如何将对列表转换为嵌套列表？

Question

给出如下变量：

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

我如何将其转换为嵌套的对列表？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")

给出：

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

然后，我不知道该怎么做，或者这是否是正确的步骤。

Answer 1

使用list comprehension分割每一行;我使用str.splitlines() method生成线条（因为它涵盖了更多的边角情况），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']]