Cholesky分解求系数参考： [1]冯天祥. 多元线性回归最小二乘法及其经济分析[J]. 经济师,,11:129. 还可以采用最小二乘法来估计参数：
算法设计也可以参考两种系数最终公式设计。
下面的Java代码由网友设计，采用第一种方法计算参数。
1 /** 2 * 多元线性回归分析 3 * 4 * @param x[m][n] 5 * 每一列存放m个自变量的观察值 6 * @param y[n] 7 * 存放随即变量y的n个观察值 8 * @param m 9 * 自变量的个数 10 * @param n 11 * 观察数据的组数 12 * @param a 13 * 返回回归系数a0,...,am 14 * @param dt[4] 15 * dt[0]偏差平方和q,dt[1] 平均标准偏差s dt[2]返回复相关系数r dt[3]返回回归平方和u 16 * @param v[m] 17 * 返回m个自变量的偏相关系数 18 */ 19 public static void sqt2(double[][] x, double[] y, int m, int n, double[] a, 20 double[] dt, double[] v) { 21int i, j, k, mm; 22double q, e, u, p, yy, s, r, pp; 23double[] b = new double[(m + 1) * (m + 1)]; 24mm = m + 1; 25b[mm * mm - 1] = n; 26for (j = 0; j <= m - 1; j++) { 27 p = 0.0; 28 for (i = 0; i <= n - 1; i++) 29 p = p + x[j][i]; 30 b[m * mm + j] = p; 31 b[j * mm + m] = p; 32} 33for (i = 0; i <= m - 1; i++) 34 for (j = i; j <= m - 1; j++) { 35 p = 0.0; 36 for (k = 0; k <= n - 1; k++) 37 p = p + x[i][k] * x[j][k]; 38 b[j * mm + i] = p; 39 b[i * mm + j] = p; 40 } 41a[m] = 0.0; 42for (i = 0; i <= n - 1; i++) 43 a[m] = a[m] + y[i]; 44for (i = 0; i <= m - 1; i++) { 45 a[i] = 0.0; 46 for (j = 0; j <= n - 1; j++) 47 a[i] = a[i] + x[i][j] * y[j]; 48} 49chlk(b, mm, 1, a); 50yy = 0.0; 51for (i = 0; i <= n - 1; i++) 52 yy = yy + y[i] / n; 53q = 0.0; 54e = 0.0; 55u = 0.0; 56for (i = 0; i <= n - 1; i++) { 57 p = a[m]; 58 for (j = 0; j <= m - 1; j++) 59 p = p + a[j] * x[j][i]; 60 q = q + (y[i] - p) * (y[i] - p); 61 e = e + (y[i] - yy) * (y[i] - yy); 62 u = u + (yy - p) * (yy - p); 63} 64s = Math.sqrt(q / n); 65r = Math.sqrt(1.0 - q / e); 66for (j = 0; j <= m - 1; j++) { 67 p = 0.0; 68 for (i = 0; i <= n - 1; i++) { 69 pp = a[m]; 70 for (k = 0; k <= m - 1; k++) 71 if (k != j) 72 pp = pp + a[k] * x[k][i]; 73 p = p + (y[i] - pp) * (y[i] - pp); 74 } 75 v[j] = Math.sqrt(1.0 - q / p); 76} 77dt[0] = q; 78dt[1] = s; 79dt[2] = r; 80dt[3] = u; 81 } 82 private static int chlk(double[] a, int n, int m, double[] d) { 83int i, j, k, u, v; 84if ((a[0] + 1.0 == 1.0) || (a[0] < 0.0)) { 85 System.out.println("fail\n"); 86 return (-2); 87} 88a[0] = Math.sqrt(a[0]); 89for (j = 1; j <= n - 1; j++) 90 a[j] = a[j] / a[0]; 91for (i = 1; i <= n - 1; i++) { 92 u = i * n + i; 93 for (j = 1; j <= i; j++) { 94 v = (j - 1) * n + i; 95 a[u] = a[u] - a[v] * a[v]; 96 } 97 if ((a[u] + 1.0 == 1.0) || (a[u] < 0.0)) { 98 System.out.println("fail\n"); 99 return (-2);100 }101 a[u] = Math.sqrt(a[u]);102 if (i != (n - 1)) {103 for (j = i + 1; j <= n - 1; j++) {104 v = i * n + j;105 for (k = 1; k <= i; k++)106 a[v] = a[v] - a[(k - 1) * n + i] * a[(k - 1) * n + j];107 a[v] = a[v] / a[u];108 }109 }110}111for (j = 0; j <= m - 1; j++) {112 d[j] = d[j] / a[0];113 for (i = 1; i <= n - 1; i++) {114 u = i * n + i;115 v = i * m + j;116 for (k = 1; k <= i; k++)117 d[v] = d[v] - a[(k - 1) * n + i] * d[(k - 1) * m + j];118 d[v] = d[v] / a[u];119 }120}121for (j = 0; j <= m - 1; j++) {122 u = (n - 1) * m + j;123 d[u] = d[u] / a[n * n - 1];124 for (k = n - 1; k >= 1; k--) {125 u = (k - 1) * m + j;126 for (i = k; i <= n - 1; i++) {127 v = (k - 1) * n + i;128 d[u] = d[u] - a[v] * d[i * m + j];129 }130 v = (k - 1) * n + k - 1;131 d[u] = d[u] / a[v];132 }133}134return (2);135 }136 /**137 * @param args138 */139 public static void main(String[] args) {140// TODO Auto-generated method stub141/**142* 一元回归143*/144// int i;145// double[] dt=new double[6];146// double[] a=new double[2];147// double[] x={ 0.0,0.1,0.2,0.3,0.4,0.5,148// 0.6,0.7,0.8,0.9,1.0};149// double[] y={ 2.75,2.84,2.965,3.01,3.20,150// 3.25,3.38,3.43,3.55,3.66,3.74};151// SPT.SPT1(x,y,11,a,dt);152// System.out.println("");153// System.out.println("a="+a[1]+" b="+a[0]);154// System.out.println("q="+dt[0]+" s="+dt[1]+" p="+dt[2]);155// System.out.println(" umax="+dt[3]+" umin="+dt[4]+" u="+dt[5]);156/**157* 多元回归158*/159int i;160double[] a = new double[4];161double[] v = new double[3];162double[] dt = new double[4];163double[][] x = { { 1.1, 1.0, 1.2, 1.1, 0.9 },164 { 2.0, 2.0, 1.8, 1.9, 2.1 }, { 3.2, 3.2, 3.0, 2.9, 2.9 } };165double[] y = { 10.1, 10.2, 10.0, 10.1, 10.0 };166SPT.sqt2(x, y, 3, 5, a, dt, v);167for (i = 0; i <= 3; i++)168 System.out.println("a(" + i + ")=" + a[i]);169System.out.println("q=" + dt[0] + " s=" + dt[1] + " r=" + dt[2]);170for (i = 0; i <= 2; i++)171 System.out.println("v(" + i + ")=" + v[i]);172System.out.println("u=" + dt[3]);173 }
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下面的C++代码由网友提供，采用第二中方法计算系数。
#include<iostream>#include<fstream>#include<iomanip>using namespace std;void transpose(double **p1,double **p2,int m,int n);void multipl(double **p1,double **p2,double **p3,int m,int n,int p);void Inver(double **p1,double **p2,int n);double SD(double **p1,double **p2,double **p3,double **p4,int m,int n);double ST(double **p1,int m);void de_allocate(double **data,int m);int main() {int row,col;char filename[30];double SDsum,STsum,F,R2;cout<<"Input original data file: \n";ifstream infile; //打开文件cin>>filename;infile.open(filename);if(!infile) {cout<<"Opening the file failed!\n";exit(1);}infile>>row>>col; //读入文件中的行数和列数double **matrix=new double*[row]; //为动态二维数组分配内存double **X=new double*[row];double **Y=new double*[row];double **XT=new double*[col];double **XTX=new double*[col];double **XTXInv=new double*[col];double **XTXInvXT=new double*[col];double **B=new double*[col];double **YE=new double*[row];for(int i=0;i<row;i++) {matrix[i]=new double[col];X[i]=new double[col];Y[i]=new double[1];Y[i]=new double[1];YE[i]=new double[1];}for(int i=0;i<col;i++) {XT[i]=new double[row];XTX[i]=new double[2*col];/为什么必须分配2*col列空间而不是col？在矩阵求逆时，XTX变增广矩阵，列数变为原来2吧倍，跟求逆算法有关。XTXInv[i]=new double[col];XTXInvXT[i]=new double[row];B[i]=new double[1];}for(int i=0;i<row;i++)for(int j=0;j<col;j++)infile>>matrix[i][j];infile.close();for(int i=0;i<row;i++) { //提取1X和Y数组列X[i][0]=1;Y[i][0]=matrix[i][col-1];for(int j=0;j<col-1;j++)X[i][j+1]=matrix[i][j];}transpose(X,XT,row,col);multipl(XT,X,XTX,col,row,col);Inver(XTX,XTXInv,col);multipl(XTXInv,XT,XTXInvXT,col,col,row);multipl(XTXInvXT,Y,B,col,row,1);SDsum=SD(Y,X,B,YE,row,col);STsum=ST(Y,row);F=((STsum-SDsum)/(col-1))/(SDsum/(row-col));R2=1/(1+(row-col)/F/(col-1));cout<<"输出B:\n"; //屏幕输出结果B，SD，ST，F，R2for(int i=0;i<col;i++)cout<<setiosflags(ios::fixed)<<setprecision(4)<<B[i][0]<<' ';cout<<endl;cout<<"SD="<<SDsum<<';'<<"ST="<<STsum<<';'<<"F="<<F<<';'<<"R2="<<R2<<endl;ofstream outfile; // 结果写入文件cout<<"Output file'name:\n";cin>>filename;outfile.open(filename);if(!outfile) {cout<<"Opening the file failed!\n";exit(1);}outfile<<"输出B:\n";for(int i=0;i<col;i++)outfile<<B[i][0]<<' ';outfile<<endl;outfile<<setiosflags(ios::fixed)<<setprecision(4)<<"SD="<<SDsum<<';'<<"ST="<<STsum<<';'<<"F="<<F<<';'<<"R2="<<R2<<endl;outfile<<"Y and YE and Y-YE's value are:\n";for(int i=0;i<row;i++)outfile<<Y[i][0]<<" "<<YE[i][0]<<" "<<Y[i][0]-YE[i][0]<<endl;outfile.close();de_allocate(matrix,row);de_allocate(X,row);de_allocate(Y,row);de_allocate(XT,col);de_allocate(XTX,col);de_allocate(XTXInv,col);de_allocate(XTXInvXT,col);de_allocate(B,col);de_allocate(YE,row);system("pause");return(0);}void de_allocate(double **data,int m) { //释放内存单元for(int i=0;i<m;i++)delete []data[i];delete []data;}double ST(double **p1,int m) { //求总离差平方和STdouble sum1=0,sum2=0,Yave=0;for(int i=0;i<m;i++)sum1+=p1[i][0];Yave=sum1/m;for(int i=0;i<m;i++)sum2+=(p1[i][0]-Yave)*(p1[i][0]-Yave);return sum2;}double SD(double **p1,double **p2,double **p3,double **p4,int m,int n) { //求偏差平方和SDdouble sum1=0,sum2=0;for(int i=0;i<m;i++) {sum1=0;for(int k=0;k<n;k++)sum1+=p2[i][k]*p3[k][0];p4[i][0]=sum1;}for(int i=0;i<m;i++)sum2+=(p1[i][0]-p4[i][0])*(p1[i][0]-p4[i][0]);return sum2;}void transpose(double **p1,double **p2,int m,int n) { //矩阵转置for(int i=0;i<n;i++)for(int j=0;j<m;j++)p2[i][j]=p1[j][i];}void multipl(double **p1,double **p2,double **p3,int m,int n,int p) { //矩阵相乘double sum;for(int i=0;i<m;i++) {for(int j=0;j<p;j++) {sum=0;for(int k=0;k<n;k++)sum+=p1[i][k]*p2[k][j];p3[i][j]=sum;}}}void Inver(double **p1,double **p2,int n) { //求逆矩阵//初始化矩阵在右侧加入单位阵for(int i=0;i<n;i++) {for(int j=0;j<n;j++) {p1[i][j+n]=0;p1[i][i+n]=1;}}//对于对角元素为0的进行换行操作for(int i=0;i<n;i++){while(p1[i][i]==0){for(int j=i+1;j<n;j++){if (p1[j][i]!=0){ double temp=0;for(int r=i;r<2*n;r++){temp=p1[j][r];p1[j][r]=p1[i][r];p1[i][r]=temp;}}break;}}//if (p1[i][i]==0) return 0;}//行变换为上三角矩阵double k=0;for(int i=0;i<n;i++) {for(int j=i+1;j<n;j++) {k=(-1)*p1[j][i]/p1[i][i];for(int r=i;r<2*n;r++)p1[j][r]+=k*p1[i][r];}}//行变换为下三角矩阵//double k=0;for(int i=n-1;i>=0;i--) {for(int j=i-1;j>=0;j--) {k=(-1)*p1[j][i]/p1[i][i];for(int r=0;r<2*n;r++)p1[j][r]+=k*p1[i][r];}}//化为单位阵for(int i=n-1;i>=0;i--) {k=p1[i][i];for(int j=0;j<2*n;j++)p1[i][j]/=k;}//拆分出逆矩阵for(int i=0;i<n;i++) {for(int j=0;j<n;j++)p2[i][j]=p1[i][n+j];}}
二元线性回归最小二乘法拟合空间直线。网友提供
#include "stdafx.h"using namespace std;using std::cout;using std::cin;using std::endl;#include<math.h>#include <windows.h>/*这是一个控制台程序，任何一个空间直线方程都能以如下的方式表示x=az+by=cz+dz=z即(x-b)/a=(y-d)/c=z/1*/#include <vector>using std::vector;/* the fitting vaule of this line are: a=2 b=3 c=3 d=0double z[10]={0.5,0.7,1,1.2,1.5,1.8,2,2.5,2.8,3};double y[10]={1.5,2.1,3,3.6,4.5,5.4,6,7.5,8.4,9};double x[10]={4,4.4,5,5.4,6,6.6,7,8,8.6,9};*/double x[]={1,1.5,2,2.5,3,3.5,4,4.5,5};double y[]={-8.1,-7.2,-6.2,-5.5,-4.8,-3.8,-3,-2.2,-1.3};double z[]={-12,-11.8,-10.7,-9.5,-8.2,-7,-6,-4.5,-3.5};int _tmain(int argc, _TCHAR* argv[]){vector<double>position_z;vector<double>position_y;vector<double>position_x;if ((sizeof(z)/sizeof(double))!=(sizeof(x)/sizeof(double))||(sizeof(z)/sizeof(double))!=(sizeof(y)/sizeof(double))||(sizeof(x)/sizeof(double))!=(sizeof(y)/sizeof(double))){::MessageBox(NULL,"请检查输入数组的长度是否相等","方程无解!",MB_OK);}//int ss=sizeof(z)/sizeof(double);for (int i=0;i<(sizeof(z)/sizeof(double));++i){position_z.push_back(z[i]);position_y.push_back(y[i]);position_x.push_back(x[i]);}//double m = position_z.size();//caculate o1,p1,c1,o2,p2,c2double o1 = 0.0;double p1 = 0.0;double x1 = 0.0;double o2 = 0.0;double p2 = 0.0;double x2 = 0.0;double y1 = 0.0;double y2 = 0.0;for (vector<double>::size_type t=0;t< position_z.size();++t){o1 += position_z[t]*position_z[t];p1 += position_z[t];x1 += position_x[t]*position_z[t];o2 += position_z[t];p2 = position_z.size();x2 += position_x[t] ;y2 += position_y[t];y1 += position_y[t]*position_z[t];}//caculate a bdouble a = 0.0 ,b = 0.0, c = 0.0, d = 0.0 ;if ((o1*p2-o2*p1)!= 0){a = (x1*p2 - x2*p1) /(o1*p2-o2*p1);if (a<1e-10){a=0;}b = (x2*o1 - x1*o2) /(o1*p2-o2*p1);if (b<1e-10){b=0;}c = (y1*p2 - y2*p1) /(o1*p2-o2*p1);if (c<1e-10){c=0;}d = (y2*o1 - y1*o2) /(o1*p2-o2*p1);if (d<1e-10){d=0;}}else{::MessageBox(NULL,"OK","方程无解!",MB_OK);}cout<<a<<" " <<b<<" "<<c<<" "<<d<<endl;system ("pause");return 0;}
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