如何根据列表中的索引创建两个新列表？

Question

所以我觉得这是很容易解释，如果我告诉一个例子：

things = ["black", 7, "red', 10, "white", 15] 

两个基于索引是否为偶数或奇数的新名单。

color = ["black", "red","white"] 
size = [7,10,15] 

Answer 1

In [4]: things = ["black", 7, "red", 10, "white", 15] 

In [5]: color = things[::2] 

In [6]: color 
Out[6]: ['black', 'red', 'white'] 

In [7]: size = things[1::2] 

In [8]: size 
Out[8]: [7, 10, 15] 

Answer 2

things = ["black", 7, "red", 10, "white", 15] 
two_lists = zip(*[(x,things[things.index(x)+1]) for x in things[::2]]) 

>>> two_lists[0] 
('black', 'red', 'white') 
>>> two_lists[1] 
(7, 10, 15) 

列表MANIP部分[(x,things[things.index(x)+1]) for x in things[::2]]方式隔开分成3个列表对（就像它准备一本字典什么的...所以克雷Cray公司）

的zip(*部分，圈整个阵列在它旁边

如果你只是做了[(x,things[things.index(x)+1]) for x in things[::2]] withotu zip部分，那么你可以调用dict()就可以了，并返回一个字典，whic h可能更有用

>>> a = [(x,things[things.index(x)+1]) for x in things[::2]] 
>>> a 
[('black', 7), ('red', 10), ('white', 15)] 
>>> cool_dict = dict(a) 
>>> cool_dict['black'] 
7