在numpy数组中找到第n个最小的元素[duplicate]

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python+numpy: efficient way to take the min/max n values and indices from a matrix 2 answers
python+numpy:从矩阵2中取最小/最大n值和指标的有效方法

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

我需要找到1D numpi .array中最小的第n个元素。

For example:

例如:

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

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

我想要得到第5个最小的元素，所以我想要的输出是40。

My current solution is this:

我现在的解决方案是:

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

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?

然而，找到5个最小的元素，然后取最大的元素，对我来说似乎有点笨拙。有更好的方法吗?我是否遗漏了一个能够实现我目标的功能?

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

对于这个问题，也有类似的题目，但我没有看到任何回答我问题的东西。

Edit:

编辑:

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

我本来应该提到它，但是性能对我来说很重要;因此，heapq解虽然不错，但对我不起作用。

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)

Result:

结果:

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

3 个解决方案

#1

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

除非我漏掉了什么，否则你想做的是:

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

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.

np。分区(a, k)将k最小的元素a在[k]中，较小的值在a[:k]中，更大的值在a[k+1:]中。唯一需要注意的是，由于0索引，第五个元素位于索引4。

#2

You can use heapq.nsmallest:

您可以使用heapq.nsmallest:

>>> import numpy as np
>>> import heapq
>>> 
>>> a = np.array([90,10,30,40,80,70,20,50,60,0])
>>> heapq.nsmallest(5, a)[-1]
40

#3

you don't need call numpy.max():

您不需要叫numpy.max():

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

#1