How do I create a incrementing filename in Python?

I'm creating a program that will create a file and save it to the directory with the filename sample.xml. Once the file is saved when i try to run the program again it overwrites the old file into the new one because they do have the same file name. How do I increment the file names so that whenever I try to run the code again it will going to increment the file name. and will not overwrite the existing one. I am thinking of checking the filename first on the directory and if they are the same the code will generate a new filename:

fh = open("sample.xml", "w")

rs = [blockresult]

fh.writelines(rs)

fh.close()

edited Apr 2 '17 at 11:11

Martin Thoma

43.4k61307532

asked Aug 1 '13 at 3:31

Oliver Ven Quilnet

106114

add a comment |

fh = open("sample.xml", "w")

rs = [blockresult]

fh.writelines(rs)

fh.close()

edited Apr 2 '17 at 11:11

Martin Thoma

43.4k61307532

asked Aug 1 '13 at 3:31

Oliver Ven Quilnet

106114

add a comment |

fh = open("sample.xml", "w")

rs = [blockresult]

fh.writelines(rs)

fh.close()

edited Apr 2 '17 at 11:11

Martin Thoma

43.4k61307532

asked Aug 1 '13 at 3:31

Oliver Ven Quilnet

106114

fh = open("sample.xml", "w")

rs = [blockresult]

fh.writelines(rs)

fh.close()

python file-io

edited Apr 2 '17 at 11:11

Martin Thoma

43.4k61307532

asked Aug 1 '13 at 3:31

Oliver Ven Quilnet

106114

edited Apr 2 '17 at 11:11

Martin Thoma

43.4k61307532

asked Aug 1 '13 at 3:31

Oliver Ven Quilnet

106114

edited Apr 2 '17 at 11:11

Martin Thoma

43.4k61307532

edited Apr 2 '17 at 11:11

Martin Thoma

43.4k61307532

edited Apr 2 '17 at 11:11

Martin Thoma

43.4k61307532

asked Aug 1 '13 at 3:31

Oliver Ven Quilnet

106114

asked Aug 1 '13 at 3:31

Oliver Ven Quilnet

106114

asked Aug 1 '13 at 3:31

Oliver Ven Quilnet

106114

add a comment |

10 Answers
10

active

oldest

votes

I would iterate through sample[int].xml for example and grab the next available name that is not used by a file or directory.

import os



i = 0

while os.path.exists("sample%s.xml" % i):

    i += 1



fh = open("sample%s.xml" % i, "w")

....

That should give you sample0.xml initially, then sample1.xml, etc.

Note that the relative file notation by default relates to the file directory/folder you run the code from. Use absolute paths if necessary. Use os.getcwd() to read your current dir and os.chdir(path_to_dir) to set a new current dir.

edited Aug 1 '13 at 8:16

Eric O Lebigot

56.9k36167218

answered Aug 1 '13 at 3:46

bossi

962914

1

Kindly asking what is non-useful or unconstructive here? Voting down without leaving a (constructive) comment seems to be more unconstructive to me.

– bossi
Aug 1 '13 at 5:18

isfile() is not correct: a directory will match. You want exists() instead, but this is @Eiyrioü von Kauyf's answer. Furthermore, relative paths are not exactly "relative to the directory where the code is run from". They are instead more generally relative to the "current directory" (which is by default the directory that the code is run from). The current directory can be changed within the program, for instance.

– Eric O Lebigot
Aug 1 '13 at 5:24

The fact that os.path.isfile() matches directories is new to me (and doesn't do as you describe for me on Python 3.3/win), isn't that why there is os.path.isdir() in place to differentiate between the two? In regards to the comment in my post towards the relative path notation neither Oliver Ven Quilnet's nor my example explicitly changes the current directory and I thought I briefly point it out to make it clear for the given context.

– bossi
Aug 1 '13 at 5:48

1

You are right, I should have been clearer. I meant that isfile() will make your loop exit when the name matches a directory, and your code tries then to open the directory in write mode, which fails with IOError. This is why isfile() is not the correct test, and should be replaced by the exists() of @Eiyrioü von Kauyf. As for relative paths, I really think that the current "the relative file notation always relates to the file directory/folder you run the code from" is misleading (because of "always").

– Eric O Lebigot
Aug 1 '13 at 6:41

@EOL: That's a good point, I honestly wasn't aware that identical names between a file and a folder in the same directory are illegal under Windows; thanks for pointing that out. I agree with you, the remark about the relative path did sound misleading, it should sound clearer now.

– bossi
Aug 1 '13 at 7:52

|
show 4 more comments

def get_nonexistant_path(fname_path):

    """

    Get the path to a filename which does not exist by incrementing path.



    Examples

    --------

    >>> get_nonexistant_path('/etc/issue')

    '/etc/issue-1'

    >>> get_nonexistant_path('whatever/1337bla.py')

    'whatever/1337bla.py'

    """

    if not os.path.exists(fname_path):

        return fname_path

    filename, file_extension = os.path.splitext(fname_path)

    i = 1

    new_fname = "{}-{}{}".format(filename, i, file_extension)

    while os.path.exists(new_fname):

        i += 1

        new_fname = "{}-{}{}".format(filename, i, file_extension)

    return new_fname

Before you open the file, call

fname = get_nonexistant_path("sample.xml")

This will either give you 'sample.xml' or - if this alreay exists - 'sample-i.xml' where i is the lowest positive integer such that the file does not already exist.

I recommend using os.path.abspath("sample.xml"). If you have ~ as home directory, you might need to expand it first.

Please note that race conditions might occur with this simple code if you have multiple instances running at the same time. If this might be a problem, please check this question.

edited May 23 '17 at 12:34

Community♦

answered Apr 2 '17 at 11:09

Martin Thoma

43.4k61307532

add a comment |

Try setting a count variable, and then incrementing that variable nested inside the same loop you write your file in. Include the count loop inside the name of the file with an escape character, so every loop ticks +1 and so does the number in the file.

Some code from a project I just finished:

numberLoops = #some limit determined by the user

currentLoop = 1

while currentLoop < numberLoops:

    currentLoop = currentLoop + 1



    fileName = ("log%d_%d.txt" % (currentLoop, str(now())))

For reference:

from time import mktime, gmtime



def now(): 

   return mktime(gmtime())

which is probably irrelevant in your case but i was running multiple instances of this program and making tons of files. Hope this helps!

edited Mar 31 '14 at 5:51

zanetu

2,23311514

answered Aug 1 '13 at 4:33

ford

1301211

2

Python has for loops for this, they are much faster to read and comprehend than the while loops that emulate them. Furthermore, the % operator is deprecated. No downvote, though, because it does the job—it just does not do it in the preferred Python way.

– Eric O Lebigot
Aug 1 '13 at 5:19

There is a problem with your format string: you format a string with %d, and this raises an exception.

– Eric O Lebigot
Aug 1 '13 at 5:31

Thanks for catching that. It should be a %s, I retyped this rather hastily instead of copying from my source. Thanks!

– ford
Aug 1 '13 at 5:36

add a comment |

Sequentially checking each file name to find the next available one works fine with small numbers of files, but quickly becomes slower as the number of files increases.

Here is a version that finds the next available file name in log(n) time:

import os



def next_path(path_pattern):

    """

    Finds the next free path in an sequentially named list of files



    e.g. path_pattern = 'file-%s.txt':



    file-1.txt

    file-2.txt

    file-3.txt



    Runs in log(n) time where n is the number of existing files in sequence

    """

    i = 1



    # First do an exponential search

    while os.path.exists(path_pattern % i):

        i = i * 2



    # Result lies somewhere in the interval (i/2..i]

    # We call this interval (a..b] and narrow it down until a + 1 = b

    a, b = (i / 2, i)

    while a + 1 < b:

        c = (a + b) / 2 # interval midpoint

        a, b = (c, b) if os.path.exists(path_pattern % c) else (a, c)



    return path_pattern % b

To measure the speed improvement I wrote a small test function that creates 10,000 files:

for i in range(1,10000):

    with open(next_path('file-%s.foo'), 'w'):

        pass

And implemented the naive approach:

def next_path_naive(path_pattern):

    """

    Naive (slow) version of next_path

    """

    i = 1

    while os.path.exists(path_pattern % i):

        i += 1

    return path_pattern % i

And here are the results:

Fast version:

real    0m2.132s

user    0m0.773s

sys 0m1.312s

Naive version:

real    2m36.480s

user    1m12.671s

sys 1m22.425s

Finally, note that either approach is susceptible to race conditions if multiple actors are trying to create files in the sequence at the same time.

answered Nov 3 '17 at 2:20

James

2,0772128

add a comment |

Without storing state data in an extra file, a quicker solution to the ones presented here would be to do the following:

from glob import glob

import os



files = glob("somedir/sample*.xml")

files = files.sorted()

cur_num = int(os.path.basename(files[-1])[6:-4])

cur_num += 1

fh = open("somedir/sample%s.xml" % cur_num, 'w')

rs = [blockresult]

fh.writelines(rs)

fh.close()

This will also keep incrementing, even if some of the lower numbered files disappear.

The other solution here that I like (pointed out by Eiyrioü) is the idea of keeping a temporary file that contains your most recent number:

temp_fh = open('somedir/curr_num.txt', 'r')

curr_num = int(temp_fh.readline().strip())

curr_num += 1

fh = open("somedir/sample%s.xml" % cur_num, 'w')

rs = [blockresult]

fh.writelines(rs)

fh.close()

edited Aug 1 '13 at 5:24

answered Aug 1 '13 at 5:06

Vorticity

2,30321837

Your cur_num calculation only works for 1-digit numbers, it is not general enough.

– Eric O Lebigot
Aug 1 '13 at 5:21

Good point, updated.

– Vorticity
Aug 1 '13 at 5:25

add a comment |

Another example using recursion

import os

def checkFilePath(testString, extension, currentCount):

    if os.path.exists(testString + str(currentCount) +extension):

        return checkFilePath(testString, extension, currentCount+1)

    else:

        return testString + str(currentCount) +extension

Use:

checkFilePath("myfile", ".txt" , 0)

answered Feb 15 '17 at 3:04

chumbaloo

1409

add a comment |

You can use a while loop with a counter which checks if a file with a name and the counter's value exists if it does then move on else break and make a file.

I have done it in this way for one of my projects:`

from os import path

import os



i = 0

flnm = "Directory\Filename" + str(i) + ".txt"

while path.exists(flnm) :

    flnm = "Directory\Filename" + str(i) + ".txt"

    i += 1

f = open(flnm, "w") #do what you want to with that file...

f.write(str(var))

f.close() # make sure to close it.

Here the counter i starts from 0 and a while loop checks everytime if the file exists, if it does it moves on else it breaks out and creates a file from then you can customize. Also make sure to close it else it will result in the file being open which can cause problems while deleting it.
I used path.exists() to check if a file exists.
Don't do from os import * it can cause problem when we use open() method as there is another os.open() method too and it can give the error. TypeError: Integer expected. (got str)
Else wish u a Happy New Year and to all.

answered Jan 1 at 7:26

Matrix Programmer

1112

Thanks For Rating!

– Matrix Programmer
Jan 31 at 10:41

add a comment |

The two ways to do it are:

Check for the existence of the old file and if it exists try the next file name +1

save state data somewhere

an easy way to do it off the bat would be:

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(pth.abspath(filename+str(filenum)+".py")):

    filenum+=1

my_next_file = open(filename+str(filenum)+".py",'w')

as a design thing, while True slows things down and isn't a great thing for code readability

edited: @EOL contributions/ thoughts

so I think not having .format is more readable at first glance - but using .format is better for generality and convention so.

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(pth.abspath(filename+str(filenum)+".py")):

    filenum+=1

my_next_file = open("{}{}.py".format(filename, filenum),'w')

# or 

my_next_file = open(filename + "{}.py".format(filenum),'w')

and you don't have to use abspath - you can use relative paths if you prefer, I prefer abs path sometimes because it helps to normalize the paths passed :).

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(filename+str(filenum)+".py"):

    filenum+=1

##removed for conciseness

edited Aug 1 '13 at 5:33

answered Aug 1 '13 at 3:33

Eiyrioü von Kauyf

1,72132336

The format() method is much more legible than string concatenation, here. I think that the while loop is fine, here. On another topic, why use abspath()?

– Eric O Lebigot
Aug 1 '13 at 5:23

format is more legible, but then he would have to look at string formatting; this is easier to understand on first glance imho. and abspath because i'm ignoring symlinks :/ .... that could lead to confusing errors

– Eiyrioü von Kauyf
Aug 1 '13 at 5:26

While I understand your point, I believe that even beginners should be shown Pythonic examples, so that they take good habits. The behavior of format() is really quite simple to understand and even guess: "{}{}.py".format(filename, filenum). It's even simpler than the algorithm presented here. :)

– Eric O Lebigot
Aug 1 '13 at 5:29

hmm added :) i agree with you on this point.

– Eiyrioü von Kauyf
Aug 1 '13 at 5:31

@EOL whatcha think ;) do I have your approval

– Eiyrioü von Kauyf
Aug 1 '13 at 5:34

add a comment |

Another solution that avoids the use of while loop is to use os.listdir() function which returns a list of all the files and directories contained in a directory whose path is taken as an argument.

To answer the example in the question, supposing that the directory you are working in only contains "sample_i.xlm" files indexed starting at 0, you can easily obtain the next index for the new file with the following code.

import os



new_index = len(os.listdir('path_to_file_containing_only_sample_i_files'))

new_file = open('path_to_file_containing_only_sample_i_files/sample_%s.xml' % new_index, 'w')

answered Aug 30 '18 at 9:43

Malo Pocheau

264

add a comment |

I needed to do something similar, but for output directories in a data processing pipeline. I was inspired by Vorticity's answer, but added use of regex to grab the trailing number. This method continues to increment the last directory, even if intermediate numbered output directories are deleted. It also adds leading zeros so the names will sort alphabetically (i.e. width 3 gives 001 etc.)

def get_unique_dir(path, width=3):

    # if it doesn't exist, create

    if not os.path.isdir(path):

        log.debug("Creating new directory - {}".format(path))

        os.makedirs(path)

        return path



    # if it's empty, use

    if not os.listdir(path):

        log.debug("Using empty directory - {}".format(path))

        return path



    # otherwise, increment the highest number folder in the series



    def get_trailing_number(search_text):

        serch_obj = re.search(r"([0-9]+)$", search_text)

        if not serch_obj:

            return 0

        else:

            return int(serch_obj.group(1))



    dirs = glob(path + "*")

    num_list = sorted([get_trailing_number(d) for d in dirs])

    highest_num = num_list[-1]

    next_num = highest_num + 1

    new_path = "{0}_{1:0>{2}}".format(path, next_num, width)



    log.debug("Creating new incremented directory - {}".format(new_path))

    os.makedirs(new_path)

    return new_path



get_unique_dir("output")

edited Feb 8 at 20:34

answered Feb 8 at 20:29

Woods26

add a comment |

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10 Answers
10

active

oldest

votes

10 Answers
10

active

oldest

votes

I would iterate through sample[int].xml for example and grab the next available name that is not used by a file or directory.

import os



i = 0

while os.path.exists("sample%s.xml" % i):

    i += 1



fh = open("sample%s.xml" % i, "w")

....

That should give you sample0.xml initially, then sample1.xml, etc.

edited Aug 1 '13 at 8:16

Eric O Lebigot

56.9k36167218

answered Aug 1 '13 at 3:46

bossi

962914

1

Kindly asking what is non-useful or unconstructive here? Voting down without leaving a (constructive) comment seems to be more unconstructive to me.

– bossi
Aug 1 '13 at 5:18

isfile() is not correct: a directory will match. You want exists() instead, but this is @Eiyrioü von Kauyf's answer. Furthermore, relative paths are not exactly "relative to the directory where the code is run from". They are instead more generally relative to the "current directory" (which is by default the directory that the code is run from). The current directory can be changed within the program, for instance.

– Eric O Lebigot
Aug 1 '13 at 5:24

The fact that os.path.isfile() matches directories is new to me (and doesn't do as you describe for me on Python 3.3/win), isn't that why there is os.path.isdir() in place to differentiate between the two? In regards to the comment in my post towards the relative path notation neither Oliver Ven Quilnet's nor my example explicitly changes the current directory and I thought I briefly point it out to make it clear for the given context.

– bossi
Aug 1 '13 at 5:48

1

You are right, I should have been clearer. I meant that isfile() will make your loop exit when the name matches a directory, and your code tries then to open the directory in write mode, which fails with IOError. This is why isfile() is not the correct test, and should be replaced by the exists() of @Eiyrioü von Kauyf. As for relative paths, I really think that the current "the relative file notation always relates to the file directory/folder you run the code from" is misleading (because of "always").

– Eric O Lebigot
Aug 1 '13 at 6:41

@EOL: That's a good point, I honestly wasn't aware that identical names between a file and a folder in the same directory are illegal under Windows; thanks for pointing that out. I agree with you, the remark about the relative path did sound misleading, it should sound clearer now.

– bossi
Aug 1 '13 at 7:52

|
show 4 more comments

I would iterate through sample[int].xml for example and grab the next available name that is not used by a file or directory.

import os



i = 0

while os.path.exists("sample%s.xml" % i):

    i += 1



fh = open("sample%s.xml" % i, "w")

....

That should give you sample0.xml initially, then sample1.xml, etc.

edited Aug 1 '13 at 8:16

Eric O Lebigot

56.9k36167218

answered Aug 1 '13 at 3:46

bossi

962914

1

Kindly asking what is non-useful or unconstructive here? Voting down without leaving a (constructive) comment seems to be more unconstructive to me.

– bossi
Aug 1 '13 at 5:18

isfile() is not correct: a directory will match. You want exists() instead, but this is @Eiyrioü von Kauyf's answer. Furthermore, relative paths are not exactly "relative to the directory where the code is run from". They are instead more generally relative to the "current directory" (which is by default the directory that the code is run from). The current directory can be changed within the program, for instance.

– Eric O Lebigot
Aug 1 '13 at 5:24

The fact that os.path.isfile() matches directories is new to me (and doesn't do as you describe for me on Python 3.3/win), isn't that why there is os.path.isdir() in place to differentiate between the two? In regards to the comment in my post towards the relative path notation neither Oliver Ven Quilnet's nor my example explicitly changes the current directory and I thought I briefly point it out to make it clear for the given context.

– bossi
Aug 1 '13 at 5:48

1

You are right, I should have been clearer. I meant that isfile() will make your loop exit when the name matches a directory, and your code tries then to open the directory in write mode, which fails with IOError. This is why isfile() is not the correct test, and should be replaced by the exists() of @Eiyrioü von Kauyf. As for relative paths, I really think that the current "the relative file notation always relates to the file directory/folder you run the code from" is misleading (because of "always").

– Eric O Lebigot
Aug 1 '13 at 6:41

@EOL: That's a good point, I honestly wasn't aware that identical names between a file and a folder in the same directory are illegal under Windows; thanks for pointing that out. I agree with you, the remark about the relative path did sound misleading, it should sound clearer now.

– bossi
Aug 1 '13 at 7:52

|
show 4 more comments

I would iterate through sample[int].xml for example and grab the next available name that is not used by a file or directory.

import os



i = 0

while os.path.exists("sample%s.xml" % i):

    i += 1



fh = open("sample%s.xml" % i, "w")

....

That should give you sample0.xml initially, then sample1.xml, etc.

edited Aug 1 '13 at 8:16

Eric O Lebigot

56.9k36167218

answered Aug 1 '13 at 3:46

bossi

962914

I would iterate through sample[int].xml for example and grab the next available name that is not used by a file or directory.

import os



i = 0

while os.path.exists("sample%s.xml" % i):

    i += 1



fh = open("sample%s.xml" % i, "w")

....

That should give you sample0.xml initially, then sample1.xml, etc.

edited Aug 1 '13 at 8:16

Eric O Lebigot

56.9k36167218

answered Aug 1 '13 at 3:46

bossi

962914

edited Aug 1 '13 at 8:16

Eric O Lebigot

56.9k36167218

edited Aug 1 '13 at 8:16

Eric O Lebigot

56.9k36167218

edited Aug 1 '13 at 8:16

Eric O Lebigot

56.9k36167218

answered Aug 1 '13 at 3:46

bossi

962914

answered Aug 1 '13 at 3:46

bossi

962914

answered Aug 1 '13 at 3:46

bossi

962914

1

Kindly asking what is non-useful or unconstructive here? Voting down without leaving a (constructive) comment seems to be more unconstructive to me.

– bossi
Aug 1 '13 at 5:18

isfile() is not correct: a directory will match. You want exists() instead, but this is @Eiyrioü von Kauyf's answer. Furthermore, relative paths are not exactly "relative to the directory where the code is run from". They are instead more generally relative to the "current directory" (which is by default the directory that the code is run from). The current directory can be changed within the program, for instance.

– Eric O Lebigot
Aug 1 '13 at 5:24

The fact that os.path.isfile() matches directories is new to me (and doesn't do as you describe for me on Python 3.3/win), isn't that why there is os.path.isdir() in place to differentiate between the two? In regards to the comment in my post towards the relative path notation neither Oliver Ven Quilnet's nor my example explicitly changes the current directory and I thought I briefly point it out to make it clear for the given context.

– bossi
Aug 1 '13 at 5:48

1

You are right, I should have been clearer. I meant that isfile() will make your loop exit when the name matches a directory, and your code tries then to open the directory in write mode, which fails with IOError. This is why isfile() is not the correct test, and should be replaced by the exists() of @Eiyrioü von Kauyf. As for relative paths, I really think that the current "the relative file notation always relates to the file directory/folder you run the code from" is misleading (because of "always").

– Eric O Lebigot
Aug 1 '13 at 6:41

@EOL: That's a good point, I honestly wasn't aware that identical names between a file and a folder in the same directory are illegal under Windows; thanks for pointing that out. I agree with you, the remark about the relative path did sound misleading, it should sound clearer now.

– bossi
Aug 1 '13 at 7:52

|
show 4 more comments

1

Kindly asking what is non-useful or unconstructive here? Voting down without leaving a (constructive) comment seems to be more unconstructive to me.

– bossi
Aug 1 '13 at 5:18

isfile() is not correct: a directory will match. You want exists() instead, but this is @Eiyrioü von Kauyf's answer. Furthermore, relative paths are not exactly "relative to the directory where the code is run from". They are instead more generally relative to the "current directory" (which is by default the directory that the code is run from). The current directory can be changed within the program, for instance.

– Eric O Lebigot
Aug 1 '13 at 5:24

The fact that os.path.isfile() matches directories is new to me (and doesn't do as you describe for me on Python 3.3/win), isn't that why there is os.path.isdir() in place to differentiate between the two? In regards to the comment in my post towards the relative path notation neither Oliver Ven Quilnet's nor my example explicitly changes the current directory and I thought I briefly point it out to make it clear for the given context.

– bossi
Aug 1 '13 at 5:48

1

You are right, I should have been clearer. I meant that isfile() will make your loop exit when the name matches a directory, and your code tries then to open the directory in write mode, which fails with IOError. This is why isfile() is not the correct test, and should be replaced by the exists() of @Eiyrioü von Kauyf. As for relative paths, I really think that the current "the relative file notation always relates to the file directory/folder you run the code from" is misleading (because of "always").

– Eric O Lebigot
Aug 1 '13 at 6:41

@EOL: That's a good point, I honestly wasn't aware that identical names between a file and a folder in the same directory are illegal under Windows; thanks for pointing that out. I agree with you, the remark about the relative path did sound misleading, it should sound clearer now.

– bossi
Aug 1 '13 at 7:52

Kindly asking what is non-useful or unconstructive here? Voting down without leaving a (constructive) comment seems to be more unconstructive to me.

– bossi
Aug 1 '13 at 5:18

isfile() is not correct: a directory will match. You want exists() instead, but this is @Eiyrioü von Kauyf's answer. Furthermore, relative paths are not exactly "relative to the directory where the code is run from". They are instead more generally relative to the "current directory" (which is by default the directory that the code is run from). The current directory can be changed within the program, for instance.

– Eric O Lebigot
Aug 1 '13 at 5:24

The fact that os.path.isfile() matches directories is new to me (and doesn't do as you describe for me on Python 3.3/win), isn't that why there is os.path.isdir() in place to differentiate between the two? In regards to the comment in my post towards the relative path notation neither Oliver Ven Quilnet's nor my example explicitly changes the current directory and I thought I briefly point it out to make it clear for the given context.

– bossi
Aug 1 '13 at 5:48

You are right, I should have been clearer. I meant that isfile() will make your loop exit when the name matches a directory, and your code tries then to open the directory in write mode, which fails with IOError. This is why isfile() is not the correct test, and should be replaced by the exists() of @Eiyrioü von Kauyf. As for relative paths, I really think that the current "the relative file notation always relates to the file directory/folder you run the code from" is misleading (because of "always").

– Eric O Lebigot
Aug 1 '13 at 6:41

@EOL: That's a good point, I honestly wasn't aware that identical names between a file and a folder in the same directory are illegal under Windows; thanks for pointing that out. I agree with you, the remark about the relative path did sound misleading, it should sound clearer now.

– bossi
Aug 1 '13 at 7:52

|
show 4 more comments

def get_nonexistant_path(fname_path):

    """

    Get the path to a filename which does not exist by incrementing path.



    Examples

    --------

    >>> get_nonexistant_path('/etc/issue')

    '/etc/issue-1'

    >>> get_nonexistant_path('whatever/1337bla.py')

    'whatever/1337bla.py'

    """

    if not os.path.exists(fname_path):

        return fname_path

    filename, file_extension = os.path.splitext(fname_path)

    i = 1

    new_fname = "{}-{}{}".format(filename, i, file_extension)

    while os.path.exists(new_fname):

        i += 1

        new_fname = "{}-{}{}".format(filename, i, file_extension)

    return new_fname

Before you open the file, call

fname = get_nonexistant_path("sample.xml")

This will either give you 'sample.xml' or - if this alreay exists - 'sample-i.xml' where i is the lowest positive integer such that the file does not already exist.

I recommend using os.path.abspath("sample.xml"). If you have ~ as home directory, you might need to expand it first.

Please note that race conditions might occur with this simple code if you have multiple instances running at the same time. If this might be a problem, please check this question.

edited May 23 '17 at 12:34

Community♦

answered Apr 2 '17 at 11:09

Martin Thoma

43.4k61307532

add a comment |

def get_nonexistant_path(fname_path):

    """

    Get the path to a filename which does not exist by incrementing path.



    Examples

    --------

    >>> get_nonexistant_path('/etc/issue')

    '/etc/issue-1'

    >>> get_nonexistant_path('whatever/1337bla.py')

    'whatever/1337bla.py'

    """

    if not os.path.exists(fname_path):

        return fname_path

    filename, file_extension = os.path.splitext(fname_path)

    i = 1

    new_fname = "{}-{}{}".format(filename, i, file_extension)

    while os.path.exists(new_fname):

        i += 1

        new_fname = "{}-{}{}".format(filename, i, file_extension)

    return new_fname

Before you open the file, call

fname = get_nonexistant_path("sample.xml")

This will either give you 'sample.xml' or - if this alreay exists - 'sample-i.xml' where i is the lowest positive integer such that the file does not already exist.

I recommend using os.path.abspath("sample.xml"). If you have ~ as home directory, you might need to expand it first.

Please note that race conditions might occur with this simple code if you have multiple instances running at the same time. If this might be a problem, please check this question.

edited May 23 '17 at 12:34

Community♦

answered Apr 2 '17 at 11:09

Martin Thoma

43.4k61307532

add a comment |

def get_nonexistant_path(fname_path):

    """

    Get the path to a filename which does not exist by incrementing path.



    Examples

    --------

    >>> get_nonexistant_path('/etc/issue')

    '/etc/issue-1'

    >>> get_nonexistant_path('whatever/1337bla.py')

    'whatever/1337bla.py'

    """

    if not os.path.exists(fname_path):

        return fname_path

    filename, file_extension = os.path.splitext(fname_path)

    i = 1

    new_fname = "{}-{}{}".format(filename, i, file_extension)

    while os.path.exists(new_fname):

        i += 1

        new_fname = "{}-{}{}".format(filename, i, file_extension)

    return new_fname

Before you open the file, call

fname = get_nonexistant_path("sample.xml")

This will either give you 'sample.xml' or - if this alreay exists - 'sample-i.xml' where i is the lowest positive integer such that the file does not already exist.

I recommend using os.path.abspath("sample.xml"). If you have ~ as home directory, you might need to expand it first.

Please note that race conditions might occur with this simple code if you have multiple instances running at the same time. If this might be a problem, please check this question.

edited May 23 '17 at 12:34

Community♦

answered Apr 2 '17 at 11:09

Martin Thoma

43.4k61307532

def get_nonexistant_path(fname_path):

    """

    Get the path to a filename which does not exist by incrementing path.



    Examples

    --------

    >>> get_nonexistant_path('/etc/issue')

    '/etc/issue-1'

    >>> get_nonexistant_path('whatever/1337bla.py')

    'whatever/1337bla.py'

    """

    if not os.path.exists(fname_path):

        return fname_path

    filename, file_extension = os.path.splitext(fname_path)

    i = 1

    new_fname = "{}-{}{}".format(filename, i, file_extension)

    while os.path.exists(new_fname):

        i += 1

        new_fname = "{}-{}{}".format(filename, i, file_extension)

    return new_fname

Before you open the file, call

fname = get_nonexistant_path("sample.xml")

This will either give you 'sample.xml' or - if this alreay exists - 'sample-i.xml' where i is the lowest positive integer such that the file does not already exist.

I recommend using os.path.abspath("sample.xml"). If you have ~ as home directory, you might need to expand it first.

Please note that race conditions might occur with this simple code if you have multiple instances running at the same time. If this might be a problem, please check this question.

edited May 23 '17 at 12:34

Community♦

answered Apr 2 '17 at 11:09

Martin Thoma

43.4k61307532

edited May 23 '17 at 12:34

Community♦

edited May 23 '17 at 12:34

Community♦

edited May 23 '17 at 12:34

Community♦

answered Apr 2 '17 at 11:09

Martin Thoma

43.4k61307532

answered Apr 2 '17 at 11:09

Martin Thoma

43.4k61307532

answered Apr 2 '17 at 11:09

Martin Thoma

43.4k61307532

add a comment |

Some code from a project I just finished:

numberLoops = #some limit determined by the user

currentLoop = 1

while currentLoop < numberLoops:

    currentLoop = currentLoop + 1



    fileName = ("log%d_%d.txt" % (currentLoop, str(now())))

For reference:

from time import mktime, gmtime



def now(): 

   return mktime(gmtime())

which is probably irrelevant in your case but i was running multiple instances of this program and making tons of files. Hope this helps!

edited Mar 31 '14 at 5:51

zanetu

2,23311514

answered Aug 1 '13 at 4:33

ford

1301211

2

Python has for loops for this, they are much faster to read and comprehend than the while loops that emulate them. Furthermore, the % operator is deprecated. No downvote, though, because it does the job—it just does not do it in the preferred Python way.

– Eric O Lebigot
Aug 1 '13 at 5:19

There is a problem with your format string: you format a string with %d, and this raises an exception.

– Eric O Lebigot
Aug 1 '13 at 5:31

Thanks for catching that. It should be a %s, I retyped this rather hastily instead of copying from my source. Thanks!

– ford
Aug 1 '13 at 5:36

add a comment |

Some code from a project I just finished:

numberLoops = #some limit determined by the user

currentLoop = 1

while currentLoop < numberLoops:

    currentLoop = currentLoop + 1



    fileName = ("log%d_%d.txt" % (currentLoop, str(now())))

For reference:

from time import mktime, gmtime



def now(): 

   return mktime(gmtime())

which is probably irrelevant in your case but i was running multiple instances of this program and making tons of files. Hope this helps!

edited Mar 31 '14 at 5:51

zanetu

2,23311514

answered Aug 1 '13 at 4:33

ford

1301211

2

Python has for loops for this, they are much faster to read and comprehend than the while loops that emulate them. Furthermore, the % operator is deprecated. No downvote, though, because it does the job—it just does not do it in the preferred Python way.

– Eric O Lebigot
Aug 1 '13 at 5:19

There is a problem with your format string: you format a string with %d, and this raises an exception.

– Eric O Lebigot
Aug 1 '13 at 5:31

Thanks for catching that. It should be a %s, I retyped this rather hastily instead of copying from my source. Thanks!

– ford
Aug 1 '13 at 5:36

add a comment |

Some code from a project I just finished:

numberLoops = #some limit determined by the user

currentLoop = 1

while currentLoop < numberLoops:

    currentLoop = currentLoop + 1



    fileName = ("log%d_%d.txt" % (currentLoop, str(now())))

For reference:

from time import mktime, gmtime



def now(): 

   return mktime(gmtime())

which is probably irrelevant in your case but i was running multiple instances of this program and making tons of files. Hope this helps!

edited Mar 31 '14 at 5:51

zanetu

2,23311514

answered Aug 1 '13 at 4:33

ford

1301211

Some code from a project I just finished:

numberLoops = #some limit determined by the user

currentLoop = 1

while currentLoop < numberLoops:

    currentLoop = currentLoop + 1



    fileName = ("log%d_%d.txt" % (currentLoop, str(now())))

For reference:

from time import mktime, gmtime



def now(): 

   return mktime(gmtime())

which is probably irrelevant in your case but i was running multiple instances of this program and making tons of files. Hope this helps!

edited Mar 31 '14 at 5:51

zanetu

2,23311514

answered Aug 1 '13 at 4:33

ford

1301211

edited Mar 31 '14 at 5:51

zanetu

2,23311514

edited Mar 31 '14 at 5:51

zanetu

2,23311514

edited Mar 31 '14 at 5:51

zanetu

2,23311514

answered Aug 1 '13 at 4:33

ford

1301211

answered Aug 1 '13 at 4:33

ford

1301211

answered Aug 1 '13 at 4:33

ford

1301211

2

Python has for loops for this, they are much faster to read and comprehend than the while loops that emulate them. Furthermore, the % operator is deprecated. No downvote, though, because it does the job—it just does not do it in the preferred Python way.

– Eric O Lebigot
Aug 1 '13 at 5:19

There is a problem with your format string: you format a string with %d, and this raises an exception.

– Eric O Lebigot
Aug 1 '13 at 5:31

Thanks for catching that. It should be a %s, I retyped this rather hastily instead of copying from my source. Thanks!

– ford
Aug 1 '13 at 5:36

add a comment |

2

Python has for loops for this, they are much faster to read and comprehend than the while loops that emulate them. Furthermore, the % operator is deprecated. No downvote, though, because it does the job—it just does not do it in the preferred Python way.

– Eric O Lebigot
Aug 1 '13 at 5:19

There is a problem with your format string: you format a string with %d, and this raises an exception.

– Eric O Lebigot
Aug 1 '13 at 5:31

Thanks for catching that. It should be a %s, I retyped this rather hastily instead of copying from my source. Thanks!

– ford
Aug 1 '13 at 5:36

Python has for loops for this, they are much faster to read and comprehend than the while loops that emulate them. Furthermore, the % operator is deprecated. No downvote, though, because it does the job—it just does not do it in the preferred Python way.

– Eric O Lebigot
Aug 1 '13 at 5:19

There is a problem with your format string: you format a string with %d, and this raises an exception.

– Eric O Lebigot
Aug 1 '13 at 5:31

Thanks for catching that. It should be a %s, I retyped this rather hastily instead of copying from my source. Thanks!

– ford
Aug 1 '13 at 5:36

add a comment |

Sequentially checking each file name to find the next available one works fine with small numbers of files, but quickly becomes slower as the number of files increases.

Here is a version that finds the next available file name in log(n) time:

import os



def next_path(path_pattern):

    """

    Finds the next free path in an sequentially named list of files



    e.g. path_pattern = 'file-%s.txt':



    file-1.txt

    file-2.txt

    file-3.txt



    Runs in log(n) time where n is the number of existing files in sequence

    """

    i = 1



    # First do an exponential search

    while os.path.exists(path_pattern % i):

        i = i * 2



    # Result lies somewhere in the interval (i/2..i]

    # We call this interval (a..b] and narrow it down until a + 1 = b

    a, b = (i / 2, i)

    while a + 1 < b:

        c = (a + b) / 2 # interval midpoint

        a, b = (c, b) if os.path.exists(path_pattern % c) else (a, c)



    return path_pattern % b

To measure the speed improvement I wrote a small test function that creates 10,000 files:

for i in range(1,10000):

    with open(next_path('file-%s.foo'), 'w'):

        pass

And implemented the naive approach:

def next_path_naive(path_pattern):

    """

    Naive (slow) version of next_path

    """

    i = 1

    while os.path.exists(path_pattern % i):

        i += 1

    return path_pattern % i

And here are the results:

Fast version:

real    0m2.132s

user    0m0.773s

sys 0m1.312s

Naive version:

real    2m36.480s

user    1m12.671s

sys 1m22.425s

Finally, note that either approach is susceptible to race conditions if multiple actors are trying to create files in the sequence at the same time.

answered Nov 3 '17 at 2:20

James

2,0772128

add a comment |

Sequentially checking each file name to find the next available one works fine with small numbers of files, but quickly becomes slower as the number of files increases.

Here is a version that finds the next available file name in log(n) time:

import os



def next_path(path_pattern):

    """

    Finds the next free path in an sequentially named list of files



    e.g. path_pattern = 'file-%s.txt':



    file-1.txt

    file-2.txt

    file-3.txt



    Runs in log(n) time where n is the number of existing files in sequence

    """

    i = 1



    # First do an exponential search

    while os.path.exists(path_pattern % i):

        i = i * 2



    # Result lies somewhere in the interval (i/2..i]

    # We call this interval (a..b] and narrow it down until a + 1 = b

    a, b = (i / 2, i)

    while a + 1 < b:

        c = (a + b) / 2 # interval midpoint

        a, b = (c, b) if os.path.exists(path_pattern % c) else (a, c)



    return path_pattern % b

To measure the speed improvement I wrote a small test function that creates 10,000 files:

for i in range(1,10000):

    with open(next_path('file-%s.foo'), 'w'):

        pass

And implemented the naive approach:

def next_path_naive(path_pattern):

    """

    Naive (slow) version of next_path

    """

    i = 1

    while os.path.exists(path_pattern % i):

        i += 1

    return path_pattern % i

And here are the results:

Fast version:

real    0m2.132s

user    0m0.773s

sys 0m1.312s

Naive version:

real    2m36.480s

user    1m12.671s

sys 1m22.425s

Finally, note that either approach is susceptible to race conditions if multiple actors are trying to create files in the sequence at the same time.

answered Nov 3 '17 at 2:20

James

2,0772128

add a comment |

Sequentially checking each file name to find the next available one works fine with small numbers of files, but quickly becomes slower as the number of files increases.

Here is a version that finds the next available file name in log(n) time:

import os



def next_path(path_pattern):

    """

    Finds the next free path in an sequentially named list of files



    e.g. path_pattern = 'file-%s.txt':



    file-1.txt

    file-2.txt

    file-3.txt



    Runs in log(n) time where n is the number of existing files in sequence

    """

    i = 1



    # First do an exponential search

    while os.path.exists(path_pattern % i):

        i = i * 2



    # Result lies somewhere in the interval (i/2..i]

    # We call this interval (a..b] and narrow it down until a + 1 = b

    a, b = (i / 2, i)

    while a + 1 < b:

        c = (a + b) / 2 # interval midpoint

        a, b = (c, b) if os.path.exists(path_pattern % c) else (a, c)



    return path_pattern % b

To measure the speed improvement I wrote a small test function that creates 10,000 files:

for i in range(1,10000):

    with open(next_path('file-%s.foo'), 'w'):

        pass

And implemented the naive approach:

def next_path_naive(path_pattern):

    """

    Naive (slow) version of next_path

    """

    i = 1

    while os.path.exists(path_pattern % i):

        i += 1

    return path_pattern % i

And here are the results:

Fast version:

real    0m2.132s

user    0m0.773s

sys 0m1.312s

Naive version:

real    2m36.480s

user    1m12.671s

sys 1m22.425s

Finally, note that either approach is susceptible to race conditions if multiple actors are trying to create files in the sequence at the same time.

answered Nov 3 '17 at 2:20

James

2,0772128

Sequentially checking each file name to find the next available one works fine with small numbers of files, but quickly becomes slower as the number of files increases.

Here is a version that finds the next available file name in log(n) time:

import os



def next_path(path_pattern):

    """

    Finds the next free path in an sequentially named list of files



    e.g. path_pattern = 'file-%s.txt':



    file-1.txt

    file-2.txt

    file-3.txt



    Runs in log(n) time where n is the number of existing files in sequence

    """

    i = 1



    # First do an exponential search

    while os.path.exists(path_pattern % i):

        i = i * 2



    # Result lies somewhere in the interval (i/2..i]

    # We call this interval (a..b] and narrow it down until a + 1 = b

    a, b = (i / 2, i)

    while a + 1 < b:

        c = (a + b) / 2 # interval midpoint

        a, b = (c, b) if os.path.exists(path_pattern % c) else (a, c)



    return path_pattern % b

To measure the speed improvement I wrote a small test function that creates 10,000 files:

for i in range(1,10000):

    with open(next_path('file-%s.foo'), 'w'):

        pass

And implemented the naive approach:

def next_path_naive(path_pattern):

    """

    Naive (slow) version of next_path

    """

    i = 1

    while os.path.exists(path_pattern % i):

        i += 1

    return path_pattern % i

And here are the results:

Fast version:

real    0m2.132s

user    0m0.773s

sys 0m1.312s

Naive version:

real    2m36.480s

user    1m12.671s

sys 1m22.425s

Finally, note that either approach is susceptible to race conditions if multiple actors are trying to create files in the sequence at the same time.

answered Nov 3 '17 at 2:20

James

2,0772128

answered Nov 3 '17 at 2:20

James

2,0772128

answered Nov 3 '17 at 2:20

James

2,0772128

answered Nov 3 '17 at 2:20

James

2,0772128

add a comment |

Without storing state data in an extra file, a quicker solution to the ones presented here would be to do the following:

from glob import glob

import os



files = glob("somedir/sample*.xml")

files = files.sorted()

cur_num = int(os.path.basename(files[-1])[6:-4])

cur_num += 1

fh = open("somedir/sample%s.xml" % cur_num, 'w')

rs = [blockresult]

fh.writelines(rs)

fh.close()

This will also keep incrementing, even if some of the lower numbered files disappear.

The other solution here that I like (pointed out by Eiyrioü) is the idea of keeping a temporary file that contains your most recent number:

temp_fh = open('somedir/curr_num.txt', 'r')

curr_num = int(temp_fh.readline().strip())

curr_num += 1

fh = open("somedir/sample%s.xml" % cur_num, 'w')

rs = [blockresult]

fh.writelines(rs)

fh.close()

edited Aug 1 '13 at 5:24

answered Aug 1 '13 at 5:06

Vorticity

2,30321837

Your cur_num calculation only works for 1-digit numbers, it is not general enough.

– Eric O Lebigot
Aug 1 '13 at 5:21

Good point, updated.

– Vorticity
Aug 1 '13 at 5:25

add a comment |

Without storing state data in an extra file, a quicker solution to the ones presented here would be to do the following:

from glob import glob

import os



files = glob("somedir/sample*.xml")

files = files.sorted()

cur_num = int(os.path.basename(files[-1])[6:-4])

cur_num += 1

fh = open("somedir/sample%s.xml" % cur_num, 'w')

rs = [blockresult]

fh.writelines(rs)

fh.close()

This will also keep incrementing, even if some of the lower numbered files disappear.

The other solution here that I like (pointed out by Eiyrioü) is the idea of keeping a temporary file that contains your most recent number:

temp_fh = open('somedir/curr_num.txt', 'r')

curr_num = int(temp_fh.readline().strip())

curr_num += 1

fh = open("somedir/sample%s.xml" % cur_num, 'w')

rs = [blockresult]

fh.writelines(rs)

fh.close()

edited Aug 1 '13 at 5:24

answered Aug 1 '13 at 5:06

Vorticity

2,30321837

Your cur_num calculation only works for 1-digit numbers, it is not general enough.

– Eric O Lebigot
Aug 1 '13 at 5:21

Good point, updated.

– Vorticity
Aug 1 '13 at 5:25

add a comment |

Without storing state data in an extra file, a quicker solution to the ones presented here would be to do the following:

from glob import glob

import os



files = glob("somedir/sample*.xml")

files = files.sorted()

cur_num = int(os.path.basename(files[-1])[6:-4])

cur_num += 1

fh = open("somedir/sample%s.xml" % cur_num, 'w')

rs = [blockresult]

fh.writelines(rs)

fh.close()

This will also keep incrementing, even if some of the lower numbered files disappear.

The other solution here that I like (pointed out by Eiyrioü) is the idea of keeping a temporary file that contains your most recent number:

temp_fh = open('somedir/curr_num.txt', 'r')

curr_num = int(temp_fh.readline().strip())

curr_num += 1

fh = open("somedir/sample%s.xml" % cur_num, 'w')

rs = [blockresult]

fh.writelines(rs)

fh.close()

edited Aug 1 '13 at 5:24

answered Aug 1 '13 at 5:06

Vorticity

2,30321837

Without storing state data in an extra file, a quicker solution to the ones presented here would be to do the following:

from glob import glob

import os



files = glob("somedir/sample*.xml")

files = files.sorted()

cur_num = int(os.path.basename(files[-1])[6:-4])

cur_num += 1

fh = open("somedir/sample%s.xml" % cur_num, 'w')

rs = [blockresult]

fh.writelines(rs)

fh.close()

This will also keep incrementing, even if some of the lower numbered files disappear.

The other solution here that I like (pointed out by Eiyrioü) is the idea of keeping a temporary file that contains your most recent number:

temp_fh = open('somedir/curr_num.txt', 'r')

curr_num = int(temp_fh.readline().strip())

curr_num += 1

fh = open("somedir/sample%s.xml" % cur_num, 'w')

rs = [blockresult]

fh.writelines(rs)

fh.close()

edited Aug 1 '13 at 5:24

answered Aug 1 '13 at 5:06

Vorticity

2,30321837

edited Aug 1 '13 at 5:24

answered Aug 1 '13 at 5:06

Vorticity

2,30321837

answered Aug 1 '13 at 5:06

Vorticity

2,30321837

answered Aug 1 '13 at 5:06

Vorticity

2,30321837

Your cur_num calculation only works for 1-digit numbers, it is not general enough.

– Eric O Lebigot
Aug 1 '13 at 5:21

Good point, updated.

– Vorticity
Aug 1 '13 at 5:25

add a comment |

Your cur_num calculation only works for 1-digit numbers, it is not general enough.

– Eric O Lebigot
Aug 1 '13 at 5:21

Good point, updated.

– Vorticity
Aug 1 '13 at 5:25

Your cur_num calculation only works for 1-digit numbers, it is not general enough.

– Eric O Lebigot
Aug 1 '13 at 5:21

Good point, updated.

– Vorticity
Aug 1 '13 at 5:25

add a comment |

Another example using recursion

import os

def checkFilePath(testString, extension, currentCount):

    if os.path.exists(testString + str(currentCount) +extension):

        return checkFilePath(testString, extension, currentCount+1)

    else:

        return testString + str(currentCount) +extension

Use:

checkFilePath("myfile", ".txt" , 0)

answered Feb 15 '17 at 3:04

chumbaloo

1409

add a comment |

Another example using recursion

import os

def checkFilePath(testString, extension, currentCount):

    if os.path.exists(testString + str(currentCount) +extension):

        return checkFilePath(testString, extension, currentCount+1)

    else:

        return testString + str(currentCount) +extension

Use:

checkFilePath("myfile", ".txt" , 0)

answered Feb 15 '17 at 3:04

chumbaloo

1409

add a comment |

Another example using recursion

import os

def checkFilePath(testString, extension, currentCount):

    if os.path.exists(testString + str(currentCount) +extension):

        return checkFilePath(testString, extension, currentCount+1)

    else:

        return testString + str(currentCount) +extension

Use:

checkFilePath("myfile", ".txt" , 0)

answered Feb 15 '17 at 3:04

chumbaloo

1409

Another example using recursion

import os

def checkFilePath(testString, extension, currentCount):

    if os.path.exists(testString + str(currentCount) +extension):

        return checkFilePath(testString, extension, currentCount+1)

    else:

        return testString + str(currentCount) +extension

Use:

checkFilePath("myfile", ".txt" , 0)

answered Feb 15 '17 at 3:04

chumbaloo

1409

answered Feb 15 '17 at 3:04

chumbaloo

1409

answered Feb 15 '17 at 3:04

chumbaloo

1409

answered Feb 15 '17 at 3:04

chumbaloo

1409

add a comment |

You can use a while loop with a counter which checks if a file with a name and the counter's value exists if it does then move on else break and make a file.

I have done it in this way for one of my projects:`

from os import path

import os



i = 0

flnm = "Directory\Filename" + str(i) + ".txt"

while path.exists(flnm) :

    flnm = "Directory\Filename" + str(i) + ".txt"

    i += 1

f = open(flnm, "w") #do what you want to with that file...

f.write(str(var))

f.close() # make sure to close it.

answered Jan 1 at 7:26

Matrix Programmer

1112

Thanks For Rating!

– Matrix Programmer
Jan 31 at 10:41

add a comment |

You can use a while loop with a counter which checks if a file with a name and the counter's value exists if it does then move on else break and make a file.

I have done it in this way for one of my projects:`

from os import path

import os



i = 0

flnm = "Directory\Filename" + str(i) + ".txt"

while path.exists(flnm) :

    flnm = "Directory\Filename" + str(i) + ".txt"

    i += 1

f = open(flnm, "w") #do what you want to with that file...

f.write(str(var))

f.close() # make sure to close it.

answered Jan 1 at 7:26

Matrix Programmer

1112

Thanks For Rating!

– Matrix Programmer
Jan 31 at 10:41

add a comment |

You can use a while loop with a counter which checks if a file with a name and the counter's value exists if it does then move on else break and make a file.

I have done it in this way for one of my projects:`

from os import path

import os



i = 0

flnm = "Directory\Filename" + str(i) + ".txt"

while path.exists(flnm) :

    flnm = "Directory\Filename" + str(i) + ".txt"

    i += 1

f = open(flnm, "w") #do what you want to with that file...

f.write(str(var))

f.close() # make sure to close it.

answered Jan 1 at 7:26

Matrix Programmer

1112

You can use a while loop with a counter which checks if a file with a name and the counter's value exists if it does then move on else break and make a file.

I have done it in this way for one of my projects:`

from os import path

import os



i = 0

flnm = "Directory\Filename" + str(i) + ".txt"

while path.exists(flnm) :

    flnm = "Directory\Filename" + str(i) + ".txt"

    i += 1

f = open(flnm, "w") #do what you want to with that file...

f.write(str(var))

f.close() # make sure to close it.

answered Jan 1 at 7:26

Matrix Programmer

1112

answered Jan 1 at 7:26

Matrix Programmer

1112

answered Jan 1 at 7:26

Matrix Programmer

1112

answered Jan 1 at 7:26

Matrix Programmer

1112

Thanks For Rating!

– Matrix Programmer
Jan 31 at 10:41

add a comment |

Thanks For Rating!

– Matrix Programmer
Jan 31 at 10:41

Thanks For Rating!

– Matrix Programmer
Jan 31 at 10:41

add a comment |

The two ways to do it are:

Check for the existence of the old file and if it exists try the next file name +1

save state data somewhere

an easy way to do it off the bat would be:

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(pth.abspath(filename+str(filenum)+".py")):

    filenum+=1

my_next_file = open(filename+str(filenum)+".py",'w')

as a design thing, while True slows things down and isn't a great thing for code readability

edited: @EOL contributions/ thoughts

so I think not having .format is more readable at first glance - but using .format is better for generality and convention so.

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(pth.abspath(filename+str(filenum)+".py")):

    filenum+=1

my_next_file = open("{}{}.py".format(filename, filenum),'w')

# or 

my_next_file = open(filename + "{}.py".format(filenum),'w')

and you don't have to use abspath - you can use relative paths if you prefer, I prefer abs path sometimes because it helps to normalize the paths passed :).

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(filename+str(filenum)+".py"):

    filenum+=1

##removed for conciseness

edited Aug 1 '13 at 5:33

answered Aug 1 '13 at 3:33

Eiyrioü von Kauyf

1,72132336

The format() method is much more legible than string concatenation, here. I think that the while loop is fine, here. On another topic, why use abspath()?

– Eric O Lebigot
Aug 1 '13 at 5:23

format is more legible, but then he would have to look at string formatting; this is easier to understand on first glance imho. and abspath because i'm ignoring symlinks :/ .... that could lead to confusing errors

– Eiyrioü von Kauyf
Aug 1 '13 at 5:26

While I understand your point, I believe that even beginners should be shown Pythonic examples, so that they take good habits. The behavior of format() is really quite simple to understand and even guess: "{}{}.py".format(filename, filenum). It's even simpler than the algorithm presented here. :)

– Eric O Lebigot
Aug 1 '13 at 5:29

hmm added :) i agree with you on this point.

– Eiyrioü von Kauyf
Aug 1 '13 at 5:31

@EOL whatcha think ;) do I have your approval

– Eiyrioü von Kauyf
Aug 1 '13 at 5:34

add a comment |

The two ways to do it are:

Check for the existence of the old file and if it exists try the next file name +1

save state data somewhere

an easy way to do it off the bat would be:

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(pth.abspath(filename+str(filenum)+".py")):

    filenum+=1

my_next_file = open(filename+str(filenum)+".py",'w')

as a design thing, while True slows things down and isn't a great thing for code readability

edited: @EOL contributions/ thoughts

so I think not having .format is more readable at first glance - but using .format is better for generality and convention so.

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(pth.abspath(filename+str(filenum)+".py")):

    filenum+=1

my_next_file = open("{}{}.py".format(filename, filenum),'w')

# or 

my_next_file = open(filename + "{}.py".format(filenum),'w')

and you don't have to use abspath - you can use relative paths if you prefer, I prefer abs path sometimes because it helps to normalize the paths passed :).

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(filename+str(filenum)+".py"):

    filenum+=1

##removed for conciseness

edited Aug 1 '13 at 5:33

answered Aug 1 '13 at 3:33

Eiyrioü von Kauyf

1,72132336

The format() method is much more legible than string concatenation, here. I think that the while loop is fine, here. On another topic, why use abspath()?

– Eric O Lebigot
Aug 1 '13 at 5:23

format is more legible, but then he would have to look at string formatting; this is easier to understand on first glance imho. and abspath because i'm ignoring symlinks :/ .... that could lead to confusing errors

– Eiyrioü von Kauyf
Aug 1 '13 at 5:26

While I understand your point, I believe that even beginners should be shown Pythonic examples, so that they take good habits. The behavior of format() is really quite simple to understand and even guess: "{}{}.py".format(filename, filenum). It's even simpler than the algorithm presented here. :)

– Eric O Lebigot
Aug 1 '13 at 5:29

hmm added :) i agree with you on this point.

– Eiyrioü von Kauyf
Aug 1 '13 at 5:31

@EOL whatcha think ;) do I have your approval

– Eiyrioü von Kauyf
Aug 1 '13 at 5:34

add a comment |

The two ways to do it are:

Check for the existence of the old file and if it exists try the next file name +1

save state data somewhere

an easy way to do it off the bat would be:

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(pth.abspath(filename+str(filenum)+".py")):

    filenum+=1

my_next_file = open(filename+str(filenum)+".py",'w')

as a design thing, while True slows things down and isn't a great thing for code readability

edited: @EOL contributions/ thoughts

so I think not having .format is more readable at first glance - but using .format is better for generality and convention so.

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(pth.abspath(filename+str(filenum)+".py")):

    filenum+=1

my_next_file = open("{}{}.py".format(filename, filenum),'w')

# or 

my_next_file = open(filename + "{}.py".format(filenum),'w')

and you don't have to use abspath - you can use relative paths if you prefer, I prefer abs path sometimes because it helps to normalize the paths passed :).

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(filename+str(filenum)+".py"):

    filenum+=1

##removed for conciseness

edited Aug 1 '13 at 5:33

answered Aug 1 '13 at 3:33

Eiyrioü von Kauyf

1,72132336

The two ways to do it are:

Check for the existence of the old file and if it exists try the next file name +1

save state data somewhere

an easy way to do it off the bat would be:

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(pth.abspath(filename+str(filenum)+".py")):

    filenum+=1

my_next_file = open(filename+str(filenum)+".py",'w')

as a design thing, while True slows things down and isn't a great thing for code readability

edited: @EOL contributions/ thoughts

so I think not having .format is more readable at first glance - but using .format is better for generality and convention so.

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(pth.abspath(filename+str(filenum)+".py")):

    filenum+=1

my_next_file = open("{}{}.py".format(filename, filenum),'w')

# or 

my_next_file = open(filename + "{}.py".format(filenum),'w')

and you don't have to use abspath - you can use relative paths if you prefer, I prefer abs path sometimes because it helps to normalize the paths passed :).

import os.path as pth

filename = "myfile"

filenum = 1

while (pth.exists(filename+str(filenum)+".py"):

    filenum+=1

##removed for conciseness

edited Aug 1 '13 at 5:33

answered Aug 1 '13 at 3:33

Eiyrioü von Kauyf

1,72132336

edited Aug 1 '13 at 5:33

answered Aug 1 '13 at 3:33

Eiyrioü von Kauyf

1,72132336

answered Aug 1 '13 at 3:33

Eiyrioü von Kauyf

1,72132336

answered Aug 1 '13 at 3:33

Eiyrioü von Kauyf

1,72132336

The format() method is much more legible than string concatenation, here. I think that the while loop is fine, here. On another topic, why use abspath()?

– Eric O Lebigot
Aug 1 '13 at 5:23

format is more legible, but then he would have to look at string formatting; this is easier to understand on first glance imho. and abspath because i'm ignoring symlinks :/ .... that could lead to confusing errors

– Eiyrioü von Kauyf
Aug 1 '13 at 5:26

While I understand your point, I believe that even beginners should be shown Pythonic examples, so that they take good habits. The behavior of format() is really quite simple to understand and even guess: "{}{}.py".format(filename, filenum). It's even simpler than the algorithm presented here. :)

– Eric O Lebigot
Aug 1 '13 at 5:29

hmm added :) i agree with you on this point.

– Eiyrioü von Kauyf
Aug 1 '13 at 5:31

@EOL whatcha think ;) do I have your approval

– Eiyrioü von Kauyf
Aug 1 '13 at 5:34

add a comment |

The format() method is much more legible than string concatenation, here. I think that the while loop is fine, here. On another topic, why use abspath()?

– Eric O Lebigot
Aug 1 '13 at 5:23

format is more legible, but then he would have to look at string formatting; this is easier to understand on first glance imho. and abspath because i'm ignoring symlinks :/ .... that could lead to confusing errors

– Eiyrioü von Kauyf
Aug 1 '13 at 5:26

While I understand your point, I believe that even beginners should be shown Pythonic examples, so that they take good habits. The behavior of format() is really quite simple to understand and even guess: "{}{}.py".format(filename, filenum). It's even simpler than the algorithm presented here. :)

– Eric O Lebigot
Aug 1 '13 at 5:29

hmm added :) i agree with you on this point.

– Eiyrioü von Kauyf
Aug 1 '13 at 5:31

@EOL whatcha think ;) do I have your approval

– Eiyrioü von Kauyf
Aug 1 '13 at 5:34

The format() method is much more legible than string concatenation, here. I think that the while loop is fine, here. On another topic, why use abspath()?

– Eric O Lebigot
Aug 1 '13 at 5:23

format is more legible, but then he would have to look at string formatting; this is easier to understand on first glance imho. and abspath because i'm ignoring symlinks :/ .... that could lead to confusing errors

– Eiyrioü von Kauyf
Aug 1 '13 at 5:26

While I understand your point, I believe that even beginners should be shown Pythonic examples, so that they take good habits. The behavior of format() is really quite simple to understand and even guess: "{}{}.py".format(filename, filenum). It's even simpler than the algorithm presented here. :)

– Eric O Lebigot
Aug 1 '13 at 5:29

hmm added :) i agree with you on this point.

– Eiyrioü von Kauyf
Aug 1 '13 at 5:31

@EOL whatcha think ;) do I have your approval

– Eiyrioü von Kauyf
Aug 1 '13 at 5:34

add a comment |

Another solution that avoids the use of while loop is to use os.listdir() function which returns a list of all the files and directories contained in a directory whose path is taken as an argument.

import os



new_index = len(os.listdir('path_to_file_containing_only_sample_i_files'))

new_file = open('path_to_file_containing_only_sample_i_files/sample_%s.xml' % new_index, 'w')

answered Aug 30 '18 at 9:43

Malo Pocheau

264

add a comment |

Another solution that avoids the use of while loop is to use os.listdir() function which returns a list of all the files and directories contained in a directory whose path is taken as an argument.

import os



new_index = len(os.listdir('path_to_file_containing_only_sample_i_files'))

new_file = open('path_to_file_containing_only_sample_i_files/sample_%s.xml' % new_index, 'w')

answered Aug 30 '18 at 9:43

Malo Pocheau

264

add a comment |

Another solution that avoids the use of while loop is to use os.listdir() function which returns a list of all the files and directories contained in a directory whose path is taken as an argument.

import os



new_index = len(os.listdir('path_to_file_containing_only_sample_i_files'))

new_file = open('path_to_file_containing_only_sample_i_files/sample_%s.xml' % new_index, 'w')

answered Aug 30 '18 at 9:43

Malo Pocheau

264

Another solution that avoids the use of while loop is to use os.listdir() function which returns a list of all the files and directories contained in a directory whose path is taken as an argument.

import os



new_index = len(os.listdir('path_to_file_containing_only_sample_i_files'))

new_file = open('path_to_file_containing_only_sample_i_files/sample_%s.xml' % new_index, 'w')

answered Aug 30 '18 at 9:43

Malo Pocheau

264

answered Aug 30 '18 at 9:43

Malo Pocheau

264

answered Aug 30 '18 at 9:43

Malo Pocheau

264

answered Aug 30 '18 at 9:43

Malo Pocheau

264

add a comment |

def get_unique_dir(path, width=3):

    # if it doesn't exist, create

    if not os.path.isdir(path):

        log.debug("Creating new directory - {}".format(path))

        os.makedirs(path)

        return path



    # if it's empty, use

    if not os.listdir(path):

        log.debug("Using empty directory - {}".format(path))

        return path



    # otherwise, increment the highest number folder in the series



    def get_trailing_number(search_text):

        serch_obj = re.search(r"([0-9]+)$", search_text)

        if not serch_obj:

            return 0

        else:

            return int(serch_obj.group(1))



    dirs = glob(path + "*")

    num_list = sorted([get_trailing_number(d) for d in dirs])

    highest_num = num_list[-1]

    next_num = highest_num + 1

    new_path = "{0}_{1:0>{2}}".format(path, next_num, width)



    log.debug("Creating new incremented directory - {}".format(new_path))

    os.makedirs(new_path)

    return new_path



get_unique_dir("output")

edited Feb 8 at 20:34

answered Feb 8 at 20:29

Woods26

add a comment |

def get_unique_dir(path, width=3):

    # if it doesn't exist, create

    if not os.path.isdir(path):

        log.debug("Creating new directory - {}".format(path))

        os.makedirs(path)

        return path



    # if it's empty, use

    if not os.listdir(path):

        log.debug("Using empty directory - {}".format(path))

        return path



    # otherwise, increment the highest number folder in the series



    def get_trailing_number(search_text):

        serch_obj = re.search(r"([0-9]+)$", search_text)

        if not serch_obj:

            return 0

        else:

            return int(serch_obj.group(1))



    dirs = glob(path + "*")

    num_list = sorted([get_trailing_number(d) for d in dirs])

    highest_num = num_list[-1]

    next_num = highest_num + 1

    new_path = "{0}_{1:0>{2}}".format(path, next_num, width)



    log.debug("Creating new incremented directory - {}".format(new_path))

    os.makedirs(new_path)

    return new_path



get_unique_dir("output")

edited Feb 8 at 20:34

answered Feb 8 at 20:29

Woods26

add a comment |

def get_unique_dir(path, width=3):

    # if it doesn't exist, create

    if not os.path.isdir(path):

        log.debug("Creating new directory - {}".format(path))

        os.makedirs(path)

        return path



    # if it's empty, use

    if not os.listdir(path):

        log.debug("Using empty directory - {}".format(path))

        return path



    # otherwise, increment the highest number folder in the series



    def get_trailing_number(search_text):

        serch_obj = re.search(r"([0-9]+)$", search_text)

        if not serch_obj:

            return 0

        else:

            return int(serch_obj.group(1))



    dirs = glob(path + "*")

    num_list = sorted([get_trailing_number(d) for d in dirs])

    highest_num = num_list[-1]

    next_num = highest_num + 1

    new_path = "{0}_{1:0>{2}}".format(path, next_num, width)



    log.debug("Creating new incremented directory - {}".format(new_path))

    os.makedirs(new_path)

    return new_path



get_unique_dir("output")

edited Feb 8 at 20:34

answered Feb 8 at 20:29

Woods26

def get_unique_dir(path, width=3):

    # if it doesn't exist, create

    if not os.path.isdir(path):

        log.debug("Creating new directory - {}".format(path))

        os.makedirs(path)

        return path



    # if it's empty, use

    if not os.listdir(path):

        log.debug("Using empty directory - {}".format(path))

        return path



    # otherwise, increment the highest number folder in the series



    def get_trailing_number(search_text):

        serch_obj = re.search(r"([0-9]+)$", search_text)

        if not serch_obj:

            return 0

        else:

            return int(serch_obj.group(1))



    dirs = glob(path + "*")

    num_list = sorted([get_trailing_number(d) for d in dirs])

    highest_num = num_list[-1]

    next_num = highest_num + 1

    new_path = "{0}_{1:0>{2}}".format(path, next_num, width)



    log.debug("Creating new incremented directory - {}".format(new_path))

    os.makedirs(new_path)

    return new_path



get_unique_dir("output")

edited Feb 8 at 20:34

answered Feb 8 at 20:29

Woods26

edited Feb 8 at 20:34

answered Feb 8 at 20:29

Woods26

answered Feb 8 at 20:29

Woods26

answered Feb 8 at 20:29

Woods26

add a comment |

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