计算矩阵A与矩阵B的欧式距离

1. 从向量的欧式距离谈起

向量x为1x3的行向量，向量y为1x3的行向量,求向量x与向量y的欧式距离
$x=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\end{array}\right)$
$y=\left(\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\end{array}\right)$

$dis{t}_{x,y}=\sqrt{(a_{11}-b_{11}{)}^2+(a_{12}-b_{12}{)}^2+(a_{13}-b_{13}{)}^2}$

变形为：
$dis{t}_{x,y}=\sqrt{a_{11}^2+a_{12}^2+a_{13}^2+b_{11}^2+b_{12}^2+b_{13}^2-2a_{11}{b}_{11}-2a_{12}{b}_{12}-2a_{13}{b}_{13}}$

从变形后的公式我们可以看出相当于$(a-b{)}^2=a^2+b^2-2ab$

我们可以变形为下列计算(两个向量的L2范数)：
$dis{t}_{x,y}={\Vert x-y\Vert }_2$
第一种方式

import numpy as np

x = np.mat([1.0, 2.0, 3.0])
y = np.mat([4.0, 5.0, 6.0])

dist = np.sqrt(np.sum(np.power(x - y, 2)))
print(dist)
# 5.196152422706632
dist = np.sqrt(np.power(x, 2).sum() + np.power(y, 2).sum() - 2 * np.dot(x, y.T)).sum()
print(dist)
# 5.196152422706632

2. 从向量的欧式距离扩展到矩阵的欧式距离

向量A为2x3的行向量，向量B为3x3的行向量,求向量A与向量B的欧式距离
$A=\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right)$

$B=\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right)=\left(\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right)$
矩阵A与矩阵B的欧式距离，会形成矩阵C，其维度为2x3
$C=\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)=\left(\begin{array}{ccc}{c}_{11}& c_{12}& c_{13}\\ c_{21}& c_{22}& c_{23}\end{array}\right)$

$dis{t}_{c_{11}}=\sqrt{(a_{11}-b_{11}{)}^2+(a_{12}-b_{12}{)}^2+(a_{13}-b_{13}{)}^2}={\Vert a_1-b_1\Vert }_2$
$dis{t}_{c_{12}}=\sqrt{(a_{11}-b_{21}{)}^2+(a_{12}-b_{22}{)}^2+(a_{13}-b_{23}{)}^2}={\Vert a_1-b_2\Vert }_2$
$dis{t}_{c_{13}}=\sqrt{(a_{11}-b_{31}{)}^2+(a_{12}-b_{32}{)}^2+(a_{13}-b_{33}{)}^2}={\Vert a_1-b_3\Vert }_2$

$dis{t}_{c_{21}}=\sqrt{(a_{21}-b_{11}{)}^2+(a_{22}-b_{12}{)}^2+(a_{23}-b_{13}{)}^2}={\Vert a_2-b_1\Vert }_2$
$dis{t}_{c_{22}}=\sqrt{(a_{21}-b_{21}{)}^2+(a_{22}-b_{22}{)}^2+(a_{23}-b_{23}{)}^2}={\Vert a_2-b_2\Vert }_2$
$dis{t}_{c_{23}}=\sqrt{(a_{21}-b_{31}{)}^2+(a_{22}-b_{32}{)}^2+(a_{23}-b_{33}{)}^2}={\Vert a_2-b_3\Vert }_2$

可以看出A中的每个行向量使用了3次(B的行数)和B中的每一行列向量使用了2次(A的行数)：
参照向量的欧式距离计算，转化成$(a-b{)}^2=a^2+b^2-2ab$
$C=sum(A^2,axis=1)\ast ones((1,3))+ones((2,1))\ast sum(B^2,axis=1{)}^T-2A{B}^T$

# 矩阵欧式距离计算
# 矩阵计算方法
A = np.mat([[1.0, 2.0, 3.0], [2.0, 3.0, 4.0]])
B = np.mat([[3.0, 4.0, 5.0], [4.0, 5.0, 6.0], [5.0, 6.0, 7.0]])
C = np.sum(np.power(A, 2), 1) * np.ones((1, 3)) + np.ones((2, 1)) * np.sum(np.power(B, 2), 1).T        - 2 * A * B.T
print(C)
# [[12. 27. 48.] [ 3. 12. 27.]]
 # 逐行向量计算方法（用于验证）
C = np.mat(np.zeros((2, 3), dtype=float))
for i in range(2):
    for j in range(3):
        C[i, j] = np.sum(np.power(A[i] - B[j], 2))
print(C)

#[[12. 27. 48.] [ 3. 12. 27.]]