在numpy数组中查找第n个最小元素 [英] Find nth smallest element in numpy array

问题描述

我只需要找到1D numpy.array中最小的第n个元素.

I need to find just the the smallest nth element in a 1D numpy.array.

例如:

a = np.array([90,10,30,40,80,70,20,50,60,0])

我想获得第五个最小的元素，所以我想要的输出是40.

I want to get 5th smallest element, so my desired output is 40.

我当前的解决方案是:

result = np.max(np.partition(a, 5)[:5])

但是，找到5个最小的元素，然后选择最大的5个元素，对我来说似乎有点笨拙.有更好的方法吗?我是否缺少一个可以实现目标的功能?

However, finding 5 smallest elements and then taking the largest one them seems little clumsy to me. Is there a better way to do it? Am I missing a single function that would achieve my goal?

有些问题的标题与此相似，但是我没有看到任何答案.

There are questions with similar titles to this one, but I did not see anything that answered my question.

我本来应该提到它，但是性能对我来说非常重要；因此，heapq解决方案虽然不错，但对我来说不起作用.

I should've mentioned it originally, but performance is very important for me; therefore, heapq solution though nice would not work for me.

import numpy as np
import heapq

def find_nth_smallest_old_way(a, n):
    return np.max(np.partition(a, n)[:n])

# Solution suggested by Jaime and HYRY    
def find_nth_smallest_proper_way(a, n):
    return np.partition(a, n-1)[n-1]

def find_nth_smallest_heapq(a, n):
    return heapq.nsmallest(n, a)[-1]
#    
n_iterations = 10000

a = np.arange(1000)
np.random.shuffle(a)

t1 = timeit('find_nth_smallest_old_way(a, 100)', 'from __main__ import find_nth_smallest_old_way, a', number = n_iterations)
print 'time taken using partition old_way: {}'.format(t1)    
t2 = timeit('find_nth_smallest_proper_way(a, 100)', 'from __main__ import find_nth_smallest_proper_way, a', number = n_iterations)
print 'time taken using partition proper way: {}'.format(t2) 
t3 = timeit('find_nth_smallest_heapq(a, 100)', 'from __main__ import find_nth_smallest_heapq, a', number = n_iterations)  
print 'time taken using heapq : {}'.format(t3)

结果:

time taken using partition old_way: 0.255564928055
time taken using partition proper way: 0.129678010941
time taken using heapq : 7.81094002724

推荐答案

除非我丢失了某些内容，否则您要做的是:

Unless I am missing something, what you want to do is:

>>> a = np.array([90,10,30,40,80,70,20,50,60,0])
>>> np.partition(a, 4)[4]
40

np.partition(a, k)将a的k最小元素放置在a[k]，a[:k]中的值较小，而a[k+1:]中的值较大.唯一需要注意的是，由于索引为0，因此第五个元素位于索引4.

np.partition(a, k) will place the k-th smallest element of a at a[k], smaller values in a[:k] and larger values in a[k+1:]. The only thing to be aware of is that, because of the 0 indexing, the fifth element is at index 4.

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