将成对的列表/元组拆成两个列表/元组 [英] Unpacking a list / tuple of pairs into two lists / tuples

问题描述

可能重复:
Python中的Transpose/Unzip函数

Possible Duplicate:
A Transpose/Unzip Function in Python

我有一个看起来像这样的列表:

I have a list that looks like this:

list = (('1','a'),('2','b'),('3','c'),('4','d'))

我想将列表分为2个列表.

I want to separate the list in 2 lists.

list1 = ('1','2','3','4')
list2 = ('a','b','c','d')

例如，我可以这样做:

list1 = []
list2 = []
for i in list:
   list1.append(i[0])
   list2.append(i[1])

但是我想知道是否有更优雅的解决方案.

But I want to know if there is a more elegant solution.

推荐答案

>>> source_list = ('1','a'),('2','b'),('3','c'),('4','d')
>>> list1, list2 = zip(*source_list)
>>> list1
('1', '2', '3', '4')
>>> list2
('a', 'b', 'c', 'd')

请注意，zip(*iterable)是其自身的反函数:

Note that zip(*iterable) is its own inverse:

>>> list(source_list) == zip(*zip(*source_list))
True

解包成两个列表时，它变成:

When unpacking into two lists, this becomes:

>>> list1, list2 = zip(*source_list)
>>> list(source_list) == zip(list1, list2)
True

rocksportrocker建议添加.

Addition suggested by rocksportrocker.

这篇关于将成对的列表/元组拆成两个列表/元组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！