虽然这个问题很久以前就有答案了。我想推荐使用pathlib模块，因为这是在Windows和Unix操作系统上工作的一种健壮的方式。

要获取特定目录下的所有路径，包括子目录:

from pathlib import Path
paths = list(Path('myhomefolder', 'folder').glob('**/*.txt'))

# all sorts of operations
file = paths[0]
file.name
file.stem
file.parent
file.suffix

etc.

全路径，计算路径为。，..， \\， ..\\…\ \文件夹等:

import os, pprint
pprint.pprint([os.path.join(os.path.abspath(path), x[0]) \
    for x in os.walk(os.path.abspath(path))])

只列出目录

print("\nWe are listing out only the directories in current directory -")
directories_in_curdir = list(filter(os.path.isdir, os.listdir(os.curdir)))
print(directories_in_curdir)

只列出当前目录中的文件

files = list(filter(os.path.isfile, os.listdir(os.curdir)))
print("\nThe following are the list of all files in the current directory -")
print(files)

如果您需要一个递归的解决方案来查找子目录中的所有子目录，请使用前面建议的walk。

如果您只需要当前目录的子目录，请将os. xml目录合并在一起。使用os.path.isdir

下面这个类将能够获得一个给定目录中的文件，文件夹和所有子文件夹的列表

import os
import json

class GetDirectoryList():
    def __init__(self, path):
        self.main_path = path
        self.absolute_path = []
        self.relative_path = []


    def get_files_and_folders(self, resp, path):
        all = os.listdir(path)
        resp["files"] = []
        for file_folder in all:
            if file_folder != "." and file_folder != "..":
                if os.path.isdir(path + "/" + file_folder):
                    resp[file_folder] = {}
                    self.get_files_and_folders(resp=resp[file_folder], path= path + "/" + file_folder)
                else:
                    resp["files"].append(file_folder)
                    self.absolute_path.append(path.replace(self.main_path + "/", "") + "/" + file_folder)
                    self.relative_path.append(path + "/" + file_folder)
        return resp, self.relative_path, self.absolute_path

    @property
    def get_all_files_folder(self):
        self.resp = {self.main_path: {}}
        all = self.get_files_and_folders(self.resp[self.main_path], self.main_path)
        return all

if __name__ == '__main__':
    mylib = GetDirectoryList(path="sample_folder")
    file_list = mylib.get_all_files_folder
    print (json.dumps(file_list))

而样本目录看起来像

sample_folder/
    lib_a/
        lib_c/
            lib_e/
                __init__.py
                a.txt
            __init__.py
            b.txt
            c.txt
        lib_d/
            __init__.py
        __init__.py
        d.txt
    lib_b/
        __init__.py
        e.txt
    __init__.py

结果

[
  {
    "files": [
      "__init__.py"
    ],
    "lib_b": {
      "files": [
        "__init__.py",
        "e.txt"
      ]
    },
    "lib_a": {
      "files": [
        "__init__.py",
        "d.txt"
      ],
      "lib_c": {
        "files": [
          "__init__.py",
          "c.txt",
          "b.txt"
        ],
        "lib_e": {
          "files": [
            "__init__.py",
            "a.txt"
          ]
        }
      },
      "lib_d": {
        "files": [
          "__init__.py"
        ]
      }
    }
  },
  [
    "sample_folder/lib_b/__init__.py",
    "sample_folder/lib_b/e.txt",
    "sample_folder/__init__.py",
    "sample_folder/lib_a/lib_c/lib_e/__init__.py",
    "sample_folder/lib_a/lib_c/lib_e/a.txt",
    "sample_folder/lib_a/lib_c/__init__.py",
    "sample_folder/lib_a/lib_c/c.txt",
    "sample_folder/lib_a/lib_c/b.txt",
    "sample_folder/lib_a/lib_d/__init__.py",
    "sample_folder/lib_a/__init__.py",
    "sample_folder/lib_a/d.txt"
  ],
  [
    "lib_b/__init__.py",
    "lib_b/e.txt",
    "sample_folder/__init__.py",
    "lib_a/lib_c/lib_e/__init__.py",
    "lib_a/lib_c/lib_e/a.txt",
    "lib_a/lib_c/__init__.py",
    "lib_a/lib_c/c.txt",
    "lib_a/lib_c/b.txt",
    "lib_a/lib_d/__init__.py",
    "lib_a/__init__.py",
    "lib_a/d.txt"
  ]
]

下面是基于@Blair Conrad的例子的几个简单函数

import os

def get_subdirs(dir):
    "Get a list of immediate subdirectories"
    return next(os.walk(dir))[1]

def get_subfiles(dir):
    "Get a list of immediate subfiles"
    return next(os.walk(dir))[2]