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：

 

       public static void LeastSquare(List<Point> points, out double a, out double b)
        {
            int N = points.Count;
            if (N < 2)
            {
                a = 0;
                b = 0;
                return;
            }
            double X1 = 0, Y1 = 0, X2 = 0, X1Y1 = 0;
            for (int i = 0; i < N; ++i)
            {
                X2 += points[i].X * points[i].X;
                X1 += points[i].X;
                X1Y1 += points[i].X * points[i].Y;
                Y1 += points[i].Y;
            }
            a = (X1Y1 * N - X1 * Y1) / (X2 * N - X1 * X1);
            b = (X2 * Y1 - X1 * X1Y1) / (X2 * N - X1 * X1);
        }

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：

  

 

        public static double[] Gradient_Descent(List<Point> points, double starting_a, double starting_b, double learningRate, int num_iterations)
        {
            double a = starting_a;
            double b = starting_b;
            int iter = 0;
            while (iter < num_iterations)
            {
                double lasta = a;
                double lastb = b;
                double a_gradient = 0;
                double b_gradient = 0;
                int N = points.Count;
                double x, y;
                for (int i = 0; i < N; i++)
                {
                    x = (double)points[i].X;
                    y = (double)points[i].Y;

                    a_gradient += -(2.00 / N) * x * (y - ((a * x) + b));
                    b_gradient += -(2.00 / N) * (y - ((a * x) + b));
                }
                a -= (learningRate * a_gradient);
                b -= (learningRate * b_gradient);
                if (Math.Abs(b- lastb) <0.00000001&& Math.Abs(a - lasta) < 0.00000001)
                    break;
                iter++;
            }
            double totalError = 0;
            for (int i = 0; i < points.Count; i++)
            {
                double x = (double)points[i].X;
                double y = (double)points[i].Y;
                totalError += Math.Pow((y - (a * x + b)), 2);
            }
            double error = totalError / points.Count;

            return new double[] { a, b ,(double)iter,error};
        }