我建议你使用fnmatch和上面的方法。通过这种方式，你可以找到以下任何一个:

Name.txt; Name.TXT; Name.Txt

.

import fnmatch
import os

    for file in os.listdir("/Users/Johnny/Desktop/MyTXTfolder"):
        if fnmatch.fnmatch(file.upper(), '*.TXT'):
            print(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"))))

这里有一个extend()

types = ('*.jpg', '*.png')
images_list = []
for files in types:
    images_list.extend(glob.glob(os.path.join(path, files)))

这样的东西就可以了

for root, dirs, files in os.walk(directory):
    for file in files:
        if file.endswith('.txt'):
            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')]

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