如何使每个元素（列表）的二维列表是由另外两个列表的并集创建的？

Question

我有两个列表：

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

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

D=[]

我想制作一个列表 D 所以，D[i]=D1[i] + D2[i]，例如第一个元素（列表）如下所示：

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

请帮助我，我是新来的 python

Answer 1

如果不想修改D1，先把D1复制到D。然后在python中使用extend方法。它将 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

Answer 2

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。如果你想要相反的，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。

Answer 3

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。假设 len(D1) == len(D2)，：

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

会完成任务的。

如何使每个元素（列表）的二维列表是由另外两个列表的并集创建的？

How to make a 2D list with every element(list) is created by union of two other lists?

python

list

nested-lists

python-3.x