如何在python中迭代列表中的两个连续元素？

Question

我正在尝试迭代列表的两个连续元素。

 mentionedlist=[[1,2,3,4][1,2,3,4][2,3,4,5][3,4,5,5][1,2,3,4][1,2,3,4][]] 

现在我想在列表的天气第一个元素比较在mentionedlist一张表的mentionedlist和第一个元素是一样的，我也希望做整个列表中的这些比较。

[1,2,3,4][1,2,3,4] is example of answer that i am expecting. 

Answer 1

你可以做这样的事情，让连续元素：

mentionedlist=[[1,2,3,4],[1,2,3,4],[2,3,4,5],[3,4,5,5],[1,2,3,4],[1,2,3,4],[]] 
for l1, l2 in zip(mentionedlist, mentionedlist[1:]): 
    print l1, l2 

输出

[1, 2, 3, 4] [1, 2, 3, 4] 
[1, 2, 3, 4] [2, 3, 4, 5] 
[2, 3, 4, 5] [3, 4, 5, 5] 
[3, 4, 5, 5] [1, 2, 3, 4] 
[1, 2, 3, 4] [1, 2, 3, 4] 
[1, 2, 3, 4] [] 

要做到两两比较：

for l1, l2 in zip(mentionedlist, mentionedlist[1:]): 
    if len(l1) == len(l2) and sum(x != y for x,y in zip(l1, l2)) == 0: 
     print l1, l2 

它给你：

[1, 2, 3, 4] [1, 2, 3, 4] 
[1, 2, 3, 4] [1, 2, 3, 4] 

Answer 2

注意：你也需要把值之间的逗号在列表中，即List=[[1,..],[2,..]]

from itertools import islice 
mentionedList=[[1,2,3,4],[1,2,3,4],[2,3,4,5],[3,4,5,5],[1,2,3,4],[1,2,3,4],[]] 
for i,v in enumerate(islice(mentionedList,0,len(mentionedList)-1)): 
    print (v,mentionedList[i+1]) 

为您提供：

([1, 2, 3, 4], [1, 2, 3, 4]) 
([1, 2, 3, 4], [2, 3, 4, 5]) 
([2, 3, 4, 5], [3, 4, 5, 5]) 
([3, 4, 5, 5], [1, 2, 3, 4]) 
([1, 2, 3, 4], [1, 2, 3, 4]) 
([1, 2, 3, 4], []) 

使用此方法，您不需要复制列表。