Pythonic通告列表 [英] Pythonic Circular List

问题描述

说我有一个清单，

l = [1, 2, 3, 4, 5, 6, 7, 8]

我想获取任意元素的索引及其相邻元素的值.例如，

I want to grab the index of an arbitrary element and the values of its neighbors. For example,

i = l.index(n)
j = l[i-1]
k = l[i+1]

但是，对于i == len(l) - 1的边缘情况，这将失败.所以我想我会把它包起来，

However, for the edge case when i == len(l) - 1 this fails. So I thought I'd just wrap it around,

if i == len(l) - 1:
    k = l[0]
else:
    k = l[i+1]

是否有一种Python方式来做到这一点?

Is there a pythonic way to do this?

推荐答案

您可以使用模运算符！

i = len(l) - 1
jIndex = (i - 1) % len(l)
kIndex = (i + 1) % len(l)

j = l[jIndex]
k = l[kIndex]

或者，不那么冗长:

k = l[(i + 1) % len(l)]

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