///<summary> ///用最小二乘法拟合二元多次曲线 ///</summary> ///<param name="arrX">已知点的x坐标集合</param> ///<param name="arrY">已知点的y坐标集合</param> ///<param name="length">已知点的个数</param> ///<param name="dimension">方程的最高次数</param> public static double[] MultiLine(double[] arrX, double[] arrY, int length, int dimension)//二元多次线性方程拟合曲线 { int n = dimension + 1; //dimension次方程需要求 dimension+1个 系数 double[,] Guass=new double[n,n+1]; //高斯矩阵 例如:y=a0+a1*x+a2*x*x for(int i=0;i<n;i++) { int j; for(j=0;j<n;j++) { Guass[i,j] = SumArr(arrX, j + i, length); } Guass[i,j] = SumArr(arrX,i,arrY,1,length); } return ComputGauss(Guass,n); } public static double SumArr(double[] arr, int n, int length) //求数组的元素的n次方的和 { double s = 0; for (int i = 0; i < length; i++) { if (arr[i] != 0 || n != 0) s = s + Math.Pow(arr[i], n); else s = s + 1; } return s; } public static double SumArr(double[] arr1, int n1, double[] arr2, int n2, int length) { double s=0; for (int i = 0; i < length; i++) { if ((arr1[i] != 0 || n1 != 0) && (arr2[i] != 0 || n2 != 0)) s = s + Math.Pow(arr1[i], n1) * Math.Pow(arr2[i], n2); else s = s + 1; } return s; } public static double[] ComputGauss(double[,] Guass,int n) { int i, j; int k,m; double temp; double max; double s; double[] x = new double[n]; for (i = 0; i < n; i++) x[i] = 0.0;//初始化 for (j = 0; j < n; j++) { max = 0; k = j; for (i = j; i < n; i++) { if (Math.Abs(Guass[i, j]) > max) { max = Guass[i, j]; k = i; } } if (k != j) { for (m = j; m < n + 1; m++) { temp = Guass[j, m]; Guass[j, m] = Guass[k, m]; Guass[k, m] = temp; } } if (0 == max) { // "此线性方程为奇异线性方程" return x; } for (i = j + 1; i < n; i++) { s = Guass[i, j]; for (m = j; m < n + 1; m++) { Guass[i, m] = Guass[i, m] - Guass[j, m] * s / (Guass[j, j]); } } }//结束for (j=0;j<n;j++) for (i = n-1; i >= 0; i--) { s = 0; for (j = i + 1; j < n; j++) { s = s + Guass[i,j] * x[j]; } x[i] = (Guass[i,n] - s) / Guass[i,i]; } return x; }//返回值是函数的系数
例如:y=a0+a1*x 返回值则为a0 a1
例如:y=a0+a1*x+a2*x*x 返回值则为a0 a1 a2
已有 4901 位网友参与,快来吐槽:
发表评论