Create file but if name exists add number

Question

Does Python have any built-in functionality to add a number to a filename if it already exists?

My idea is that it would work the way certain OS\'s work - if a file

Answer 1

I ended up writing my own simple function for this. Primitive, but gets the job done:

def uniquify(path):
    filename, extension = os.path.splitext(path)
    counter = 1

    while os.path.exists(path):
        path = filename + " (" + str(counter) + ")" + extension
        counter += 1

    return path

Answer 2

As of Python 3.6 [f-strings format appeared]

A great solution is to use a function that takes 2 arguments and returns the first non-existed filename available. On the first iteration, the function tries to check if filename without any number exists. If this filename is already taken, the function tries to apply a number 0 right before the file extension. If this filename is already taken too, the function adds +1 to this number until non-existed filename will be found.

import os

def unique_filename(output_filename, file_extension):
    n = ''
    while os.path.exists(f'{output_filename}{n}{file_extension}'):
        if isinstance(n, str):
            n = -1
        n += 1
    return f'{output_filename}{n}{file_extension}'

For older Python versions you can use [%-strings format]

import os

def unique_filename(output_filename, file_extension):
    n = ''
    while os.path.exists('%s%s%s' % (output_filename, n, file_extension)):
        if isinstance(n, str):
            n = -1
        n += 1
    return '%s%s%s' % (output_filename, n, file_extension)

Another option is a function that takes file_path as an input argument.

file_path will be split into file path with filename and file extension (f.e. ".jpg")

import os

def unique_filename_path(file_path):
    output_filename, file_extension = os.path.splitext(file_path)
    n = ''
    while os.path.exists(f'{output_filename}{n}{file_extension}'):
      if isinstance(n, str):
          n = -1
      n += 1
    return f'{output_filename}{n}{file_extension}'

Answer 3

If all files being numbered isn't a problem, and you know beforehand the name of the file to be written, you could simply do:

import os

counter = 0
filename = "file{}.pdf"
while os.path.isfile(filename.format(counter)):
    counter += 1
filename = filename.format(counter)

Answer 4

import os

class Renamer():
    def __init__(self, name):
        self.extension = name.split('.')[-1]
        self.name = name[:-len(self.extension)-1]
        self.filename = self.name
    def rename(self):
        i = 1
        if os.path.exists(self.filename+'.'+self.extension):
            while os.path.exists(self.filename+'.'+self.extension):
                self.filename = '{} ({})'.format(self.name,i)
                i += 1
        return self.filename+'.'+self.extension

Answer 5

Since the tempfile hack A) is a hack and B) still requires a decent amount of code anyway, I went with a manual implementation. You basically need:

1. A way to Safely create a file if and only if it does not exist (this is what the tempfile hack affords us).
2. A generator for filenames.
3. A wrapping function to hide the mess.

I defined a safe_open that can be used just like open:

def iter_incrementing_file_names(path):
    """
    Iterate incrementing file names. Start with path and add " (n)" before the
    extension, where n starts at 1 and increases.

    :param path: Some path
    :return: An iterator.
    """
    yield path
    prefix, ext = os.path.splitext(path)
    for i in itertools.count(start=1, step=1):
        yield prefix + ' ({0})'.format(i) + ext


def safe_open(path, mode):
    """
    Open path, but if it already exists, add " (n)" before the extension,
    where n is the first number found such that the file does not already
    exist.

    Returns an open file handle. Make sure to close!

    :param path: Some file name.

    :return: Open file handle... be sure to close!
    """
    flags = os.O_CREAT | os.O_EXCL | os.O_WRONLY

    if 'b' in mode and platform.system() == 'Windows':
        flags |= os.O_BINARY

    for filename in iter_incrementing_file_names(path):
        try:
            file_handle = os.open(filename, flags)
        except OSError as e:
            if e.errno == errno.EEXIST:
                pass
            else:
                raise
        else:
            return os.fdopen(file_handle, mode)

# Example
with safe_open("some_file.txt", "w") as fh:
    print("Hello", file=fh)

Answer 6

I haven't tested this yet but it should work, iterating over possible filenames until the file in question does not exist at which point it breaks.

def increment_filename(fn):
    fn, extension = os.path.splitext(path)

    n = 1
    yield fn + extension
    for n in itertools.count(start=1, step=1)
        yield '%s%d.%s' % (fn, n, extension)

for filename in increment_filename(original_filename):
    if not os.isfile(filename):
        break