Python 如何以列表的形式获取代码的输出？

我只能打印子目录。如何将子目录的名称作为列表获取？

声明一个列表以收集

for

循环之外的名称<代码>将附加到循环内的列表中

import os

for dirname, dirnames, filenames in os.walk('.'):
    # print path to all subdirectories first.
    for subdirname in dirnames:
        print os.path.join(dirname, subdirname)

    # print path to all filenames.
    for filename in filenames:
        print os.path.join(dirname, filename)

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声明一个列表以收集

for

循环外部的名称<代码>将附加到循环内的列表中。

import os

subdirs = []

for dirname, dirnames, filenames in os.walk('.'):
    # collect path to all subdirectories first.
    for subdirname in dirnames:
        name = os.path.join(dirname, subdirname)
        subdirs.append(name)

    # print path to all filenames.
    for filename in filenames:
        print(os.path.join(dirname, filename))

print(subdirs)