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"))))

以python方式获取“dataPath”文件夹中所有的“。txt”文件名:

from os import listdir
from os.path import isfile, join
path = "/dataPath/"
onlyTxtFiles = [f for f in listdir(path) if isfile(join(path, f)) and  f.endswith(".txt")]
print onlyTxtFiles

使用Python OS模块查找具有特定扩展名的文件。

简单的例子如下:

import os

# This is the path where you want to search
path = r'd:'  

# this is extension you want to detect
extension = '.txt'   # this can be : .jpg  .png  .xls  .log .....

for root, dirs_list, files_list in os.walk(path):
    for file_name in files_list:
        if os.path.splitext(file_name)[-1] == extension:
            file_name_path = os.path.join(root, file_name)
            print file_name
            print file_name_path   # This is the full path of the filter file

Python有所有工具可以做到这一点:

import os

the_dir = 'the_dir_that_want_to_search_in'
all_txt_files = filter(lambda x: x.endswith('.txt'), os.listdir(the_dir))

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')]

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']