Path ordering based on particular criteria

I have four files (or any number of files for that matter) named

file_V2023.2.2_0.txt
file_V2023.2.2_1.txt
file_V2023.2.3_0.txt
file_V2023.2.3_1.txt

If I do

from pathlib import Path
output_path = Path("./")
for video_path in sorted(output_path.glob("*.txt")):
    print(video_path)

I get the order above.

Is there a way I can get the following order:

file_V2023.2.2_0.txt
file_V2023.2.3_0.txt
file_V2023.2.2_1.txt
file_V2023.2.3_1.txt

The sorted() function has a key parameter, and you give that a function that provides a sort key for the things you're sorting.

So:

import re


names = [
    'file_V2023.2.2_0.txt',
    'file_V2023.2.3_0.txt',
    'file_V2023.2.2_1.txt',
    'file_V2023.2.3_1.txt'
]


name_pattern = re.compile('.*\.(\d+)\.(\d+)_(\d+)\.txt')
def get_key(name):
    a, b, c = re.match(name_pattern, name).groups()
    return int(a), int(c), int(b)  # reordering here


print(sorted(names, key=get_key))

Output:

['file_V2023.2.2_0.txt', 'file_V2023.2.3_0.txt', 'file_V2023.2.2_1.txt', 'file_V2023.2.3_1.txt']

The regular expression is really useful to break up the name, since it will also work for names like file_V2023.10.2_99.txt. The parts in the regex that are enclosed in parentheses ((, )) are matched as separate groups, which are then retrieved with .groups() and since there's three of them, they can be spread over a, b, c.

The reason the matched numerical strings (e.g., for 'file_V2023.2.3_1.txt' they would be '2', '3', and '1') are converted to int, is to make sure that something like '19' would end up after '2' instead of before it because the '1' that '19' starts with comes before '2' alphanumerically.

Note that the call to re.compile is there only for efficiency. This way, the regex only has to be compiled once, instead of every time the function is called. But if you want the code to be shorter, or avoid accessing the global, this would do the same:

def get_key(name):
    a, b, c = re.match('.*\.(\d+)\.(\d+)_(\d+)\.txt', name).groups()
    return int(a), int(c), int(b)  # reordering here

Also note that this example assumes the numbers are the only thing you're sorting by, so only a 3-tuple of integer values is returned as a sorting key. If you have names like 'afile_V2023.2.3_0.txt' and 'bfile_V2023.2.2_0.txt', and you want the 'afile' to come before the 'bfile' in spite of their numbering, this works:

name_pattern = re.compile('(.*)\.(\d+)\.(\d+)_(\d+)\.txt')
def get_key(name):
    t, a, b, c = re.match(name_pattern, name).groups()
    return t, int(a), int(c), int(b)

That is, you can of course mix types in the sort key, as long as Python knows how to order them.

Actually given the file name design you have, there is no need to use regular expressions.

Simple:

names = [
    'file_V2023.2.2_0.txt',
    'file_V2023.2.3_1.txt',
    'file_V2023.2.3_0.txt',
    'file_V2023.2.2_1.txt',
]


names.sort(key=lambda name: float(name[11:-4]))

print(names)

Examples of casting:

>>> float("2.2_0")
>>> 2.2
>>> float("2.2_1")
>>> 2.21