利用最小二乘法拟合任意次函数曲线（C#）

原文:

利用最小二乘法拟合任意次函数曲线（C#）

///<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>

name="arrY">已知点的y坐标集合</param>

name="length">已知点的个数</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);

    }

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;

static double SumArr(double[] arr1, int n1, double[] arr2, int n2,

int length)

double s=0;

if ((arr1[i] != 0 || n1 != 0) &&

(arr2[i] != 0 || n2 != 0))

s = s + Math.Pow(arr1[i], n1) * Math.Pow(arr2[i], n2);

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];

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];

}//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

剩下的就不用写了吧