查找列表的x个最小值的索引

Question

I have a list of length n.

I want to find the indices that hold the 5 minimum values of this list.

I know how to find the index holding the minimum value using operator

min_index,min_value = min(enumerate(list), key=operator.itemgetter(1))

Can this code be altered to get a list of the 5 indices I am after?

Answer 1

尽管这需要对整个列表进行排序，但是您可以从sorted列表中得到一部分：

data = sorted(enumerate(list), key=operator.itemgetter(1))[:5]

Answer 2

what about something like this?

    map(lambda x: [a.index(x),x],sorted(list)[:5])

that will return a list of lists where list[x][0] = the index and list[x][1] = the value

EDIT:

This assumes the list doesn't have repeated minimum values. As adhg McDonald-Jensen pointed out, it this will only return the first instance of the give value.

Answer 3

如果使用的软件包heapq ，它可以这样做nsamllest ：

heapq.nsmallest(5, enumerate(list), key=operator.itemgetter(1))