列表中最小的n个数字 [英] Smallest n numbers from a list

问题描述

如果我有一个列表 list1 = [1,15,9,3,6,21,10,11] 如何从中获得最小的2个整数?

If I have a list list1=[1,15,9,3,6,21,10,11] how do I obtain the smallest 2 integers from that?

min()给我一个数字，但是2呢?

min() gives me one number, but what about 2?

推荐答案

您可以导入 heapq

import heapq

list1=[1,15,9,3,6,21,10,11]

print(heapq.nsmallest(2,list1))

限制是，如果您有一个重复的值，例如 l = [1,3,5,1] ，则两个最小值将是 [1,1] .

The limitation with that is if you have a repeated value let's say l=[1,3,5,1], the two smallest values will be [1,1].

In [2]:
    list1=[1,15,9,3,6,21,10,11]

In [3]:

    %timeit sorted(list1)[:2]
    1000000 loops, best of 3: 1.58 µs per loop
In [5]:

    import heapq
    %timeit heapq.nsmallest(2,list1)
    100000 loops, best of 3: 4.18 µs per loop

从这两者看来，对较小的集合排序列表似乎更快.

From the two, it seems sorting the list is faster for smaller sets.

In [14]:

    import random
    list1=[[random.random() for i in range(100)] for j in range(100)]
In [15]:

    %timeit sorted(list1)[:2]
    10000 loops, best of 3: 55.6 µs per loop
In [16]:

    import heapq
    %timeit heapq.nsmallest(2,list1)
    10000 loops, best of 3: 27.7 µs per loop

感谢Padraic Cunningham， heapq 在设置较大的集合时更快

Thanks to Padraic Cunningham, heapq is faster with larger sets

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