上述答案的 Python 3.6 版本，使用os-
假设您将目录路径作为一个str名为的变量中的对象directory_in_str：

import os

directory = os.fsencode(directory_in_str)

for file in os.listdir(directory):
     filename = os.fsdecode(file)
     if filename.endswith(".asm") or filename.endswith(".py"): 
         # print(os.path.join(directory, filename))
         continue
     else:
         continue

或递归地，使用pathlib：

from pathlib import Path

pathlist = Path(directory_in_str).glob('**/*.asm')
for path in pathlist:
     # because path is object not string
     path_in_str = str(path)
     # print(path_in_str)

• 用于rglob替换glob('**/*.asm')_rglob('*.asm')

  • 这就像在给定的相对模式之前调用Path.glob()with '**/'added：

  from pathlib import Path

  pathlist = Path(directory_in_str).rglob(‘*.asm’)
  for path in pathlist:
  # because path is object not string
  path_in_str = str(path)
  # print(path_in_str)

原答案：

import os

for filename in os.listdir("/path/to/dir/"):
    if filename.endswith(".asm") or filename.endswith(".py"): 
         # print(os.path.join(directory, filename))
        continue
    else:
        continue