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?

Solution 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.

Solution 2

I'd suggest something plain and simple - use os.path.splitext to retrieve basename and extension, and after that simple merge all result components via str.format method.

import os
import random
import string

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

def append_id(filename):
    name, ext = os.path.splitext(filename)
    return "{name}_{uid}{ext}".format(name=name, uid=generate_id(), ext=ext)

Some testcases:

append_id("hello.txt")
append_id("hello")
append_id("multiple.dots.in.file")

Solution 3

Since Python 3.4, a good tool for working with paths is pathlib. This treats paths as objects instead of strings and offers a powerful interface.

Since Python 3.9, a new method - with_stem was added, so your function can be:

from pathlib import Path

def append_id(filename):
    path = Path(filename)
    return path.with_stem(f"{path.stem}_{generate_id()}")

For older versions you can still use the with_name method:

from pathlib import Path

def append_id(filename):
    path = Path(filename)
    return path.with_name(f"{path.stem}_{generate_id()}{path.suffix}")

Those functions are generalized and will work also on full/relative paths and not just file names:

>>> append_id("sample.text")
PosixPath('sample_PBMQBFD.text')

>>> append_id("/a/b/c/sample.text")
PosixPath('/a/b/c/sample_UMTYXCP.text')

Note that a platform-specific Path object is returned. You can always just convert it back to a string with str():

>>> str(append_id("/a/b/c/sample.text"))
'/a/b/c/sample_O1LPTEC.text'

Solution 4

def append_id(filename):
    parts = filename.split('.')
    return "".join(parts[:-1])+ '_' + generate_id() + '.' + parts[-1]

Solution 5

your one line answer with the random generation -

map(lambda x :x.split(".")[0] + "_" + hashlib.md5(("%.15f" % time.time())+"_"+''.join(random.choice(string.ascii_uppercase + string.digits) for x in range(5))).hexdigest()[:7]+"."+x.split(".")[1], filenames)

here you can input filenames as a list

just used a random id generation function which takes time and and random string of 5 characters md5's it and takes the first 7 characters from that.

It's not very readable but since you asked for a one line solution, I couldn't think of a more elaborate way.

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Comments

[BartoszKP]

for name, ext in filename.split('.') won't give you two successive elements of a collection in name and ext, but will try to unpack each collection's element into two variables: name and ext.

[ss7]

just to add a string to files that have the same name when being moved

[Stevoisiak]

Will this work if filename is a path like C:/path/filename.txt

[DV82XL]

I can't live without pathlib. Thanks for posting this!

[user1330734]

As of June 2021, this is for me is the most up to date and simplest answer.