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'))

或者:

from glob import glob
...
for files in glob('path/*.mp3'): 
  do something
for files in glob('path/*.flac'): 
  do something else

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

在检查扩展名之前，您应该确保“文件”实际上不是一个文件夹。上面的一些答案没有说明带句点的文件夹名称。(folder.mp3是一个有效的文件夹名)。

检查文件扩展名:

import os

file_path = "C:/folder/file.mp3"
if os.path.isfile(file_path):
    file_extension = os.path.splitext(file_path)[1]
    if file_extension.lower() == ".mp3":
        print("It's an mp3")
    if file_extension.lower() == ".flac":
        print("It's a flac")

输出:

It's an mp3

检查文件夹中所有文件的扩展名:

import os

directory = "C:/folder"
for file in os.listdir(directory):
    file_path = os.path.join(directory, file)
    if os.path.isfile(file_path):
        file_extension = os.path.splitext(file_path)[1]
        print(file, "ends in", file_extension)

输出:

abc.txt ends in .txt
file.mp3 ends in .mp3
song.flac ends in .flac

比较多种类型的文件扩展名:

import os

file_path = "C:/folder/file.mp3"
if os.path.isfile(file_path):
    file_extension = os.path.splitext(file_path)[1]
    if file_extension.lower() in {'.mp3', '.flac', '.ogg'}:
        print("It's a music file")
    elif file_extension.lower() in {'.jpg', '.jpeg', '.png'}:
        print("It's an image file")

输出:

It's a music file

我很惊讶没有一个答案建议使用pathlib库。

当然，它的使用是视情况而定的，但当涉及到文件处理或统计pathlib是黄金。

下面是一个片段:

import pathlib


def get_parts(p: str or pathlib.Path) -> None:
    p_ = pathlib.Path(p).expanduser().resolve()
    print(p_)
    print(f"file name: {p_.name}")
    print(f"file extension: {p_.suffix}")
    print(f"file extensions: {p_.suffixes}\n")


if __name__ == '__main__':
    file_path = 'conf/conf.yml'
    arch_file_path = 'export/lib.tar.gz'

    get_parts(p=file_path)
    get_parts(p=arch_file_path)

输出:

/Users/hamster/temp/src/pro1/conf/conf.yml
file name: conf.yml
file extension: .yml
file extensions: ['.yml']

/Users/hamster/temp/src/pro1/conf/lib.tar.gz
file name: lib.tar.gz
file extension: .gz
file extensions: ['.tar', '.gz']