如何获取NumPy数组中N个最大值的索引？

本文翻译自：How do I get indices of N maximum values in a NumPy array?

NumPy proposes a way to get the index of the maximum value of an array vianp.argmax .NumPy提出了一种通过np.argmax数组最大值的索引的np.argmax 。

I would like a similar thing, but returning the indexes of theN maximum values.我想要类似的事情，但是返回N最大值的索引。

For instance, if I have an array,[1, 3, 2, 4, 5] ,function(array, n=3) would return the indices[4, 3, 1] which correspond to the elements[5, 4, 3] .例如，如果我有一个数组[1, 3, 2, 4, 5] ，则function(array, n=3)将返回对应于元素[5, 4, 3]的索引[4, 3, 1] 4，3，1[4, 3, 1][5, 4, 3] 。

#1楼

参考：https://stackoom.com/question/Szlx/如何获取NumPy数组中N个最大值的索引

#2楼

Use:采用：

>>> import heapq
>>> import numpy
>>> a = numpy.array([1, 3, 2, 4, 5])
>>> heapq.nlargest(3, range(len(a)), a.take)
[4, 3, 1]

For regular Python lists:对于常规的Python列表：

>>> a = [1, 3, 2, 4, 5]
>>> heapq.nlargest(3, range(len(a)), a.__getitem__)
[4, 3, 1]

If you use Python 2, usexrange instead ofrange .如果使用Python 2，请使用xrange而不是range 。

Source:heapq — Heap queue algorithm来源：heapq —堆队列算法

#3楼

Newer NumPy versions (1.8 and up) have a function calledargpartition for this.较新的NumPy版本（1.8及更高版本）具有一个称为argpartition的函数。To get the indices of the four largest elements, do要获取四个最大元素的索引，请执行

>>> a = np.array([9, 4, 4, 3, 3, 9, 0, 4, 6, 0])
>>> a
array([9, 4, 4, 3, 3, 9, 0, 4, 6, 0])
>>> ind = np.argpartition(a, -4)[-4:]
>>> ind
array([1, 5, 8, 0])
>>> a[ind]
array([4, 9, 6, 9])

Unlikeargsort , this function runs in linear time in the worst case, but the returned indices are not sorted, as can be seen from the result of evaluatinga[ind] .与argsort不同，此函数在最坏的情况下以线性时间运行，但是返回的索引未排序，从评估a[ind]的结果可以看出。If you need that too, sort them afterwards:如果您也需要它，请对它们进行排序：

>>> ind[np.argsort(a[ind])]
array([1, 8, 5, 0])

To get the top-k elements in sorted order in this way takes O(n +k logk ) time.要以这种方式获得排序前k个元素，需要O（n +k logk ）时间。

#4楼

Simpler yet:更简单了：

idx = (-arr).argsort()[:n]

wheren is the number of maximum values.其中，n是最大值的数量。

#5楼

Use:采用：

from operator import itemgetter
from heapq import nlargest
result = nlargest(N, enumerate(your_list), itemgetter(1))

Now theresult list would containN tuples (index ,value ) wherevalue is maximized.现在result列表将包含N个元组（index ，value ），其中value最大化。

#6楼

If you don't care about theorder of the K-th largest elements you can useargpartition , which should perform better than a full sort throughargsort .如果您不关心第K个最大元素的顺序 ，则可以使用argpartition ，它的性能要比通过argsort进行完整排序argsort 。

K = 4 # We want the indices of the four largest values
a = np.array([0, 8, 0, 4, 5, 8, 8, 0, 4, 2])
np.argpartition(a,-K)[-K:]
array([4, 1, 5, 6])

Credits go tothis question .学分到这个问题 。

I ran a few tests and it looks likeargpartition outperformsargsort as the size of the array and the value of K increase.我进行了一些测试，随着数组的大小和K值的增加，argpartition表现优于argsort 。