Path.py是另一种选择:https://github.com/jaraco/path.py

from path import path
p = path('/path/to/the/directory')
for f in p.files(pattern='*.txt'):
    print f

使用 fnmatch： https://docs.python.org/2/library/fnmatch.html

import fnmatch
import os

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

试试这个，这将递归地找到你所有的文件:

import glob, os
os.chdir("H:\\wallpaper")# use whatever directory you want

#double\\ no single \

for file in glob.glob("**/*.txt", recursive = True):
    print(file)

如果文件夹包含大量文件或内存受限，可以考虑使用生成器:

def yield_files_with_extensions(folder_path, file_extension):
   for _, _, files in os.walk(folder_path):
       for file in files:
           if file.endswith(file_extension):
               yield file

选项A:迭代

for f in yield_files_with_extensions('.', '.txt'): 
    print(f)

选项B:全部获取

files = [f for f in yield_files_with_extensions('.', '.txt')]

子目录的功能性解决方案:

from fnmatch import filter
from functools import partial
from itertools import chain
from os import path, walk

print(*chain(*(map(partial(path.join, root), filter(filenames, "*.txt")) for root, _, filenames in walk("mydir"))))

用一团。

>>> import glob
>>> glob.glob('./*.txt')
['./outline.txt', './pip-log.txt', './test.txt', './testingvim.txt']