如何将成对列表转换为嵌套列表？

Question

给定一个像这样的变量：

stories = """adress:address
alot:a lot
athiest:atheist
noone:no one
beleive:believe
fivety:fifty
wierd:weird
writen:written"""

我将如何将其转换为嵌套的配对列表？ 即：

new_stories = [['adress', 'address'], ['alot', 'a lot'], ['athiest', 'atheist'], ['noone', 'no one']] # etc.

我目前正在这样做：

s = stories.split("\n")

这使：

['adress:address', 'alot:a lot', 'athiest:atheist', 'noone:no one']  # etc.

然后，我不确定该怎么做，或者这是否完全正确。

Answer 1

使用列表推导拆分每行； 我将使用str.splitlines()方法生成线（因为它涵盖了更多的str.split()情况），然后使用str.split()生成每线对：

[s.split(':') for s in stories.splitlines()]

演示：

>>> stories = """adress:address
... alot:a lot
... athiest:atheist
... noone:no one
... beleive:believe
... fivety:fifty
... wierd:weird
... writen:written"""
>>> [s.split(':') for s in stories.splitlines()]
[['adress', 'address'], ['alot', 'a lot'], ['athiest', 'atheist'], ['noone', 'no one'], ['beleive', 'believe'], ['fivety', 'fifty'], ['wierd', 'weird'], ['writen', 'written']]