Question

我想得到一个主numpy 2d数组A的交叉行的索引，另一个得到B.

A = array([[1,2],
           [1,3],
           [2,3],
           [2,4],
           [2,5],
           [3,4]
           [4,5]])

B = array([[1,2],
           [3,2],
           [2,4]])

result=[0, -2, 3]  
##Note that the intercept 3,2 must assign (-) because it is the opposite

这应该根据数组A的索引返回[0，-2,3]。

谢谢！

Answer 1

您可以参考代码。

import numpy as np
A = np.array([[1,2],
           [1,3],
           [2,3],
           [2,4],
           [2,5],
           [3,4],
           [4,5]])

B = np.array([[1,2],
           [3,2],
           [2,4]])

result=[]

for i in range(0, len(A)):
    for j in range(0, len(B)):
        if A[i].tolist() == B[j].tolist():
            result.append(i)
        if A[i].tolist()[::-1] == B[j].tolist():
            result.append(-i)
print(result)

输出结果为：

[0, -2, 3]

Answer 2

numpy_indexed包（免责声明：我是其作者）具有有效解决此类问题的功能。

import numpy_indexed as npi
A = np.sort(A, axis=1)
B = np.sort(B, axis=1)
result = npi.indices(A, B)
result *= (A[:, 0] == B[:, 0]) * 2 - 1

Numpy 2d阵列相交行的索引

2 个答案: