获取列表的前N个值的索引

本文介绍了获取列表的前N个值的索引的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个列表，说a = [5,3,1,4,10].我需要获取列表的前两个值的索引，即对于5和10我将得到[0, 4]. Python在这种情况下是否提供一线服务?

I have a list say a = [5,3,1,4,10]. I need to get indices of the top two values of the list, that is for 5 and 10 I would get [0, 4]. Is there a one-liner that Python offers for such a case?

推荐答案

sorted(range(len(a)), key=lambda i: a[i])[-2:]

或

sorted(range(len(a)), key=lambda i: a[i], reverse=True)[:2]

或

zip(*sorted(enumerate(a), key=operator.itemgetter(1)))[0][-2:]

或(对于长列表)，请考虑使用 heapq.nlargest

or (for long lists), consider using heapq.nlargest

zip(*heapq.nlargest(2, enumerate(a), key=operator.itemgetter(1)))[0]