二维直线回归(拟合)算法(一)

一、最小二乘法(Least Squares)：

假设样本集单个样本的线性函数为：

为解二元一次方程中的a,b，给上式两边乘以xi得到(也可以两边乘yi)：

方程组：

解样本集方程组：

矩阵形式：

对矩阵求逆：

则：

 1        public static void LeastSquare(List<Point> points, out double a, out double b)
 2         {
 3             int N = points.Count;
 4             if (N < 2)
 5             {
 6                 a = 0;
 7                 b = 0;
 8                 return;
 9             }
10             double X1 = 0, Y1 = 0, X2 = 0, X1Y1 = 0;
11             for (int i = 0; i < N; ++i)
12             {
13                 X2 += points[i].X * points[i].X;
14                 X1 += points[i].X;
15                 X1Y1 += points[i].X * points[i].Y;
16                 Y1 += points[i].Y;
17             }
18             a = (X1Y1 * N - X1 * Y1) / (X2 * N - X1 * X1);
19             b = (X2 * Y1 - X1 * X1Y1) / (X2 * N - X1 * X1);
20         }

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二、梯度下降法(Gradient Descent)：

针对离散数据样本集的最小化函数为：

样本集中单一数据的Q函数对a、b偏导数：

批样本集梯度：

 1         public static double[] Gradient_Descent(List<Point> points, double starting_a, double starting_b, double learningRate, int num_iterations)
 2         {
 3             double a = starting_a;
 4             double b = starting_b;
 5             int iter = 0;
 6             while (iter < num_iterations)
 7             {
 8                 double lasta = a;
 9                 double lastb = b;
10                 double a_gradient = 0;
11                 double b_gradient = 0;
12                 int N = points.Count;
13                 double x, y;
14                 for (int i = 0; i < N; i++)
15                 {
16                     x = (double)points[i].X;
17                     y = (double)points[i].Y;
18 
19                     a_gradient += -(2.00 / N) * x * (y - ((a * x) + b));
20                     b_gradient += -(2.00 / N) * (y - ((a * x) + b));
21                 }
22                 a -= (learningRate * a_gradient);
23                 b -= (learningRate * b_gradient);
24                 if (Math.Abs(b- lastb) <0.00000001&& Math.Abs(a - lasta) < 0.00000001)
25                     break;
26                 iter++;
27             }
28             double totalError = 0;
29             for (int i = 0; i < points.Count; i++)
30             {
31                 double x = (double)points[i].X;
32                 double y = (double)points[i].Y;
33                 totalError += Math.Pow((y - (a * x + b)), 2);
34             }
35             double error = totalError / points.Count;
36 
37             return new double[] { a, b ,(double)iter,error};
38         }

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