最小二乘法算法问题。

[ksq2010]

我做了一个最小二乘法的算法，是用C#写的。但是这个算法是有截距的，我希望截距等于0.求高人指教如何修改，或者提供一个新的算法，不要截距。类似y=ax2+bx。

///<summary>
        ///用最小二乘法拟合二元多次曲线
        ///例如y=ax+b
        ///其中MultiLine将返回a，b两个参数。
        ///a对应MultiLine[1]
        ///b对应MultiLine[0]
        ///</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);

        }
        private 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;
        }
        private 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;

        }
        private 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

[fishfish1901]

只要简单修改就可以了 思路：让最低次的系数，不参与循环，直接使用默认值0. 改两个地方： 1. MultiLine第六行： for (j = 0; j < n; j++) 改为 for (j = 1; j < n; j++) 2. ComputGauss第八行（不计算空行）： for (j = 0; j < n; j++) 改为 for (j = 1; j < n; j++)