一个简单的方法是:

import os

if os.path.splitext(file)[1] == ".mp3":
    # do something

Os.path.splitext (file)将返回一个包含两个值的元组(没有扩展名的文件名+只有扩展名的文件名)。因此，第二个索引([1])只提供扩展名。最酷的是，如果需要的话，这样你也可以很容易地访问文件名!

操作系统。Path提供了许多操作路径/文件名的函数。(文档)

Os.path.splitext接受一个路径，并将文件扩展名从路径的末尾分离出来。

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."

给:

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

假设m是一个字符串，你可以使用endswith:

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

不区分大小写，并消除可能较大的else-if链:

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

查看模块fnmatch。它会达到你想要的效果。

import fnmatch
import os

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

#!/usr/bin/python

import shutil, os

source = ['test_sound.flac','ts.mp3']

for files in source:
  fileName,fileExtension = os.path.splitext(files)

  if fileExtension==".flac" :
    print 'This file is flac file %s' %files
  elif  fileExtension==".mp3":
    print 'This file is mp3 file %s' %files
  else:
    print 'Format is not valid'

import os
source = ['test_sound.flac','ts.mp3']

for files in source:
   fileName,fileExtension = os.path.splitext(files)
   print fileExtension   # Print File Extensions
   print fileName   # It print file name