python - 如何以列表形式获取代码的输出？

Question

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)

我只能打印子目录。如何以列表形式获取子目录的名称？

Answer 1

声明一个列表来收集 for 循环之外的名称。 append 到循环内的该列表。

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)