Least squares fitting two-dimensional straight line

Crawler websites are self-respecting.

1. Principle overview

  The expression of a plane line is:
$$y=kx+b\text{\hspace{1em}( 1 )}$$
  Suppose there are$n$个点$(x_i,y_i)（0\le i<n)$ , so the objective function is:
$$f(x)=\sum _{i=1}^{n}k{x}_i+b-y_i\text{\hspace{1em}( 2 )}$$ Calculate f ( x ) f(x)
  respectively$f\left(x\right)$ for$k、b$的偏导数：
$$\begin{gathered} \frac{f}{\delta k}=\sum _{i=1}^{n}2(k{x}_i+b-y_i)x_i=2\sum _{i=1}^{n}k{x}_i^2+b{x}_i-x_i{y}_i \\ =2(k\sum _{i=1}^n{x}_i^2+b\sum _{i=1}^n{x}_i-\sum _{i=1}^n{x}_i{y}_i)\text{\hspace{1em}(3)} \end{gathered}$$

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

  Let the partial derivatives of formula (3) and formula (4) be 0, and get a system of binary linear equations:

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

$$\{\begin{array}{l}Aq+Bb=C\\ Bk+bn=D\end{array}\text{\hspace{1em}( 6 )}$$
Solve the above equations to get$k、b$
$$\{\begin{array}{l}k=\frac{nC-BD}{nA-B^2}\\ b=\frac{DA-}{nA-B^2}\end{array}\text{\hspace{1em}(7)}$$

2. Python implementation

# -----------------------------------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 implementation

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++ implementation

    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);