numpy:在较大的数组B中搜索具有相同模式的数组A [英] Numpy: Search for an array A with same pattern in a larger array B

问题描述

我有两个一维numpy数组A(小)和B(大)

I have two 1D numpy array A(small) and B(large)

A=np.array([6,7,8,9,10])

B=np.array([0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,10])

我想检查数组A中的元素是否与数组B中检测到的顺序相同. 从我们检测到数组A起始处的位置获取数组B的索引值

I want to check if we have elements of the array A in the same order being detected in the array B. Get the index value of array B from where the we detect the starting of array A

Index Value returned = 6

我们有内置的numpy函数来执行这种操作吗?

Do we have any inbuilt numpy function to perform such an operation?

推荐答案

我找到了一个不错的解决方案.

I found a nice solution.

由@EdSmith在在一个Numpy数组中查找模式给出

Given by @EdSmith in Finding Patterns in a Numpy Array

简而言之，就是这个过程

In short this is the process

• 缩短要搜索的数组的长度.(我的示例A)
• 使用np.where和np.all检查要搜索的数组的整个长度(我的示例B)

这不是我的代码，而是可以在about链接中找到的代码，简单易行.我将对其稍作修改，以使其适合上面的示例，希望它对某人有帮助:)

This is not my code but the code that can be found in the about link, Simple and easy. I'll just alter it a bit to fit my example above Hope it helps someone :)

感谢@EdSmith

Thanks to @EdSmith

import numpy as np

A=np.array([6,7,8,9,10])

B=np.array([0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,10])

N = len(A)
possibles = np.where(B == A[0])[0]

solns = []
for p in possibles:
    check = B[p:p+N]
if np.all(check == A):
    solns.append(p)

print(solns)

输出

[6]

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