug 11 '16 at 21:57

|
show 12 more comments

1163

If you mean the directory of the script being run:

import os

os.path.dirname(os.path.abspath(__file__))

If you mean the current working directory:

import os

os.getcwd()

Note that before and after file is two underscores, not just one.

edited Jan 3 at 22:42

answered Aug 7 '10 at 12:24

Bryan Oakley

221k22274434

If you mean the directory of the script being run:

import os

os.path.dirname(os.path.abspath(__file__))

If you mean the current working directory:

import os

os.getcwd()

Note that before and after file is two underscores, not just one.

edited Jan 3 at 22:42

answered Aug 7 '10 at 12:24

Bryan Oakley

221k22274434

edited Jan 3 at 22:42

answered Aug 7 '10 at 12:24

Bryan Oakley

221k22274434

answered Aug 7 '10 at 12:24

Bryan Oakley

221k22274434

answered Aug 7 '10 at 12:24

Bryan Oakley

221k22274434

34

abspath() is mandatory if you do not want to discover weird behaviours on windows, where dirname(file) may return an empty string!

– sorin
Oct 25 '11 at 10:10

4

should be os.path.dirname(os.path.abspath(os.__file__))?

– DrBailey
Mar 27 '14 at 12:28

2

@DrBailey: no, there's nothing special about ActivePython. __file__ (note that it's two underscores on either side of the word) is a standard part of python. It's not available in C-based modules, for example, but it should always be available in a python script.

– Bryan Oakley
Apr 17 '14 at 21:32

7

I would recommend using realpath instead of abspath to resolve a possible symbolic link.

– TTimo
Jan 9 '15 at 21:37

7

@cph2117: this will only work if you run it in a script. There is no __file__ if running from an interactive prompt.

– Bryan Oakley
Aug 11 '16 at 21:57

|
show 12 more comments

34

abspath() is mandatory if you do not want to discover weird behaviours on windows, where dirname(file) may return an empty string!

– sorin
Oct 25 '11 at 10:10

4

should be os.path.dirname(os.path.abspath(os.__file__))?

– DrBailey
Mar 27 '14 at 12:28

2

@DrBailey: no, there's nothing special about ActivePython. __file__ (note that it's two underscores on either side of the word) is a standard part of python. It's not available in C-based modules, for example, but it should always be available in a python script.

– Bryan Oakley
Apr 17 '14 at 21:32

7

I would recommend using realpath instead of abspath to resolve a possible symbolic link.

– TTimo
Jan 9 '15 at 21:37

7

@cph2117: this will only work if you run it in a script. There is no __file__ if running from an interactive prompt.

– Bryan Oakley
Aug 11 '16 at 21:57

abspath() is mandatory if you do not want to discover weird behaviours on windows, where dirname(file) may return an empty string!

– sorin
Oct 25 '11 at 10:10

should be os.path.dirname(os.path.abspath(os.__file__))?

– DrBailey
Mar 27 '14 at 12:28

@DrBailey: no, there's nothing special about ActivePython. __file__ (note that it's two underscores on either side of the word) is a standard part of python. It's not available in C-based modules, for example, but it should always be available in a python script.

– Bryan Oakley
Apr 17 '14 at 21:32

I would recommend using realpath instead of abspath to resolve a possible symbolic link.

– TTimo
Jan 9 '15 at 21:37

@cph2117: this will only work if you run it in a script. There is no __file__ if running from an interactive prompt.

– Bryan Oakley
Aug 11 '16 at 21:57

|
show 12 more comments

In Python 3:

from pathlib import Path



mypath = Path().absolute()

print(mypath)

edited May 15 '18 at 20:56

A-B-B

24.2k66370

answered Apr 30 '18 at 10:51

Ron Kalian

806515

7

I had to do Path(__file__).parent to get the folder that is containing the file

– YellowPillow
Jun 6 '18 at 4:09

That is correct @YellowPillow, Path(__file__) gets you the file. .parent gets you one level above ie the containing directory. You can add more .parent to that to go up as many directories as you require.

– Ron Kalian
Jun 6 '18 at 8:18

Sorry I should've have made this clearer, but if Path().absolute() exists in some module located at path/to/module and you're calling the module from some script located at path/to/script then would return path/to/script instead of path/to/module

– YellowPillow
Jun 6 '18 at 12:22

@YellowPillow Path(__file__).cwd() is more explicit

– Charles
Feb 8 at 16:00

Path(__file__) doesn't always work, for example, it doesn't work in Jupyter Notebook. Path().absolute() solves that problem.

– Ron Kalian
Feb 8 at 16:09

add a comment |

In Python 3:

from pathlib import Path



mypath = Path().absolute()

print(mypath)

edited May 15 '18 at 20:56

A-B-B

24.2k66370

answered Apr 30 '18 at 10:51

Ron Kalian

806515

7

I had to do Path(__file__).parent to get the folder that is containing the file

– YellowPillow
Jun 6 '18 at 4:09

That is correct @YellowPillow, Path(__file__) gets you the file. .parent gets you one level above ie the containing directory. You can add more .parent to that to go up as many directories as you require.

– Ron Kalian
Jun 6 '18 at 8:18

Sorry I should've have made this clearer, but if Path().absolute() exists in some module located at path/to/module and you're calling the module from some script located at path/to/script then would return path/to/script instead of path/to/module

– YellowPillow
Jun 6 '18 at 12:22

@YellowPillow Path(__file__).cwd() is more explicit

– Charles
Feb 8 at 16:00

Path(__file__) doesn't always work, for example, it doesn't work in Jupyter Notebook. Path().absolute() solves that problem.

– Ron Kalian
Feb 8 at 16:09

add a comment |

In Python 3:

from pathlib import Path



mypath = Path().absolute()

print(mypath)

edited May 15 '18 at 20:56

A-B-B

24.2k66370

answered Apr 30 '18 at 10:51

Ron Kalian

806515

In Python 3:

from pathlib import Path



mypath = Path().absolute()

print(mypath)

edited May 15 '18 at 20:56

A-B-B

24.2k66370

answered Apr 30 '18 at 10:51

Ron Kalian

806515

edited May 15 '18 at 20:56

A-B-B

24.2k66370

edited May 15 '18 at 20:56

A-B-B

24.2k66370

edited May 15 '18 at 20:56

A-B-B

24.2k66370

answered Apr 30 '18 at 10:51

Ron Kalian

806515

answered Apr 30 '18 at 10:51

Ron Kalian

806515

answered Apr 30 '18 at 10:51

Ron Kalian

806515

7

I had to do Path(__file__).parent to get the folder that is containing the file

– YellowPillow
Jun 6 '18 at 4:09

That is correct @YellowPillow, Path(__file__) gets you the file. .parent gets you one level above ie the containing directory. You can add more .parent to that to go up as many directories as you require.

– Ron Kalian
Jun 6 '18 at 8:18

Sorry I should've have made this clearer, but if Path().absolute() exists in some module located at path/to/module and you're calling the module from some script located at path/to/script then would return path/to/script instead of path/to/module

– YellowPillow
Jun 6 '18 at 12:22

@YellowPillow Path(__file__).cwd() is more explicit

– Charles
Feb 8 at 16:00

Path(__file__) doesn't always work, for example, it doesn't work in Jupyter Notebook. Path().absolute() solves that problem.

– Ron Kalian
Feb 8 at 16:09

add a comment |

7

I had to do Path(__file__).parent to get the folder that is containing the file

– YellowPillow
Jun 6 '18 at 4:09

That is correct @YellowPillow, Path(__file__) gets you the file. .parent gets you one level above ie the containing directory. You can add more .parent to that to go up as many directories as you require.

– Ron Kalian
Jun 6 '18 at 8:18

Sorry I should've have made this clearer, but if Path().absolute() exists in some module located at path/to/module and you're calling the module from some script located at path/to/script then would return path/to/script instead of path/to/module

– YellowPillow
Jun 6 '18 at 12:22

@YellowPillow Path(__file__).cwd() is more explicit

– Charles
Feb 8 at 16:00

Path(__file__) doesn't always work, for example, it doesn't work in Jupyter Notebook. Path().absolute() solves that problem.

– Ron Kalian
Feb 8 at 16:09

I had to do Path(__file__).parent to get the folder that is containing the file

– YellowPillow
Jun 6 '18 at 4:09

That is correct @YellowPillow, Path(__file__) gets you the file. .parent gets you one level above ie the containing directory. You can add more .parent to that to go up as many directories as you require.

– Ron Kalian
Jun 6 '18 at 8:18

Sorry I should've have made this clearer, but if Path().absolute() exists in some module located at path/to/module and you're calling the module from some script located at path/to/script then would return path/to/script instead of path/to/module

– YellowPillow
Jun 6 '18 at 12:22

@YellowPillow Path(__file__).cwd() is more explicit

– Charles
Feb 8 at 16:00

Path(__file__) doesn't always work, for example, it doesn't work in Jupyter Notebook. Path().absolute() solves that problem.

– Ron Kalian
Feb 8 at 16:09

add a comment |

import os

print os.path.dirname(__file__)

answered Aug 7 '10 at 12:24

chefsmart

3,84283446

19

Sorry but this answer is incorrect, the correct one is the one made by Bryan `dirname(abspath(file)). See comments for details.

– sorin
Oct 25 '11 at 10:11

1

It will give / as output

– Tripathi29
Sep 24 '15 at 6:31

1

@sorin Actually on Python 3.6 they are both the same

– Akshay L Aradhya
Apr 2 '18 at 12:25

add a comment |

import os

print os.path.dirname(__file__)

answered Aug 7 '10 at 12:24

chefsmart

3,84283446

19

Sorry but this answer is incorrect, the correct one is the one made by Bryan `dirname(abspath(file)). See comments for details.

– sorin
Oct 25 '11 at 10:11

1

It will give / as output

– Tripathi29
Sep 24 '15 at 6:31

1

@sorin Actually on Python 3.6 they are both the same

– Akshay L Aradhya
Apr 2 '18 at 12:25

add a comment |

import os

print os.path.dirname(__file__)

answered Aug 7 '10 at 12:24

chefsmart

3,84283446

import os

print os.path.dirname(__file__)

answered Aug 7 '10 at 12:24

chefsmart

3,84283446

answered Aug 7 '10 at 12:24

chefsmart

3,84283446

answered Aug 7 '10 at 12:24

chefsmart

3,84283446

answered Aug 7 '10 at 12:24

chefsmart

3,84283446

19

Sorry but this answer is incorrect, the correct one is the one made by Bryan `dirname(abspath(file)). See comments for details.

– sorin
Oct 25 '11 at 10:11

1

It will give / as output

– Tripathi29
Sep 24 '15 at 6:31

1

@sorin Actually on Python 3.6 they are both the same

– Akshay L Aradhya
Apr 2 '18 at 12:25

add a comment |

19

Sorry but this answer is incorrect, the correct one is the one made by Bryan `dirname(abspath(file)). See comments for details.

– sorin
Oct 25 '11 at 10:11

1

It will give / as output

– Tripathi29
Sep 24 '15 at 6:31

1

@sorin Actually on Python 3.6 they are both the same

– Akshay L Aradhya
Apr 2 '18 at 12:25

Sorry but this answer is incorrect, the correct one is the one made by Bryan `dirname(abspath(file)). See comments for details.

– sorin
Oct 25 '11 at 10:11

It will give / as output

– Tripathi29
Sep 24 '15 at 6:31

@sorin Actually on Python 3.6 they are both the same

– Akshay L Aradhya
Apr 2 '18 at 12:25

add a comment |

You can use os and os.path library easily as follows

import os

os.chdir(os.path.dirname(os.getcwd()))

os.path.dirname returns upper directory from current one.
It lets us change to an upper level without passing any file argument and without knowing absolute path.

edited Apr 2 '17 at 18:38

Paolo

10.3k1467103

answered Oct 16 '14 at 13:05

mulg0r

1,51311222

6

This does not give the directory of the current file. It returns the directory of the current working directory which could be completely different. What you suggest only works if the file is in the current working directory.

– Bryan Oakley
Jul 1 '16 at 19:34

add a comment |

You can use os and os.path library easily as follows

import os

os.chdir(os.path.dirname(os.getcwd()))

os.path.dirname returns upper directory from current one.
It lets us change to an upper level without passing any file argument and without knowing absolute path.

edited Apr 2 '17 at 18:38

Paolo

10.3k1467103

answered Oct 16 '14 at 13:05

mulg0r

1,51311222

6

This does not give the directory of the current file. It returns the directory of the current working directory which could be completely different. What you suggest only works if the file is in the current working directory.

– Bryan Oakley
Jul 1 '16 at 19:34

add a comment |

You can use os and os.path library easily as follows

import os

os.chdir(os.path.dirname(os.getcwd()))

os.path.dirname returns upper directory from current one.
It lets us change to an upper level without passing any file argument and without knowing absolute path.

edited Apr 2 '17 at 18:38

Paolo

10.3k1467103

answered Oct 16 '14 at 13:05

mulg0r

1,51311222

You can use os and os.path library easily as follows

import os

os.chdir(os.path.dirname(os.getcwd()))

os.path.dirname returns upper directory from current one.
It lets us change to an upper level without passing any file argument and without knowing absolute path.

edited Apr 2 '17 at 18:38

Paolo

10.3k1467103

answered Oct 16 '14 at 13:05

mulg0r

1,51311222

edited Apr 2 '17 at 18:38

Paolo

10.3k1467103

edited Apr 2 '17 at 18:38

Paolo

10.3k1467103

edited Apr 2 '17 at 18:38

Paolo

10.3k1467103

answered Oct 16 '14 at 13:05

mulg0r

1,51311222

answered Oct 16 '14 at 13:05

mulg0r

1,51311222

answered Oct 16 '14 at 13:05

mulg0r

1,51311222

6

This does not give the directory of the current file. It returns the directory of the current working directory which could be completely different. What you suggest only works if the file is in the current working directory.

– Bryan Oakley
Jul 1 '16 at 19:34

add a comment |

6

This does not give the directory of the current file. It returns the directory of the current working directory which could be completely different. What you suggest only works if the file is in the current working directory.

– Bryan Oakley
Jul 1 '16 at 19:34

This does not give the directory of the current file. It returns the directory of the current working directory which could be completely different. What you suggest only works if the file is in the current working directory.

– Bryan Oakley
Jul 1 '16 at 19:34

add a comment |

In Python 3.x I do:

from pathlib import Path



path = Path(__file__).parent.absolute()

Explanation:

Path(__file__) is the path to the current file.

.parent gives you the directory the file is in.

.absolute() gives you the full absolute path to it.

Using pathlib is the modern way to work with paths. If you need it as a string later for some reason, just do str(path).

answered Feb 26 at 18:36

Arminius

452514

add a comment |

In Python 3.x I do:

from pathlib import Path



path = Path(__file__).parent.absolute()

Explanation:

Path(__file__) is the path to the current file.

.parent gives you the directory the file is in.

.absolute() gives you the full absolute path to it.

Using pathlib is the modern way to work with paths. If you need it as a string later for some reason, just do str(path).

answered Feb 26 at 18:36

Arminius

452514

add a comment |

In Python 3.x I do:

from pathlib import Path



path = Path(__file__).parent.absolute()

Explanation:

Path(__file__) is the path to the current file.

.parent gives you the directory the file is in.

.absolute() gives you the full absolute path to it.

Using pathlib is the modern way to work with paths. If you need it as a string later for some reason, just do str(path).

answered Feb 26 at 18:36

Arminius

452514

In Python 3.x I do:

from pathlib import Path



path = Path(__file__).parent.absolute()

Explanation:

Path(__file__) is the path to the current file.

.parent gives you the directory the file is in.

.absolute() gives you the full absolute path to it.

Using pathlib is the modern way to work with paths. If you need it as a string later for some reason, just do str(path).

answered Feb 26 at 18:36

Arminius

452514

answered Feb 26 at 18:36

Arminius

452514

answered Feb 26 at 18:36

Arminius

452514

answered Feb 26 at 18:36

Arminius

452514

add a comment |

IPython has a magic command %pwd to get the present working directory. It can be used in following way:

from IPython.terminal.embed import InteractiveShellEmbed



ip_shell = InteractiveShellEmbed()



present_working_directory = ip_shell.magic("%pwd")

On IPython Jupyter Notebook %pwd can be used directly as following:

present_working_directory = %pwd

answered Mar 7 '18 at 5:50

Nafeez Quraishi

595717

2

The question isn't about IPython

– Kiro
Apr 10 '18 at 8:07

1

@Kiro, my solution answers the question using IPython. For example If some one answers a question with a solution using a new library then also imho it remains a pertinent answer to the question.

– Nafeez Quraishi
Apr 11 '18 at 9:36

add a comment |

IPython has a magic command %pwd to get the present working directory. It can be used in following way:

from IPython.terminal.embed import InteractiveShellEmbed



ip_shell = InteractiveShellEmbed()



present_working_directory = ip_shell.magic("%pwd")

On IPython Jupyter Notebook %pwd can be used directly as following:

present_working_directory = %pwd

answered Mar 7 '18 at 5:50

Nafeez Quraishi

595717

2

The question isn't about IPython

– Kiro
Apr 10 '18 at 8:07

1

@Kiro, my solution answers the question using IPython. For example If some one answers a question with a solution using a new library then also imho it remains a pertinent answer to the question.

– Nafeez Quraishi
Apr 11 '18 at 9:36

add a comment |

IPython has a magic command %pwd to get the present working directory. It can be used in following way:

from IPython.terminal.embed import InteractiveShellEmbed



ip_shell = InteractiveShellEmbed()



present_working_directory = ip_shell.magic("%pwd")

On IPython Jupyter Notebook %pwd can be used directly as following:

present_working_directory = %pwd

answered Mar 7 '18 at 5:50

Nafeez Quraishi

595717

IPython has a magic command %pwd to get the present working directory. It can be used in following way:

from IPython.terminal.embed import InteractiveShellEmbed



ip_shell = InteractiveShellEmbed()



present_working_directory = ip_shell.magic("%pwd")

On IPython Jupyter Notebook %pwd can be used directly as following:

present_working_directory = %pwd

answered Mar 7 '18 at 5:50

Nafeez Quraishi

595717

answered Mar 7 '18 at 5:50

Nafeez Quraishi

595717

answered Mar 7 '18 at 5:50

Nafeez Quraishi

595717

answered Mar 7 '18 at 5:50

Nafeez Quraishi

595717

2

The question isn't about IPython

– Kiro
Apr 10 '18 at 8:07

1

@Kiro, my solution answers the question using IPython. For example If some one answers a question with a solution using a new library then also imho it remains a pertinent answer to the question.

– Nafeez Quraishi
Apr 11 '18 at 9:36

add a comment |

2

The question isn't about IPython

– Kiro
Apr 10 '18 at 8:07

1

@Kiro, my solution answers the question using IPython. For example If some one answers a question with a solution using a new library then also imho it remains a pertinent answer to the question.

– Nafeez Quraishi
Apr 11 '18 at 9:36

The question isn't about IPython

– Kiro
Apr 10 '18 at 8:07

@Kiro, my solution answers the question using IPython. For example If some one answers a question with a solution using a new library then also imho it remains a pertinent answer to the question.

– Nafeez Quraishi
Apr 11 '18 at 9:36

add a comment |

To keep the migration consistency across platforms (macOS/Windows/Linux), try:

path = r'%s' % os.getcwd().replace('\','/')

edited Apr 12 '18 at 7:33

answered Apr 12 '18 at 5:16

Qiao Zhang

11716

add a comment |

To keep the migration consistency across platforms (macOS/Windows/Linux), try:

path = r'%s' % os.getcwd().replace('\','/')

edited Apr 12 '18 at 7:33

answered Apr 12 '18 at 5:16

Qiao Zhang

11716

add a comment |

To keep the migration consistency across platforms (macOS/Windows/Linux), try:

path = r'%s' % os.getcwd().replace('\','/')

edited Apr 12 '18 at 7:33

answered Apr 12 '18 at 5:16

Qiao Zhang

11716

To keep the migration consistency across platforms (macOS/Windows/Linux), try:

path = r'%s' % os.getcwd().replace('\','/')

edited Apr 12 '18 at 7:33

answered Apr 12 '18 at 5:16

Qiao Zhang

11716

edited Apr 12 '18 at 7:33

answered Apr 12 '18 at 5:16

Qiao Zhang

11716

answered Apr 12 '18 at 5:16

Qiao Zhang

11716

answered Apr 12 '18 at 5:16

Qiao Zhang

11716

add a comment |

I have made a function to use when running python under IIS in CGI in order to get the current folder:

import os 

   def getLocalFolder():

        path=str(os.path.dirname(os.path.abspath(__file__))).split('\')

        return path[len(path)-1]

answered Dec 27 '18 at 10:35

Gil Allen

882920

add a comment |

I have made a function to use when running python under IIS in CGI in order to get the current folder:

import os 

   def getLocalFolder():

        path=str(os.path.dirname(os.path.abspath(__file__))).split('\')

        return path[len(path)-1]

answered Dec 27 '18 at 10:35

Gil Allen

882920

add a comment |

I have made a function to use when running python under IIS in CGI in order to get the current folder:

import os 

   def getLocalFolder():

        path=str(os.path.dirname(os.path.abspath(__file__))).split('\')

        return path[len(path)-1]

answered Dec 27 '18 at 10:35

Gil Allen

882920

I have made a function to use when running python under IIS in CGI in order to get the current folder:

import os 

   def getLocalFolder():

        path=str(os.path.dirname(os.path.abspath(__file__))).split('\')

        return path[len(path)-1]

answered Dec 27 '18 at 10:35

Gil Allen

882920

answered Dec 27 '18 at 10:35

Gil Allen

882920

answered Dec 27 '18 at 10:35

Gil Allen

882920

answered Dec 27 '18 at 10:35

Gil Allen

882920

add a comment |

System: MacOS

Version: Python 3.6 w/ Anaconda

import os rootpath = os.getcwd() os.chdir(rootpath)

answered Jan 30 at 1:42

Suyang Xu

112

add a comment |

System: MacOS

Version: Python 3.6 w/ Anaconda

import os rootpath = os.getcwd() os.chdir(rootpath)

answered Jan 30 at 1:42

Suyang Xu

112

add a comment |

System: MacOS

Version: Python 3.6 w/ Anaconda

import os rootpath = os.getcwd() os.chdir(rootpath)

answered Jan 30 at 1:42

Suyang Xu

112

System: MacOS

Version: Python 3.6 w/ Anaconda

import os rootpath = os.getcwd() os.chdir(rootpath)

answered Jan 30 at 1:42

Suyang Xu

112

answered Jan 30 at 1:42

Suyang Xu

112

answered Jan 30 at 1:42

Suyang Xu

112

answered Jan 30 at 1:42

Suyang Xu

112

add a comment |

USEFUL PATH PROPERTIES IN PYTHON:

 from pathlib import Path



    #Returns the path of the directory, where your script file is placed

    mypath = Path().absolute()

    print('Absolute path : {}'.format(mypath))



    #if you want to go to any other file inside the subdirectories of the directory path got from above method

    filePath = mypath/'data'/'fuel_econ.csv'

    print('File path : {}'.format(filePath))



    #To check if file present in that directory or Not

    isfileExist = filePath.exists()

    print('isfileExist : {}'.format(isfileExist))



    #To check if the path is a directory or a File

    isadirectory = filePath.is_dir()

    print('isadirectory : {}'.format(isadirectory))



    #To get the extension of the file

    fileExtension = mypath/'data'/'fuel_econ.csv'

    print('File extension : {}'.format(filePath.suffix))

OUTPUT:
ABSOLUTE PATH IS THE PATH WHERE YOUR PYTHON FILE IS PLACED

Absolute path : D:StudyMachine LearningJupitor NotebookJupytorNotebookTest2Udacity_ScriptsMatplotlib and seaborn Part2

File path : D:StudyMachine LearningJupitor NotebookJupytorNotebookTest2Udacity_ScriptsMatplotlib and seaborn Part2datafuel_econ.csv

isfileExist : True

isadirectory : False

File extension : .csv

edited Mar 10 at 17:21

answered Mar 10 at 4:06

Arpan Saini

428519

add a comment |

USEFUL PATH PROPERTIES IN PYTHON:

 from pathlib import Path



    #Returns the path of the directory, where your script file is placed

    mypath = Path().absolute()

    print('Absolute path : {}'.format(mypath))



    #if you want to go to any other file inside the subdirectories of the directory path got from above method

    filePath = mypath/'data'/'fuel_econ.csv'

    print('File path : {}'.format(filePath))



    #To check if file present in that directory or Not

    isfileExist = filePath.exists()

    print('isfileExist : {}'.format(isfileExist))



    #To check if the path is a directory or a File

    isadirectory = filePath.is_dir()

    print('isadirectory : {}'.format(isadirectory))



    #To get the extension of the file

    fileExtension = mypath/'data'/'fuel_econ.csv'

    print('File extension : {}'.format(filePath.suffix))

OUTPUT:
ABSOLUTE PATH IS THE PATH WHERE YOUR PYTHON FILE IS PLACED

Absolute path : D:StudyMachine LearningJupitor NotebookJupytorNotebookTest2Udacity_ScriptsMatplotlib and seaborn Part2

File path : D:StudyMachine LearningJupitor NotebookJupytorNotebookTest2Udacity_ScriptsMatplotlib and seaborn Part2datafuel_econ.csv

isfileExist : True

isadirectory : False

File extension : .csv

edited Mar 10 at 17:21

answered Mar 10 at 4:06

Arpan Saini

428519

add a comment |

USEFUL PATH PROPERTIES IN PYTHON:

 from pathlib import Path



    #Returns the path of the directory, where your script file is placed

    mypath = Path().absolute()

    print('Absolute path : {}'.format(mypath))



    #if you want to go to any other file inside the subdirectories of the directory path got from above method

    filePath = mypath/'data'/'fuel_econ.csv'

    print('File path : {}'.format(filePath))



    #To check if file present in that directory or Not

    isfileExist = filePath.exists()

    print('isfileExist : {}'.format(isfileExist))



    #To check if the path is a directory or a File

    isadirectory = filePath.is_dir()

    print('isadirectory : {}'.format(isadirectory))



    #To get the extension of the file

    fileExtension = mypath/'data'/'fuel_econ.csv'

    print('File extension : {}'.format(filePath.suffix))

OUTPUT:
ABSOLUTE PATH IS THE PATH WHERE YOUR PYTHON FILE IS PLACED

Absolute path : D:StudyMachine LearningJupitor NotebookJupytorNotebookTest2Udacity_ScriptsMatplotlib and seaborn Part2

File path : D:StudyMachine LearningJupitor NotebookJupytorNotebookTest2Udacity_ScriptsMatplotlib and seaborn Part2datafuel_econ.csv

isfileExist : True

isadirectory : False

File extension : .csv

edited Mar 10 at 17:21

answered Mar 10 at 4:06

Arpan Saini

428519

USEFUL PATH PROPERTIES IN PYTHON:

 from pathlib import Path



    #Returns the path of the directory, where your script file is placed

    mypath = Path().absolute()

    print('Absolute path : {}'.format(mypath))



    #if you want to go to any other file inside the subdirectories of the directory path got from above method

    filePath = mypath/'data'/'fuel_econ.csv'

    print('File path : {}'.format(filePath))



    #To check if file present in that directory or Not

    isfileExist = filePath.exists()

    print('isfileExist : {}'.format(isfileExist))



    #To check if the path is a directory or a File

    isadirectory = filePath.is_dir()

    print('isadirectory : {}'.format(isadirectory))



    #To get the extension of the file

    fileExtension = mypath/'data'/'fuel_econ.csv'

    print('File extension : {}'.format(filePath.suffix))

OUTPUT:
ABSOLUTE PATH IS THE PATH WHERE YOUR PYTHON FILE IS PLACED

Absolute path : D:StudyMachine LearningJupitor NotebookJupytorNotebookTest2Udacity_ScriptsMatplotlib and seaborn Part2

File path : D:StudyMachine LearningJupitor NotebookJupytorNotebookTest2Udacity_ScriptsMatplotlib and seaborn Part2datafuel_econ.csv

isfileExist : True

isadirectory : False

File extension : .csv

edited Mar 10 at 17:21

answered Mar 10 at 4:06

Arpan Saini

428519

edited Mar 10 at 17:21

answered Mar 10 at 4:06

Arpan Saini

428519

answered Mar 10 at 4:06

Arpan Saini

428519

answered Mar 10 at 4:06

Arpan Saini

428519

add a comment |

-1

## IMPORT MODULES

import os



## CALCULATE FILEPATH VARIABLE

filepath = os.path.abspath('') ## ~ os.getcwd()

## TEST TO MAKE SURE os.getcwd() is EQUIVALENT ALWAYS..

## ..OR DIFFERENT IN SOME CIRCUMSTANCES

edited Jan 6 at 8:47

answered Jan 2 at 13:13

Jerusalem Programmer

162212

2

The current working directory might not be the same as the directory of the current file. Your solution will only work if the current working directory is the same as the directory that holds the file. That can be fairly common during development, but is usually pretty rare for a deployed script.

– Bryan Oakley
Jan 3 at 22:45

Thanks for the info. I updated my answer to reflect the possibility that these may not be equivalent returning the same filepath. Such is reason why I adhere to philosophy of coding called "incremental programming", testing, and lots of print() function calls to test output and/or values return at each step of the way. I hope my code now better reflects the reality of os.path.abspath() vs. os.getcwd() since I do not claim to know everything. However you can see that I put os.getcwd() in comments as not the main solution, but rather as an alt piece of code worth remembering and checking.

– Jerusalem Programmer
Jan 6 at 8:45

P.S. I think someone edited and erased the majority of my answer that I posted here... :( Such discourages people from participating if their answers are going to be censored and chopped.

– Jerusalem Programmer
Jan 6 at 8:49

You're the only person to have edited this answer. Also, your edits still don't properly answer the question. The question is specifically asking about the path to the file's directory, and your answer only mentions getting the current directory. That is not a correct answer, since the current directory is very often not the same as the directory the script is in. This has nothing to do with abspath.

– Bryan Oakley
Jan 6 at 14:49

Most certainly the code I shared gives path to the file's directory if run from the directory and/or file from where you want to know the absolute file path. Most certainly it is a correct answer. Please don't be a troll.

– Jerusalem Programmer
Jan 7 at 4:48

|
show 2 more comments

-1

## IMPORT MODULES

import os



## CALCULATE FILEPATH VARIABLE

filepath = os.path.abspath('') ## ~ os.getcwd()

## TEST TO MAKE SURE os.getcwd() is EQUIVALENT ALWAYS..

## ..OR DIFFERENT IN SOME CIRCUMSTANCES

edited Jan 6 at 8:47

answered Jan 2 at 13:13

Jerusalem Programmer

162212

2

The current working directory might not be the same as the directory of the current file. Your solution will only work if the current working directory is the same as the directory that holds the file. That can be fairly common during development, but is usually pretty rare for a deployed script.

– Bryan Oakley
Jan 3 at 22:45

Thanks for the info. I updated my answer to reflect the possibility that these may not be equivalent returning the same filepath. Such is reason why I adhere to philosophy of coding called "incremental programming", testing, and lots of print() function calls to test output and/or values return at each step of the way. I hope my code now better reflects the reality of os.path.abspath() vs. os.getcwd() since I do not claim to know everything. However you can see that I put os.getcwd() in comments as not the main solution, but rather as an alt piece of code worth remembering and checking.

– Jerusalem Programmer
Jan 6 at 8:45

P.S. I think someone edited and erased the majority of my answer that I posted here... :( Such discourages people from participating if their answers are going to be censored and chopped.

– Jerusalem Programmer
Jan 6 at 8:49

You're the only person to have edited this answer. Also, your edits still don't properly answer the question. The question is specifically asking about the path to the file's directory, and your answer only mentions getting the current directory. That is not a correct answer, since the current directory is very often not the same as the directory the script is in. This has nothing to do with abspath.

– Bryan Oakley
Jan 6 at 14:49

Most certainly the code I shared gives path to the file's directory if run from the directory and/or file from where you want to know the absolute file path. Most certainly it is a correct answer. Please don't be a troll.

– Jerusalem Programmer
Jan 7 at 4:48

|
show 2 more comments

-1

## IMPORT MODULES

import os



## CALCULATE FILEPATH VARIABLE

filepath = os.path.abspath('') ## ~ os.getcwd()

## TEST TO MAKE SURE os.getcwd() is EQUIVALENT ALWAYS..

## ..OR DIFFERENT IN SOME CIRCUMSTANCES

edited Jan 6 at 8:47

answered Jan 2 at 13:13

Jerusalem Programmer

162212

## IMPORT MODULES

import os



## CALCULATE FILEPATH VARIABLE

filepath = os.path.abspath('') ## ~ os.getcwd()

## TEST TO MAKE SURE os.getcwd() is EQUIVALENT ALWAYS..

## ..OR DIFFERENT IN SOME CIRCUMSTANCES

edited Jan 6 at 8:47

answered Jan 2 at 13:13

Jerusalem Programmer

162212

edited Jan 6 at 8:47

answered Jan 2 at 13:13

Jerusalem Programmer

162212

answered Jan 2 at 13:13

Jerusalem Programmer

162212

answered Jan 2 at 13:13

Jerusalem Programmer

162212

2

The current working directory might not be the same as the directory of the current file. Your solution will only work if the current working directory is the same as the directory that holds the file. That can be fairly common during development, but is usually pretty rare for a deployed script.

– Bryan Oakley
Jan 3 at 22:45

Thanks for the info. I updated my answer to reflect the possibility that these may not be equivalent returning the same filepath. Such is reason why I adhere to philosophy of coding called "incremental programming", testing, and lots of print() function calls to test output and/or values return at each step of the way. I hope my code now better reflects the reality of os.path.abspath() vs. os.getcwd() since I do not claim to know everything. However you can see that I put os.getcwd() in comments as not the main solution, but rather as an alt piece of code worth remembering and checking.

– Jerusalem Programmer
Jan 6 at 8:45

P.S. I think someone edited and erased the majority of my answer that I posted here... :( Such discourages people from participating if their answers are going to be censored and chopped.

– Jerusalem Programmer
Jan 6 at 8:49

You're the only person to have edited this answer. Also, your edits still don't properly answer the question. The question is specifically asking about the path to the file's directory, and your answer only mentions getting the current directory. That is not a correct answer, since the current directory is very often not the same as the directory the script is in. This has nothing to do with abspath.

– Bryan Oakley
Jan 6 at 14:49

Most certainly the code I shared gives path to the file's directory if run from the directory and/or file from where you want to know the absolute file path. Most certainly it is a correct answer. Please don't be a troll.

– Jerusalem Programmer
Jan 7 at 4:48

|
show 2 more comments

2

The current working directory might not be the same as the directory of the current file. Your solution will only work if the current working directory is the same as the directory that holds the file. That can be fairly common during development, but is usually pretty rare for a deployed script.

– Bryan Oakley
Jan 3 at 22:45

Thanks for the info. I updated my answer to reflect the possibility that these may not be equivalent returning the same filepath. Such is reason why I adhere to philosophy of coding called "incremental programming", testing, and lots of print() function calls to test output and/or values return at each step of the way. I hope my code now better reflects the reality of os.path.abspath() vs. os.getcwd() since I do not claim to know everything. However you can see that I put os.getcwd() in comments as not the main solution, but rather as an alt piece of code worth remembering and checking.

– Jerusalem Programmer
Jan 6 at 8:45

P.S. I think someone edited and erased the majority of my answer that I posted here... :( Such discourages people from participating if their answers are going to be censored and chopped.

– Jerusalem Programmer
Jan 6 at 8:49

You're the only person to have edited this answer. Also, your edits still don't properly answer the question. The question is specifically asking about the path to the file's directory, and your answer only mentions getting the current directory. That is not a correct answer, since the current directory is very often not the same as the directory the script is in. This has nothing to do with abspath.

– Bryan Oakley
Jan 6 at 14:49

Most certainly the code I shared gives path to the file's directory if run from the directory and/or file from where you want to know the absolute file path. Most certainly it is a correct answer. Please don't be a troll.

– Jerusalem Programmer
Jan 7 at 4:48

The current working directory might not be the same as the directory of the current file. Your solution will only work if the current working directory is the same as the directory that holds the file. That can be fairly common during development, but is usually pretty rare for a deployed script.

– Bryan Oakley
Jan 3 at 22:45

Thanks for the info. I updated my answer to reflect the possibility that these may not be equivalent returning the same filepath. Such is reason why I adhere to philosophy of coding called "incremental programming", testing, and lots of print() function calls to test output and/or values return at each step of the way. I hope my code now better reflects the reality of os.path.abspath() vs. os.getcwd() since I do not claim to know everything. However you can see that I put os.getcwd() in comments as not the main solution, but rather as an alt piece of code worth remembering and checking.

– Jerusalem Programmer
Jan 6 at 8:45

P.S. I think someone edited and erased the majority of my answer that I posted here... :( Such discourages people from participating if their answers are going to be censored and chopped.

– Jerusalem Programmer
Jan 6 at 8:49

You're the only person to have edited this answer. Also, your edits still don't properly answer the question. The question is specifically asking about the path to the file's directory, and your answer only mentions getting the current directory. That is not a correct answer, since the current directory is very often not the same as the directory the script is in. This has nothing to do with abspath.

– Bryan Oakley
Jan 6 at 14:49

Most certainly the code I shared gives path to the file's directory if run from the directory and/or file from where you want to know the absolute file path. Most certainly it is a correct answer. Please don't be a troll.

– Jerusalem Programmer
Jan 7 at 4:48

|
show 2 more comments

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