用一团。

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

我喜欢os.walk():

import os

for root, dirs, files in os.walk(dir):
    for f in files:
        if os.path.splitext(f)[1] == '.txt':
            fullpath = os.path.join(root, f)
            print(fullpath)

或者使用生成器:

import os

fileiter = (os.path.join(root, f)
    for root, _, files in os.walk(dir)
    for f in files)
txtfileiter = (f for f in fileiter if os.path.splitext(f)[1] == '.txt')
for txt in txtfileiter:
    print(txt)

import os
import sys 

if len(sys.argv)==2:
    print('no params')
    sys.exit(1)

dir = sys.argv[1]
mask= sys.argv[2]

files = os.listdir(dir); 

res = filter(lambda x: x.endswith(mask), files); 

print res

要从同一个目录中名为“data”的文件夹中获取一个“。txt”文件名的数组，我通常使用以下简单的代码行:

import os
fileNames = [fileName for fileName in os.listdir("data") if fileName.endswith(".txt")]

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

这里有更多相同的版本，产生略微不同的结果:

glob.iglob ()

import glob
for f in glob.iglob("/mydir/*/*.txt"): # generator, search immediate subdirectories 
    print f

glob.glob1（）

print glob.glob1("/mydir", "*.tx?")  # literal_directory, basename_pattern

fnmatch.filter（）

import fnmatch, os
print fnmatch.filter(os.listdir("/mydir"), "*.tx?") # include dot-files