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的新手

Answer 1

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。它将list2的所有元素添加到list1。

这是一个简单的代码：尽管此代码的时间复杂度为O（n ^ 2），但可以改进。

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

另外一点：请按照@Patrick的说明进行操作 Artner。否则，您的问题更有可能被否决，可能会导致您无法提出进一步的问题。

Answer 2

尝试一下：

RewriteEngine On
RewriteBase /

RewriteRule ^images/background\.png$ includes/adaptive-images.php

# redirects non-www to www
RewriteCond %{HTTP_HOST} ^my-first.domain\.com [NC]
RewriteRule (.*) http://www.my-first-domain.com/$1 [R=301,L]

# redirects /index.php to /
RewriteCond %{IS_SUBREQ} false
RewriteRule ^index\.php$ http://www.my-first-domain.com/ [R=301,L]

# redirects missing files and directories to home (instead of error document)
RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-d
RewriteRule ^(.*)$ http://www.my-first-domain.com/ [R=301,L]

如果长度不同，它将削减其余部分并前进到D = [i+j for i,j in zip(D1,D2)]，D1的最小值。如果您要相反，请使用D2，如下所示：

zip_longest

但是如果from itertools import zip_longest D = [i+j for i,j in zip_longest(D1,D2)]和D1具有相同的长度，则两者都可以工作。

Answer 3

直截了当（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)) ]

会做。

如何通过其他两个列表的组合创建每个元素（列表）的2D列表？

3 个答案: