通过从这里加入多个解决方案，这是我最终使用的:

import os
import glob

def list_dirs(path):
    return [os.path.basename(x) for x in filter(
        os.path.isdir, glob.glob(os.path.join(path, '*')))]

import os
path = "test/"
files = [x[0] + "/" + y for x in os.walk(path) if len(x[-1]) > 0 for y in x[-1]]

这个答案似乎并不存在。

directories = [ x for x in os.listdir('.') if os.path.isdir(x) ]

以Eli Bendersky的解决方案为基础，使用以下示例:

import os
test_directory = <your_directory>
for child in os.listdir(test_directory):
    test_path = os.path.join(test_directory, child)
    if os.path.isdir(test_path):
        print test_path
        # Do stuff to the directory "test_path"

>是要遍历的目录的路径。

对于像我这样只需要一个目录中直接文件夹名称的人来说，这在Windows上是可行的。

import os

for f in os.scandir(mypath):
    print(f.name)

函数返回给定文件路径内所有子目录的List。将搜索整个文件树。

import os

def get_sub_directory_paths(start_directory, sub_directories):
    """
    This method iterates through all subdirectory paths of a given 
    directory to collect all directory paths.

    :param start_directory: The starting directory path.
    :param sub_directories: A List that all subdirectory paths will be 
        stored to.
    :return: A List of all sub-directory paths.
    """

    for item in os.listdir(start_directory):
        full_path = os.path.join(start_directory, item)

        if os.path.isdir(full_path):
            sub_directories.append(full_path)

            # Recursive call to search through all subdirectories.
            get_sub_directory_paths(full_path, sub_directories)

return sub_directories