这篇文章给大家分享的是有关怎么获取numpy array前N个最大值的内容。小编觉得挺实用的,因此分享给大家做个参考,一起跟随小编过来看看吧。
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_likeArray to sort. axis : int or None, optionalAxis along which to sort. The default is -1 (the last axis). If None, the flattened array is used. kind : {‘quicksort', ‘mergesort', ‘heapsort', ‘stable'}, optionalSorting algorithm. order : str or list of str, optionalWhen 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, intArray 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 nptop_k=3arr = 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 npa = 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]
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