如何制作每个元素（列表）的二维列表是由两个其他列表的联合创建的？答案

【问题标题】：How to make a 2D list with every element(list) is created by union of two other lists?如何制作每个元素（列表）的二维列表是由两个其他列表的联合创建的？
【发布时间】：2018-12-09 10:58:28
【问题描述】：

我有两个列表：

D1=[["a "," "," "," "," "," "],["b "," ","o"," "," "," "],["c ","x"," "," "," "," "],["d "," "," "," "," "," "],["e "," "," "," "," "," "]]

D2=[["a "," ","o"," ","x"," "],
["b "," "," "," "," "," "],["c "," "," "," "," "," "],["d "," "," "," "," "," "],["e "," "," "," "," "," "]]

D=[]

我想创建一个列表D 所以，D[i]=D1[i] + D2[i]，例如第一个元素（列表）看起来像这样：

D=[["a "," "," "," "," "," ","a "," ","o"," ","x"," "],...]

请帮助我，我是 python 新手

【问题讨论】：

D= [a+b for a,b in zip(D1, D2)]
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标签： python python-3.x list nested-lists

【解决方案1】：

试试这个：

D = [i+j for i,j in zip(D1,D2)]

如果长度不同，它将削减其余部分并前进到D1、D2 的最小值。如果你想要相反，使用zip_longest，像这样：

from itertools import zip_longest

D = [i+j for i,j in zip_longest(D1,D2)]

但如果D1 和D2 的长度相同，两者都可以工作。

【讨论】：

【解决方案2】：

根据您的意愿 (D[i] = D1[i] + D2[i])，最简单的方法是使用理解列表。总结len(D1) == len(D2)，：

 D = [ D1[i] + D2[i] for i in range(len(D1)) ]

会做的。

【讨论】：

【解决方案3】：

如果您不想更改 D1，请先将 D1 复制到 D。然后在python中使用extend方法。它会将list2的所有元素添加到list1。

这里是一个简单的代码：虽然这段代码的时间复杂度是O(n^2)，但是可以改进。

D1=[["a "," "," "," "," "," "],["b "," ","o"," "," "," "],["c ","x"," "," "," "," "],["d "," "," "," "," "," "],["e "," "," "," "," "," "]]

D2=[["a "," ","o"," ","x"," "],
["b "," "," "," "," "," "],["c "," "," "," "," "," "],["d "," "," "," "," "," "],["e "," "," "," "," "," "]]

D = D1 [:]
for i in range (len (D)):
    D[i].extend (D2 [i])
print D

【讨论】：

@B.M. extend () 方法需要 O (n)。因此总共是 O (n^2)。欲了解更多信息，请参阅此处link
我的理解是复杂度是 O[ len(D1)*len(D1[0])]。这里没有二次复杂度。