这么多的答案都建议全局替换和扩展的数量一样多，我更喜欢只替换一次:

from pathlib import Path

files = (p.resolve() for p in Path(path).glob("**/*") if p.suffix in {".c", ".cc", ".cpp", ".hxx", ".h"})

您可以尝试制作一个手动列表，将现有扩展与所需扩展进行比较。

ext_list = ['gif','jpg','jpeg','png'];
file_list = []
for file in glob.glob('*.*'):
  if file.rsplit('.',1)[1] in ext_list :
    file_list.append(file)

根据我从经验测试中得到的结果，那一团东西。Glob并不是通过扩展名过滤文件的更好方法。其中一些原因是:

通配符“语言”不允许对多个扩展进行完美的规范。 前一点导致根据文件扩展名获得不正确的结果。 经经验证明，通配符法比大多数其他方法都要慢。 即使这很奇怪，甚至其他文件系统对象也可以有“扩展名”，文件夹也是如此。

我已经测试了以下4种不同的方法，通过扩展名过滤文件，并将它们放在一个列表中:

from glob import glob, iglob
from re import compile, findall
from os import walk


def glob_with_storage(args):

    elements = ''.join([f'[{i}]' for i in args.extensions])
    globs = f'{args.target}/**/*{elements}'
    results = glob(globs, recursive=True)

    return results


def glob_with_iteration(args):

    elements = ''.join([f'[{i}]' for i in args.extensions])
    globs = f'{args.target}/**/*{elements}'
    results = [i for i in iglob(globs, recursive=True)]

    return results


def walk_with_suffixes(args):

    results = []
    for r, d, f in walk(args.target):
        for ff in f:
            for e in args.extensions:
                if ff.endswith(e):
                    results.append(path_join(r,ff))
                    break
    return results


def walk_with_regs(args):

    reg = compile('|'.join([f'{i}$' for i in args.extensions]))

    results = []
    for r, d, f in walk(args.target):
        for ff in f:
            if len(findall(reg,ff)):
                results.append(path_join(r, ff))

    return results

通过在我的笔记本电脑上运行上面的代码，我得到了以下自动解释的结果。

Elapsed time for '7 times glob_with_storage()':  0.365023 seconds.
mean   : 0.05214614
median : 0.051861
stdev  : 0.001492152
min    : 0.050864
max    : 0.054853

Elapsed time for '7 times glob_with_iteration()':  0.360037 seconds.
mean   : 0.05143386
median : 0.050864
stdev  : 0.0007847381
min    : 0.050864
max    : 0.052859

Elapsed time for '7 times walk_with_suffixes()':  0.26529 seconds.
mean   : 0.03789857
median : 0.037899
stdev  : 0.0005759071
min    : 0.036901
max    : 0.038896

Elapsed time for '7 times walk_with_regs()':  0.290223 seconds.
mean   : 0.04146043
median : 0.040891
stdev  : 0.0007846776
min    : 0.04089
max    : 0.042885

Results sizes:
0 2451
1 2451
2 2446
3 2446

Differences between glob() and walk():
0 E:\x\y\z\venv\lib\python3.7\site-packages\Cython\Includes\numpy
1 E:\x\y\z\venv\lib\python3.7\site-packages\Cython\Utility\CppSupport.cpp
2 E:\x\y\z\venv\lib\python3.7\site-packages\future\moves\xmlrpc
3 E:\x\y\z\venv\lib\python3.7\site-packages\Cython\Includes\libcpp
4 E:\x\y\z\venv\lib\python3.7\site-packages\future\backports\xmlrpc

Elapsed time for 'main':  1.317424 seconds.

通过扩展名过滤文件的最快方法，甚至是最丑陋的方法。也就是说，使用endswith()方法进行嵌套for循环和字符串比较。

此外，正如您所看到的，配色算法(使用模式E:\x\y\z\**/*[py][pyc])即使只给出两个扩展名(py和pyc)也会返回不正确的结果。

这对我很管用!

split('.')[-1]

上面的代码分开了文件名后缀(*.xxx)，所以它可以帮助你

    for filename in glob.glob(folder + '*.*'):
        print(folder+filename)
        if  filename.split('.')[-1] != 'tif' and \
            filename.split('.')[-1] != 'tiff' and \
            filename.split('.')[-1] != 'bmp' and \
            filename.split('.')[-1] != 'jpg' and \
            filename.split('.')[-1] != 'jpeg' and \
            filename.split('.')[-1] != 'png':
                continue
        # Your code

例如:

import glob
lst_img = []
base_dir = '/home/xy/img/'

# get all the jpg file in base_dir 
lst_img += glob.glob(base_dir + '*.jpg')
print lst_img
# ['/home/xy/img/2.jpg', '/home/xy/img/1.jpg']

# append all the png file in base_dir to lst_img
lst_img += glob.glob(base_dir + '*.png')
print lst_img
# ['/home/xy/img/2.jpg', '/home/xy/img/1.jpg', '/home/xy/img/3.png']

一个函数:

import glob
def get_files(base_dir='/home/xy/img/', lst_extension=['*.jpg', '*.png']):
    """
    :param base_dir:base directory
    :param lst_extension:lst_extension: list like ['*.jpg', '*.png', ...]
    :return:file lists like ['/home/xy/img/2.jpg','/home/xy/img/3.png']
    """
    lst_files = []
    for ext in lst_extension:
        lst_files += glob.glob(base_dir+ext)
    return lst_files

与@BPL相同的答案(计算效率高)，但它可以处理任何glob模式，而不是扩展:

import os
from fnmatch import fnmatch

folder = "path/to/folder/"
patterns = ("*.txt", "*.md", "*.markdown")

files = [f.path for f in os.scandir(folder) if any(fnmatch(f, p) for p in patterns)]

这种解决方案既高效又方便。它还与glob的行为紧密匹配(请参阅文档)。

注意，使用内置包pathlib会更简单:

from pathlib import Path

folder = Path("/path/to/folder")
patterns = ("*.txt", "*.md", "*.markdown")

files = [f for f in folder.iterdir() if any(f.match(p) for p in patterns)]