How can I check the extension of a file?

Question

Assuming m is a string, you can use endswith:

if m.endswith('.mp3'):
...
elif m.endswith('.flac'):
...

To be case-insensitive, and to eliminate a potentially large else-if chain:

m.lower().endswith(('.png', '.jpg', '.jpeg'))

os.path provides many functions for manipulating paths/filenames. (docs)

os.path.splitext takes a path and splits the file extension from the end of it.

import os

filepaths = ["/folder/soundfile.mp3", "folder1/folder/soundfile.flac"]

for fp in filepaths:
    # Split the extension from the path and normalise it to lowercase.
    ext = os.path.splitext(fp)[-1].lower()

    # Now we can simply use == to check for equality, no need for wildcards.
    if ext == ".mp3":
        print fp, "is an mp3!"
    elif ext == ".flac":
        print fp, "is a flac file!"
    else:
        print fp, "is an unknown file format."

Gives:

/folder/soundfile.mp3 is an mp3!
folder1/folder/soundfile.flac is a flac file!

Use pathlib From Python3.4 onwards.

from pathlib import Path
Path('my_file.mp3').suffix == '.mp3'

Look at module fnmatch. That will do what you're trying to do.

import fnmatch
import os

for file in os.listdir('.'):
    if fnmatch.fnmatch(file, '*.txt'):
        print file