Is this the best way to get unique version of filename w/ Python?

Question

Still 'diving in' to Python, and want to make sure I'm not overlooking something. I wrote a script that extracts files from several zip files, and saves the extracted files together in one directory. To prevent duplicate filenames from being over-written, I wrote this little function - and I'm just wondering if there is a better way to do this? Thanks!

def unique_filename(file_name):
counter = 1
file_name_parts = os.path.splitext(file_name) # returns ('/path/file', '.ext')
while os.path.isfile(file_name): 
    file_name = file_name_parts[0] + '_' + str(counter) + file_name_parts[1]
    counter += 1
return file_name

I really do require the files to be in a single directory, and numbering duplicates is definitely acceptable in my case, so I'm not looking for a more robust method (tho' I suppose any pointers are welcome), but just to make sure that what this accomplishes is getting done the right way.

Answer 1

One issue is that there is a race condition in your above code, since there is a gap between testing for existance, and creating the file. There may be security implications to this (think about someone maliciously inserting a symlink to a sensitive file which they wouldn't be able to overwrite, but your program running with a higher privilege could) Attacks like these are why things like os.tempnam() are deprecated.

To get around it, the best approach is to actually try create the file in such a way that you'll get an exception if it fails, and on success, return the actually opened file object. This can be done with the lower level os.open functions, by passing both the os.O_CREAT and os.O_EXCL flags. Once opened, return the actual file (and optionally filename) you create. Eg, here's your code modified to use this approach (returning a (file, filename) tuple):

def unique_file(file_name):
    counter = 1
    file_name_parts = os.path.splitext(file_name) # returns ('/path/file', '.ext')
    while 1:
        try:
            fd = os.open(file_name, os.O_CREAT | os.O_EXCL | os.O_RDRW)
            return os.fdopen(fd), file_name
        except OSError:
            pass
        file_name = file_name_parts[0] + '_' + str(counter) + file_name_parts[1]
        counter += 1

[Edit] Actually, a better way, which will handle the above issues for you, is probably to use the tempfile module, though you may lose some control over the naming. Here's an example of using it (keeping a similar interface):

def unique_file(file_name):
    dirname, filename = os.path.split(file_name)
    prefix, suffix = os.path.splitext(filename)

    fd, filename = tempfile.mkstemp(suffix, prefix+"_", dirname)
    return os.fdopen(fd), filename

>>> f, filename=unique_file('/home/some_dir/foo.txt')
>>> print filename
/home/some_dir/foo_z8f_2Z.txt

The only downside with this approach is that you will always get a filename with some random characters in it, as there's no attempt to create an unmodified file (/home/some_dir/foo.txt) first. You may also want to look at tempfile.TemporaryFile and NamedTemporaryFile, which will do the above and also automatically delete from disk when closed.