28、（向量）歐幾里得距離計算

基本思想:求一組樣本點到一組向量的最短距離，並且輸出其距離資訊

樣本向量點
$$\begin{gathered} 已知的樣本：A=\left\{\begin{array}{c}{x}_1{y}_1\\ x_2{y}_2\end{array}\right\} \\ 已知的向量：B=\left\{\begin{array}{c}{x}_3{y}_3\\ x_4{y}_4\end{array}\right\} \end{gathered}$$

先推導一下基本公式：
$$d_{13}=\sqrt{(x_1-x_3{)}^2+(y_1-y_3{)}^2}=\sqrt{x_1^2+x_3^2-2x_1{x}_2+y_1^2+y_3^2-2y_1{y}_2}$$

$$d_{14}=\sqrt{(x_1-x_4{)}^2+(y_1-y_4{)}^2}=\sqrt{x_1^2+x_4^2-2x_1{x}_4+y_1^2+y_4^2-2y_1{y}_4}$$

$$d_{23}=\sqrt{(x_2-x_3{)}^2+(y_2-y_3{)}^2}=\sqrt{x_2^2+x_3^2-2x_3{x}_2+y_2^2+y_3^2-2y_3{y}_2}$$

$$d_{24}=\sqrt{(x_2-x_4{)}^2+(y_2-y_4{)}^2}=\sqrt{x_4^2+x_2^2-2x_4{x}_2+y_4^2+y_2^2-2y_4{y}_2}$$

尋找樣本點對應的的最小距離
$$min(d_1,d_2,d_3,d_4)$$

如果使用python 的numpy計算：
第一步，計算平方和矩陣A、B,其實類似
$$A=\left\{\begin{array}{c}{x}_1{y}_1\\ x_2{y}_2\end{array}\right\}=>A\ast A^T=>A^{\prime }=\left\{\begin{array}{c}{x}_1^2+y_1^2{x}_2^2+y_2^2\\ x_1^2+y_1^2{x}_2^2+y_2^2\end{array}\right\}$$

$$B=\left\{\begin{array}{c}{x}_3{y}_3\\ x_4{y}_4\end{array}\right\}=>B\ast B^T=>B^{\prime }=\left\{\begin{array}{c}{x}_3^2+y_3^2{x}_4^2+y_4^2\\ x_3^2+y_3^2{x}_4^2+y_4^2\end{array}\right\}$$

 a2, b2 = np.square(a).sum(axis=1), np.square(b).sum(axis=1)

關於軸axis的變化參考：https://blog.csdn.net/sxj731533730/article/details/108831284
第二步：計算出$$A^{\prime }\ast B^{\prime T}$$的乘積矩陣
$$A\ast B^T=\left\{\begin{array}{c}{x}_1{y}_1\\ x_2{y}_2\end{array}\right\}\ast {\left\{\begin{array}{c}{x}_3{y}_3\\ x_4{y}_4\end{array}\right\}}^T=>C=\left\{\begin{array}{c}{x}_1\ast x_3+y_1\ast y_3{x}_1\ast x_4+y_1\ast y_4\\ x_2\ast x_3+y_2\ast y_3{x}_2\ast x_4+y_2\ast y_4\end{array}\right\}$$
第三步進行計算距離
$$D=A^{\prime }+B^{\prime T}-2C^T=\left\{\begin{array}{c}{x}_1^2+y_1^2{x}_2^2+y_2^2\\ x_1^2+y_1^2{x}_2^2+y_2^2\end{array}\right\}+{\left\{\begin{array}{c}{x}_3^2+y_3^2{x}_4^2+y_4^2\\ x_3^2+y_3^2{x}_4^2+y_4^2\end{array}\right\}}^T-2\ast {\left\{\begin{array}{c}{x}_1\ast x_3+y_1\ast y_3{x}_1\ast x_4+y_1\ast y_4\\ x_2\ast x_3+y_2\ast y_3{x}_2\ast x_4+y_2\ast y_4\end{array}\right\}}^T$$

$$D=A^{\prime }+B^{\prime T}-2C^T=\left\{\begin{array}{c}{x}_1^2+y_1^2+x_3^2+y_3^2{x}_2^2+y_2^2+x_3^2+y_3^2\\ x_1^2+y_1^2+x_4^2+y_4^2{x}_2^2+y_2^2+x_4^2+y_4^2\end{array}\right\}-2\ast \left\{\begin{array}{c}{x}_1\ast x_3+y_1\ast y_3{x}_2\ast x_3+y_2\ast y_3\\ x_1\ast x_4+y_1\ast y_4{x}_2\ast x_4+y_2\ast y_4\end{array}\right\}$$
$$D=A^{\prime }+B^{\prime T}-2C^T=\left\{\begin{array}{c}{d}_{13}{d}_{23}\\ d_{14}{d}_{24}\end{array}\right\}$$

import numpy as np
def _pdist(a, b):
    a, b = np.asarray(a), np.asarray(b)
    if len(a) == 0 or len(b) == 0:
        return np.zeros((len(a), len(b)))
    a2, b2 = np.square(a).sum(axis=1), np.square(b).sum(axis=1)
    r2 = -2. * np.dot(a, b.T) + a2[:, None] + b2[None, :]
    r2 = np.clip(r2, 0., float(np.inf))
    return r2
def _nn_euclidean_distance(x, y):
    distances = _pdist(x, y)
    return np.maximum(0.0, distances.min(axis=0))
a=np.asarray([[6,7],[4,8]])
b=np.asarray([[6,8],[7,9]])
print(_nn_euclidean_distance(a,b)

測試結果

C:\Users\sxj\venv\Scripts\python.exe E:/deepsort/deep_sort/test.py
[1. 5.]

Process finished with exit code 0

C++: