How to add an id to filename before extension?

Question

I have a filename: name.ext

I want to do the following:

name + id + '.' + ext for name, ext in filename.split()

or find a better way to take a filename and add a random 7 character string to the end before the extension.

Here is what I have so far:

def generate_id(size=7, chars=string.ascii_uppercase + string.digits):
    return ''.join(random.choice(chars) for _ in range(size))

def append_id(filename):
    return (name + '_' + generate_id() + '.' + ext for name, ext in filename.split('.'))

but it treats it as a generator expression, which is not my intended result.

What would be the correct way to write the append_id function?

Answer 1

To do it in one line you can try:

def append_id(filename):
    return "{0}_{2}.{1}".format(*filename.rsplit('.', 1) + [generate_id()])

It's not very readable, though.

Most language implementations provide functions to deal with file names, and Python is no exception. You should use os.path.splitext:

def append_id(filename):
  return "{0}_{2}{1}".format(*os.path.splitext(filename) + (generate_id(),))

Note that the second version needs two additional modifications:

• splitext returns a tuple not a list, so we need to wrap the result of generate_id with a tuple
• splitext retains the dot, so you need to remove it from the format string

Still, I wouldn't struggle to have a oneliner here - see the next answer for more readable solutions.

Python 3.4 introduces pathlib module and you can use it like this:

from pathlib import Path

def append_id(filename):
  p = Path(filename)
  return "{0}_{2}{1}".format(p.stem, p.suffix, generate_id())

This will work for filenames without preceding path only. For files with paths use this:

from pathlib import Path

def append_id(filename):
  p = Path(filename)
  return "{0}_{2}{1}".format(Path.joinpath(p.parent, p.stem), p.suffix, generate_id())

In Python 3.9 there is also with_stem, which might be the most suitable choice for this case.