实现两个点集的欧式距离和索引值寻找（含有两种解法，for循环和矩阵操作）

一，直接for循环

两个点集points1，points2，用dist来存储距离 

points1=np.array([[1,2],[3,4]])
points2 = np.array([[5, 6],[7,8]])
dist = np.zeros(shape=[points1.shape[0],points2.shape[0]])
for i in range(points1.shape[0]):
    for j in range(points2.shape[0]):
        print(points1[i, :] - points2[j, :])

遍历两个点集的索引相减值

加上这句话

print(np.square(points1[i, :] - points2[j, :]))

print(np.sum(np.square(points1[i, :] - points2[j, :])))

print(np.sqrt(np.sum(np.square(points1[i, :] - points2[j, :]))))

points1=np.array([[1,2],[3,4]])
points2 = np.array([[5, 6],[7,8]])
dist = np.zeros(shape=[points1.shape[0],points2.shape[0]])
for i in range(points1.shape[0]):
    for j in range(points2.shape[0]):
        # print(points1[i, :] - points2[j, :])
        # print('======================')
        # print(np.square(points1[i, :] - points2[j, :]))
        # print('============================')
        # print(np.sum(np.square(points1[i, :] - points2[j, :])))
        # print(np.sqrt(np.sum(np.square(points1[i, :] - points2[j, :]))))
        dist[i, j] = np.sqrt(np.sum(np.square(points1[i, :] - points2[j, :])))
print(dist)

ind = np.unravel_index(np.argmax(dist), dist.shape)
print(ind)

返回索引值

二，矩阵操作求欧式距离

假设有两个三维向量集，用矩阵表示:

要求A,B这两个矩阵中的元素两两之间的欧式距离。

先求出：

然后对和分别求其中每个向量的模平方，并扩展为2*3矩阵：

然后：

 

将上面这个矩阵一开方，就得到了A，B矩阵各个元素两两之间的欧式距离。

import numpy as np
A=np.array([[1,2],
            [3,4]])
print('A=',A)
B=np.array([[5,6],
            [7,8]])
print('B=',B)

BT=np.transpose(B)
print('BT=',BT)
A_BT=np.dot(A,BT)
print('A_BT=',A_BT)

Asq=A**2
Asq=np.tile(np.sum(Asq,axis=1,keepdims=True),(1,A_BT.shape[1]))
print('Asq=',Asq)

Bsq=BT**2
Bsq=np.tile(np.sum(Bsq,axis=0,keepdims=True),(A_BT.shape[0],1))
print('Bsq=',Bsq)

print(Asq+Bsq-2*A_BT)
ED=np.sqrt(Asq+Bsq-2*A_BT)
print('ED=',ED)
ind=np.unravel_index(np.argmax(ED),ED.shape)
print(ind)
print(ED[ind[0],ind[1])

 

来源：CSDN

作者：DL_fan

链接：https://blog.csdn.net/fanzonghao/article/details/83037747