最小二乘拟合二维直线

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1. 原理概述

  平面直线的表达式为：
$$y=kx+b\text{\hspace{1em}(1)}$$
  假设有$n$个点$(x_i,y_i)（0\le i<n）$，于是目标函数为：
$$f(x)=\sum _{i=1}^{n}k{x}_i+b-y_i\text{\hspace{1em}(2)}$$
  分别求$f(x)$对于$k、b$的偏导数：
$$\begin{gathered} \frac{\delta f}{\delta k}=\sum _{i=1}^{n}2(k{x}_i+b-y_i)x_i=2\sum _{i=1}^{n}k{x}_i^2+b{x}_i-x_i{y}_i \\ =2(k\sum _{i=1}^n{x}_i^2+b\sum _{i=1}^n{x}_i-\sum _{i=1}^n{x}_i{y}_i)\text{\hspace{1em}(3)} \end{gathered}$$

$$\begin{gathered} \frac{\delta f}{\delta b}=\sum _{i=1}^{n}2(k{x}_i+b-y_i)=2\sum _{i=1}^{n}k{x}_i+b-y_i \\ =2(k\sum _{i=1}^n{x}_i+bn-\sum _{i=1}^n{y}_i)\text{\hspace{1em}(4)} \end{gathered}$$

  令(3)式和(4)式偏导数为0，得到一个二元一次方程组：

$$\{\begin{array}{l}k{\sum }_{i=1}^n{x}_i^2+b{\sum }_{i=1}^n{x}_i={\sum }_{i=1}^n{x}_i{y}_i\\ k{\sum }_{i=1}^n{x}_i+bn={\sum }_{i=1}^n{y}_i\end{array}\text{\hspace{1em}(5)}$$
记
$$A=\sum _{i=1}^n{x}_i^2,B=\sum _{i=1}^n{x}_i,C=\sum _{i=1}^n{x}_i{y}_i,D=\sum _{i=1}^n{y}_i$$
于是有：

$$\{\begin{array}{l}Ak+Bb=C\\ Bk+bn=D\end{array}\text{\hspace{1em}(6)}$$
解以上方程组得到$k、b$
$$\{\begin{array}{l}k=\frac{nC-BD}{nA-B^2}\\ b=\frac{DA-BC}{nA-B^2}\end{array}\text{\hspace{1em}(7)}$$

2. python实现

# -----------------------------------拟合直线y=kx+b--------------------------------
n = points.shape[0]
A = sum(x * x)
B = sum(x)
C = sum(x * y)
D = sum(y)
k = (n * C - B * D) / (n * A - B * B)
b = (D * A - B * C) / (n * A - B * B)

3. matlab实现

n = pc.Count;
A = sum(x.^2);
B = sum(x);
C = sum(x.*y);
D = sum(y);
k = (n * C - B * D) / (n * A - B * B);
b = (D * A - B * C) / (n * A - B * B);

4. C++实现

    int n = cloud->points.size();
	double A = 0.0, B = 0.0, C = 0.0, D = 0.0;

	for (int i = 0; i < n; i++)
	{
		A += pow(cloud->points[i].x, 2);
		B += cloud->points[i].x;
		C += cloud->points[i].x * cloud->points[i].y;
		D += cloud->points[i].y;
		
	}

	double k = (n * C - B * D) / (n * A - B * B);
	double b = (D * A - B * C) / (n * A - B * B);