VC++的MFC实现的最小二乘法曲线拟合？

Question

要求做成类似于计算器似得东西，输入数值返回结果。本人对编程几乎是一窍不通，请高人帮忙..求详细解答...
应用最小二乘法进行二次曲线拟合，要求用MFC，输入一些数值点，返回二次曲线系数a= ，b= c= 以及相关系数γ= 。

Answer 1

BOOL YLF_Curve_CalPolyFactor(//计算多项式系数 calculate polynomial fact
/*************************输入***************************/
double *x_in,//拟合前数据点X轴坐标数组
double *y_in,//拟合前数据点Y轴坐标数组
unsigned long num_out,//拟合后输出数据点组数
unsigned int nSimulateNum,//拟合方式(可选值1-4次曲线)
/*************************输出***************************/
double *factor//factor(长度固定为5)
)
{
if (0 < num_out)
{
double dbfactor[5];//系数数组
double *pdbX,*pdbY;//x,y数组零时变量

pdbX = new double[num_out];
pdbY = new double[num_out];
memcpy(pdbX,x_in,sizeof(double)*num_out);
memcpy(pdbY,y_in,sizeof(double)*num_out);

double *b;
b = dbfactor;
int n = num_out;
int m = nSimulateNum;
int i, j, k, l;
double B[6][7];
double temp, det;
temp = det = 0.0f;
BOOL BreakFlag;
double *CurveX, *CurveY,*CurveS;
CurveX = new double[n];
CurveY = new double[n];
CurveS = new double[n];
ZeroMemory(B, 4 * 6 * 7);
ZeroMemory(CurveX, n);
ZeroMemory(CurveY, n);
ZeroMemory(CurveS, n);

for(i = 0; i < n; i++)
{
CurveX[i] = pdbX[i];
CurveY[i] = pdbY[i];
}
for(i = 0; i < 5; i++)
{
for(j = 0; j < 6; j++)
{
B[i][j] = 0;
}
}
B[0][0] = (double)n;
CurveS[0] = CurveX[0];
B[0][m + 1] = CurveY[0];
B[1][m + 1] = CurveY[0] * CurveX[0];
B[1][0] = CurveX[0];

for(k = 1; k < n; k++)
{
B[0][m + 1] = B[0][m + 1] + CurveY[k];
B[1][m + 1] = B[1][m + 1] + CurveY[k] * CurveX[k];
B[1][0] = B[1][0] + CurveX[k];
CurveS[k] = CurveX[k];
}

B[0][1] = B[1][0];
for(i = 2; i < m + 1; i++)
{
temp = 0;
B[i][m + 1] = 0;
for(k = 0; k < n; k++)
{
CurveS[k] = CurveS[k] * CurveX[k];
B[i][m + 1] = B[i][m + 1] + CurveY[k] * CurveS[k];
temp += CurveS[k];
}
for(j = 0; j < i + 1; j++)
{
l = i - j;
B[l][j] = temp;
}
}

for(j = 1; j < m + 1; j++)
{
temp = 0;
for(k = 0; k < n; k++)
{
CurveS[k] = CurveS[k] * CurveX[k];
temp = temp + CurveS[k];
}
for(i = m; i >= j; i--)
{
l = m + j - i;
B[i][l] = temp;
}
}

BreakFlag = FALSE;
for(i = 0; i < m; i++)
{
det = B[i][i];
for(j = i; j < m + 2; j++)
{
if(fabs(det) > 0.00001)
B[i][j] = B[i][j] * 1.0f / det;
else
{
BreakFlag = TRUE;
break;
}
}
if(BreakFlag == TRUE)
break;
for(j = i + 1; j < m + 1; j++)
{
det = B[j][i];
for(k = 0; k < m + 2; k++)
{
B[j][k] = B[j][k] - det * B[i][k];
}
}
}
if(BreakFlag)
{
delete []CurveX;
delete []CurveY;
delete []CurveS;
delete []pdbX;
delete []pdbY;

::MessageBox(NULL,"离散点数据数组无效!请检验有效性。","错误提示",MB_OK|MB_ICONWARNING);
return FALSE;//数据不合理
}

B[m][m + 1] = B[m][m + 1] * 1.0f / B[m][m];
B[m][m] = 1;

for(i = m - 1; i >= 0; i--)
{
for(k = 1; k < m - i + 1; k++)
{
det = B[i][i + k];
for(j = 0; j < m + 2; j++)
{
B[i][j] = B[i][j] - B[i + k][j] * det;
}
}
}
for(i = 0; i < m + 1; i++)
b[i] = B[i][m + 1];
delete []CurveX;
delete []CurveY;
delete []CurveS;
delete []pdbX;
delete []pdbY;

//输出系数数据数组
memcpy(factor,dbfactor,5*sizeof(double));

return TRUE;
}
else return FALSE;
}

追问

能把程序打包发我邮箱里面么？不胜感谢pg0612@126.com....感激不尽啊！~

追答

效果图，源码已经发送到你邮箱。有用记得给分。

Answer 2

void CYourDlg::linearReg_Y_a0_a1X(double &a0, double &a1, const CArray<double,double> & X_Con_Lin_Reg, const CArray<double,double> & Y_Volts_Lin_Reg) { double sumXiYi(0.0), sumXi(0.0), sumYi(0.0), sumXi_2(0.0), X_Average(0.0), Y_Average(0.0); INT_PTR n = X_Con_Lin_Reg.GetSize(); for (int i=0; i<n; i++) { sumXiYi += X_Con_Lin_Reg[i]*Y_Volts_Lin_Reg[i]; sumXi += X_Con_Lin_Reg[i]; sumYi += Y_Volts_Lin_Reg[i]; sumXi_2 += X_Con_Lin_Reg[i]*X_Con_Lin_Reg[i]; } /*求平均值*/X_Average = sumXi/n; Y_Average = sumYi/n; /*y=a0+a1x*/a1 = (sumXiYi-n*X_Average*Y_Average)/(sumXi_2-n*X_Average*X_Average); a0 = Y_Average – a1*X_Average; }

Answer 3

求发源码1264958401@qq.com
感激不尽

Answer 4

可以给我发一份吗？727889069@qq.com 万分感谢

Answer 5

楼主，我导师让我编这个程序，我一头雾水，您可以把收到的答案发我一份吗？感激涕零啊