如何理解最小二乘法？如何程序实现。
最小二乘法，个人觉得叫“最小平方法”更恰当。
本文是以下这篇文章的程序实现，使用了Ceres优化库。
https://mp.weixin.qq.com/s/4e9ZiiGIOWx_ZUGjzgavWw
1 #include <iostream> 2 #include <ceres/ceres.h> 3 using namespace std; 4 using namespace ceres; 5 6 struct CURVE_FITTING_COST 7 { 8 CURVE_FITTING_COST(double x, double y):_x(x),_y(y){} 9 template <typename T> 10 bool operator()(const T* const ab, T* residual) const 11 { 12 //y-0*x+a; 13 //y-ax-b 14 residual[0] = T(_y) - ab[0]*T(_x) - ab[1]; 15 return true; 16 17 } 18 const double _x,_y; 19 20 }; 21 22 int main(int argc, char** argv) 23 { 24 double x[5] = {25, 27, 31, 33, 35}; 25 double y[5] = {110, 115, 155, 160, 180}; 26 double ab[2]={0,0}; 27 28 //构建最小平方问题 29 Problem problem; 30 for(int i=0; i<5; i++) 31 { 32 problem.AddResidualBlock( 33 new AutoDiffCostFunction<CURVE_FITTING_COST,1,2>(new CURVE_FITTING_COST(x[i], y[i])), 34 nullptr, 35 ab 36 ); 37 38 } 39 40 Solver::Options options; 41 options.linear_solver_type = DENSE_QR; 42 options.minimizer_progress_to_stdout = true; 43 44 Solver::Summary summary; 45 Solve(options, &problem, &summary); 46 47 cout<<"estimated a = "; 48 for(auto a:ab) cout<<a<<" "; 49 cout<<endl; 50 }
输出结果为：
➜ build ./least_square
iter cost cost_change |gradient| |step| tr_ratio tr_radius ls_iter iter_time total_time
0 5.367500e+04 0.00e+00 2.22e+04 0.00e+00 0.00e+00 1.00e+04 0 1.27e-04 2.07e-04
1 4.723331e+01 5.36e+04 3.31e+00 7.21e+01 1.00e+00 3.00e+04 1 2.00e-04 8.96e-04
2 4.709303e+01 1.40e-01 9.81e-03 1.93e+00 1.00e+00 9.00e+04 1 1.10e-04 1.03e-03
estimated a,b = 7.20902 -73.7123
