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

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

我做了一个测试(Python 3.6.4, W7x64)，看看对于一个文件夹，没有子目录，哪种解决方案是最快的，以获得具有特定扩展名的文件的完整文件路径列表。

简而言之，对于这个任务，os.listdir()是最快的，比次优os.walk()快1.7倍，比pathlib快2.7倍，比os.scandir()快3.2倍，比glob快3.3倍。 请记住，当您需要递归结果时，这些结果将会改变。如果您复制/粘贴下面的一个方法，请添加.lower()，否则在搜索.ext时将找不到.ext。

import os
import pathlib
import timeit
import glob

def a():
    path = pathlib.Path().cwd()
    list_sqlite_files = [str(f) for f in path.glob("*.sqlite")]

def b(): 
    path = os.getcwd()
    list_sqlite_files = [f.path for f in os.scandir(path) if os.path.splitext(f)[1] == ".sqlite"]

def c():
    path = os.getcwd()
    list_sqlite_files = [os.path.join(path, f) for f in os.listdir(path) if f.endswith(".sqlite")]

def d():
    path = os.getcwd()
    os.chdir(path)
    list_sqlite_files = [os.path.join(path, f) for f in glob.glob("*.sqlite")]

def e():
    path = os.getcwd()
    list_sqlite_files = [os.path.join(path, f) for f in glob.glob1(str(path), "*.sqlite")]

def f():
    path = os.getcwd()
    list_sqlite_files = []
    for root, dirs, files in os.walk(path):
        for file in files:
            if file.endswith(".sqlite"):
                list_sqlite_files.append( os.path.join(root, file) )
        break



print(timeit.timeit(a, number=1000))
print(timeit.timeit(b, number=1000))
print(timeit.timeit(c, number=1000))
print(timeit.timeit(d, number=1000))
print(timeit.timeit(e, number=1000))
print(timeit.timeit(f, number=1000))

结果:

# Python 3.6.4
0.431
0.515
0.161
0.548
0.537
0.274

你可以使用glob:

import glob, os
os.chdir("/mydir")
for file in glob.glob("*.txt"):
    print(file)

或者简单的os.listdir:

import os
for file in os.listdir("/mydir"):
    if file.endswith(".txt"):
        print(os.path.join("/mydir", file))

或者如果你想遍历目录，使用os.walk:

import os
for root, dirs, files in os.walk("/mydir"):
    for file in files:
        if file.endswith(".txt"):
             print(os.path.join(root, file))

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"))))

这样做是可行的:

>>> import os
>>> path = '/usr/share/cups/charmaps'
>>> text_files = [f for f in os.listdir(path) if f.endswith('.txt')]
>>> text_files
['euc-cn.txt', 'euc-jp.txt', 'euc-kr.txt', 'euc-tw.txt', ... 'windows-950.txt']