C#直线的最小二乘法线性回归运算实例

/// <summary>/// 二维笛卡尔坐标系坐标/// </summary>public struct Point{  public double X;  public double Y;  public Point(double x = 0, double y = 0)  {    X = x;    Y = y;  }}

/// <summary>/// 对一组点通过最小二乘法进行线性回归/// </summary>/// <param name="parray"></param>public static void LinearRegression(Point[] parray){  //点数不能小于2  if (parray.Length < 2)  {    Console.WriteLine("点的数量小于2，无法进行线性回归");    return;  }  //求出横纵坐标的平均值  double averagex = 0, averagey = 0;  foreach (Point p in parray)  {    averagex += p.X;    averagey += p.Y;  }  averagex /= parray.Length;  averagey /= parray.Length;  //经验回归系数的分子与分母  double numerator = 0;  double denominator = 0;  foreach (Point p in parray)  {    numerator += (p.X - averagex) * (p.Y - averagey);    denominator += (p.X - averagex) * (p.X - averagex);  }  //回归系数b（Regression Coefficient）  double RCB = numerator / denominator;  //回归系数a  double RCA = averagey - RCB * averagex;  Console.WriteLine("回归系数A： " + RCA.ToString("0.0000"));  Console.WriteLine("回归系数B： " + RCB.ToString("0.0000"));  Console.WriteLine(string.Format("方程为： y = {0} + {1} * x",    RCA.ToString("0.0000"), RCB.ToString("0.0000")));  //剩余平方和与回归平方和  double residualSS = 0;  //（Residual Sum of Squares）  double regressionSS = 0; //（Regression Sum of Squares）  foreach (Point p in parray)  {    residualSS +=      (p.Y - RCA - RCB * p.X) *      (p.Y - RCA - RCB * p.X);    regressionSS +=      (RCA + RCB * p.X - averagey) *      (RCA + RCB * p.X - averagey);  }  Console.WriteLine("剩余平方和： " + residualSS.ToString("0.0000"));  Console.WriteLine("回归平方和： " + regressionSS.ToString("0.0000"));}

static void Main(string[] args){  //设置一个包含9个点的数组  Point[] array = new Point[9];  array[0] = new Point(0, 66.7);  array[1] = new Point(4, 71.0);  array[2] = new Point(10, 76.3);  array[3] = new Point(15, 80.6);  array[4] = new Point(21, 85.7);  array[5] = new Point(29, 92.9);  array[6] = new Point(36, 99.4);  array[7] = new Point(51, 113.6);  array[8] = new Point(68, 125.1);  LinearRegression(array);  Console.Read();}