检查numpy数组中的每个元素是否在另一个数组中 [英] Check if each element in a numpy array is in another array

问题描述

这个问题似乎很简单，但我无法完全找到一个不错的解决方案.我有两个numpy数组(A和B)，我想获取A的索引(其中A的元素在B中)，还要获取A的索引，其中A的元素不在B中.

This problem seems easy but I cannot quite get a nice-looking solution. I have two numpy arrays (A and B), and I want to get the indices of A where the elements of A are in B and also get the indices of A where the elements are not in B.

所以，如果

A = np.array([1,2,3,4,5,6,7])
B = np.array([2,4,6])

当前我正在使用

C = np.searchsorted(A,B)

利用了A是按顺序排列的事实，并为我[1, 3, 5]提供了A中元素的索引.太好了，但是如何获取D = [0,2,4,6](B中不在的A元素的索引)?

which takes advantage of the fact that A is in order, and gives me [1, 3, 5], the indices of the elements that are in A. This is great, but how do I get D = [0,2,4,6], the indices of elements of A that are not in B?

推荐答案

import numpy as np

A = np.array([1,2,3,4,5,6,7])
B = np.array([2,4,6])
C = np.searchsorted(A, B)

D = np.delete(np.arange(np.alen(A)), C)

D
#array([0, 2, 4, 6])

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