检查numpy数组是否是另一个数组的子集

本文介绍了检查numpy数组是否是另一个数组的子集的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

类似的问题已经在SO上提出，但是它们有更具体的限制，其答案不适用于我的问题.

Similar questions have already been asked on SO, but they have more specific constraints and their answers don't apply to my question.

一般而言，确定任意numpy数组是否为另一个数组的子集的最有效方法是什么?更具体地说，我有一个大约20000x3的数组，我需要知道完全包含在集合中的1x3元素的索引.更普遍的是，是否有更Python化的方式来编写以下内容:

Generally speaking, what is the most pythonic way to determine if an arbitrary numpy array is a subset of another array? More specifically, I have a roughly 20000x3 array and I need to know the indices of the 1x3 elements that are entirely contained within a set. More generally, is there a more pythonic way of writing the following:

master=[12,155,179,234,670,981,1054,1209,1526,1667,1853] #some indices of interest triangles=np.random.randint(2000,size=(20000,3)) #some data for i,x in enumerate(triangles): if x[0] in master and x[1] in master and x[2] in master: print i

对于我的用例，我可以安全地假设len(master)<< 20000.(因此，可以安全地假定master是有序的，因为这很便宜).

For my use case, I can safely assume that len(master) << 20000. (Consequently, it is also safe to assume that master is sorted because this is cheap).

推荐答案

您可以通过迭代列表推导中的数组来轻松实现此目的.一个玩具示例如下:

You can do this easily via iterating over an array in list comprehension. A toy example is as follows:

import numpy as np x = np.arange(30).reshape(10,3) searchKey = [4,5,8] x[[0,3,7],:] = searchKey x

给予

array([[ 4, 5, 8], [ 3, 4, 5], [ 6, 7, 8], [ 4, 5, 8], [12, 13, 14], [15, 16, 17], [18, 19, 20], [ 4, 5, 8], [24, 25, 26], [27, 28, 29]])

现在遍历元素:

ismember = [row==searchKey for row in x.tolist()]

结果是

[True, False, False, True, False, False, False, True, False, False]

您可以将其修改为问题中的子集:

You can modify it for being a subset as in your question:

searchKey = [2,4,10,5,8,9] # Add more elements for testing setSearchKey = set(searchKey) ismember = [setSearchKey.issuperset(row) for row in x.tolist()]

如果需要索引，请使用

np.where(ismember)[0]

它给出了

array([0, 3, 7])