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
。