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
$x = \begin{pmatrix}a_{11} & a_{12} & a_{13}\\\end{pmatrix}$x=(a11​​a12​​a13​​)
$y = \begin{pmatrix}b_{11} & b_{12} & b_{13}\\\end{pmatrix}$y=(b11​​b12​​b13​​)

$dist_{x,y}=\sqrt{(a_{11} - b_{11})^2 + (a_{12} - b_{12})^2 + (a_{13} - b_{13})^2}$distx,y​=(a11​−b11​)2+(a12​−b12​)2+(a13​−b13​)2​

变形为：
$dist_{x,y}=\sqrt{a_{11}^2 + a_{12}^2 + a_{13}^2 + b_{11}^2 + b_{12}^2 + b_{13}^2 - 2a_{11}b_{11} - 2a_{12}b_{12} - 2a_{13}b_{13} }$distx,y​=a112​+a122​+a132​+b112​+b122​+b132​−2a11​b11​−2a12​b12​−2a13​b13​​

从变形后的公式我们可以看出相当于$(a-b)^2=a^2+b^2-2ab$(a−b)2=a2+b2−2ab

我们可以变形为下列计算(两个向量的L2范数)：
$dist_{x,y}=\left \| x-y \right \|_2$distx,y​=∥x−y∥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
$A = \begin{pmatrix} a_{1} \\ a_{2} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \end{pmatrix}$A=(a1​a2​​)=(a11​a21​​a12​a22​​a13​a23​​)

$B = \begin{pmatrix} b_{1} \\ b_{2}\\ b_{3} \end{pmatrix} = \begin{pmatrix} b_{11} & b_{12} & b_{13}\\ b_{21} & b_{22} & b_{23}\\ b_{31} & b_{32} & b_{33} \end{pmatrix}$B=⎝⎛​b1​b2​b3​​⎠⎞​=⎝⎛​b11​b21​b31​​b12​b22​b32​​b13​b23​b33​​⎠⎞​
矩阵A与矩阵B的欧式距离，会形成矩阵C，其维度为2x3
$C = \begin{pmatrix} c_{1} \\ c_{2} \end{pmatrix}= \begin{pmatrix} c_{11} & c_{12} & c_{13}\\ c_{21} & c_{22} & c_{23} \end{pmatrix}$C=(c1​c2​​)=(c11​c21​​c12​c22​​c13​c23​​)

$dist_{c_{11}}=\sqrt{(a_{11} - b_{11})^2 + (a_{12} - b_{12})^2 + (a_{13} - b_{13})^2} = \left \| a_1-b_1 \right \|_2$distc11​​=(a11​−b11​)2+(a12​−b12​)2+(a13​−b13​)2​=∥a1​−b1​∥2​
$dist_{c_{12}}=\sqrt{(a_{11} - b_{21})^2 + (a_{12} - b_{22})^2 + (a_{13} - b_{23})^2} = \left \| a_1-b_2 \right \|_2$distc12​​=(a11​−b21​)2+(a12​−b22​)2+(a13​−b23​)2​=∥a1​−b2​∥2​
$dist_{c_{13}}=\sqrt{(a_{11} - b_{31})^2 + (a_{12} - b_{32})^2 + (a_{13} - b_{33})^2} = \left \| a_1-b_3 \right \|_2$distc13​​=(a11​−b31​)2+(a12​−b32​)2+(a13​−b33​)2​=∥a1​−b3​∥2​

$dist_{c_{21}}=\sqrt{(a_{21} - b_{11})^2 + (a_{22} - b_{12})^2 + (a_{23} - b_{13})^2} = \left \| a_2-b_1 \right \|_2$distc21​​=(a21​−b11​)2+(a22​−b12​)2+(a23​−b13​)2​=∥a2​−b1​∥2​
$dist_{c_{22}}=\sqrt{(a_{21} - b_{21})^2 + (a_{22} - b_{22})^2 + (a_{23} - b_{23})^2}= \left \| a_2-b_2 \right \|_2$distc22​​=(a21​−b21​)2+(a22​−b22​)2+(a23​−b23​)2​=∥a2​−b2​∥2​
$dist_{c_{23}}=\sqrt{(a_{21} - b_{31})^2 + (a_{22} - b_{32})^2 + (a_{23} - b_{33})^2}= \left \| a_2-b_3 \right \|_2$distc23​​=(a21​−b31​)2+(a22​−b32​)2+(a23​−b33​)2​=∥a2​−b3​∥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$(a-b)^2=a^2+b^2-2ab$(a−b)2=a2+b2−2ab
$C=sum(A^2, axis=1) * ones((1, 3)) + ones((1, 3)) * sum(B^2, axis=1)^T - 2AB^T$C=sum(A2,axis=1)∗ones((1,3))+ones((1,3))∗sum(B2,axis=1)T−2ABT

# 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.]]