如何获取numpy array前N个最大值

主要应用了argsort()函数，函数原型：

numpy.argsort(a, axis=-1, kind='quicksort', order=none)
'''
returns the indices that would sort an array.
perform an indirect sort along the given axis using the algorithm specified by the kind keyword. it returns an array of indices of the same shape as a that index data along the given axis in sorted order.
'''
parameters: 
a : array_like
array to sort.
 
axis : int or none, optional
axis along which to sort. the default is -1 (the last axis). if none, the flattened array is used.
 
kind : {‘quicksort', ‘mergesort', ‘heapsort', ‘stable'}, optional
sorting algorithm.
 
order : str or list of str, optional
when a is an array with fields defined, this argument specifies which fields to compare first, second, etc. a single field can be specified as a string, and not all fields need be specified, but unspecified fields will still be used, in the order in which they come up in the dtype, to break ties.
 
returns: 
index_array : ndarray, int
array of indices that sort a along the specified axis. if a is one-dimensional, a[index_array] yields a sorted a. more generally, np.take_along_axis(a, index_array, axis=a) always yields the sorted a, irrespective of dimensionality.
 

示例：

import numpy as np
top_k=3
arr = np.array([1, 3, 2, 4, 5])
top_k_idx=arr.argsort()[::-1][0:top_k]
print(top_k_idx)
#[4 3 1]

补充：python topn / topk 取 最大的n个数 或 最小的n个数

import numpy as np
a = np.array([1,4,3,5,2])
b = np.argsort(a)
print(b)

print结果[0 4 2 1 3]

说明a[0]最小，a[3]最大

a[0]<a[4]<a[2]<a[1]<a[3]

补充：利用python获取数组或列表中最大的n个数及其索引

看代码吧~

import heapq
 
a=[43,5,65,4,5,8,87]
re1 = heapq.nlargest(3, a) #求最大的三个元素，并排序
re2 = map(a.index, heapq.nlargest(3, a)) #求最大的三个索引    nsmallest与nlargest相反，求最小
print(re1)
print(list(re2)) #因为re2由map()生成的不是list，直接print不出来，添加list()就行了

结果：

re1：[87, 65, 43]

re2：[6, 2, 0]

以上为个人经验，希望能给大家一个参考，也希望大家多多支持。