有没有办法在Python中返回当前目录中所有子目录的列表?

我知道您可以对文件执行此操作,但我需要获得目录列表。


当前回答

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

其他回答

在ipython中复制粘贴友好:

import os
d='.'
folders = list(filter(lambda x: os.path.isdir(os.path.join(d, x)), os.listdir(d)))

从打印(文件夹)输出:

['folderA', 'folderB']

这应该可以工作,因为它还创建了一个目录树;

import os
import pathlib

def tree(directory):
    print(f'+ {directory}')
    print("There are " + str(len(os.listdir(os.getcwd()))) + \
    " folders in this directory;")
    for path in sorted(directory.glob('*')):
        depth = len(path.relative_to(directory).parts)
        spacer = '    ' * depth
        print(f'{spacer}+ {path.name}')

这应该列出使用pathlib库的文件夹中的所有目录。path.relative_to(目录)。Parts获取相对于当前工作目录的元素。

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))])

我就是这么做的。

    import os
    for x in os.listdir(os.getcwd()):
        if os.path.isdir(x):
            print(x)

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