一种访问列表中的后两个元素的优雅方式? [英] Elegant way to access last two and next two elements in a list?

问题描述

我需要访问前两个元素和后两个元素(如果存在).例如 [...，a，b，c，d，e，..] ->对于 c 我想要 abcde .

I need to access the previous two and next two elements(if exist). e.g. [..., a, b, c, d, e, ..] -> For c I want abcde.

这对于列表中间的元素很容易，但是在开始和结束时将导致索引超出范围错误.如果列表较小，它将变得更加复杂.

This is easy for elements in the middle of a list, but in the beginning and end it will cause Index out of range errors. And if the list is smaller it becomes even more complicated.

我可以用一堆 if - elif - elif - else 语句来做到这一点案件，但有没有简洁而优雅的方法来做到这一点?

I can do it with a bunch of if-elif-elif--else statements for all the cases, but is there any concise and elegant way to accomplish this?

我正在使用python 2.7

I'm using python 2.7

推荐答案

只使用切片(但要注意负索引):

Just use slicing (but be careful about negative indices):

inputlist[max(0, index - 2):index + 3]

将为您提供一个围绕着 index 最多5个元素的窗口；前2个，后2个.

will get you a window of up to 5 elements around index; 2 before and 2 after.

max()调用可确保起始索引的上限为 0 或更高.

The max() call ensures that the start index is capped to 0 or up.

这将与您用来生成索引的任何方法一起使用.假设您正在使用循环以及 enumerate()来跟踪索引:

This'll work with whatever method you used to produce your index. Say you are using a loop, together with enumerate() to keep track of the index:

def window(inputlist, index):
    return inputlist[max(0, index - 2):index + 3]

for index, elem in enumerate(inputlist):
    print window(inputlist, index)

或通过 list.index()找到了元素:

print window(inputlist, inputlist.index('ham'))

演示:

>>> def window(inputlist, index):
...     return inputlist[max(0, index - 2):index + 3]
... 
>>> window(['foo', 'bar', 'baz', 'spam', 'ham', 'eggs'], 1)
['foo', 'bar', 'baz', 'spam']
>>> window(['foo', 'bar', 'baz', 'spam', 'ham', 'eggs'], 2)
['foo', 'bar', 'baz', 'spam', 'ham']
>>> window(['foo', 'bar', 'baz', 'spam', 'ham', 'eggs'], 4)
['baz', 'spam', 'ham', 'eggs']
>>> window(['foo', 'bar', 'baz', 'spam', 'ham', 'eggs'], 5)
['spam', 'ham', 'eggs']

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