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')]

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

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

用一团。

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

import os

path = 'mypath/path' 
files = os.listdir(path)

files_txt = [i for i in files if i.endswith('.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