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  • C++:最小二乘法 拟合圆

    一、推导

    二、分享给有需要的人,代码质量勿喷。

     1 void xjLeastSquares::FitCenterByLeastSquares(std::map<int, std::vector<double>> mapPoint, std::vector<double> &centerP, double &radius)
     2 {
     3     double sumX = 0, sumY = 0;
     4     double sumXX = 0, sumYY = 0, sumXY = 0;
     5     double sumXXX = 0, sumXXY = 0, sumXYY = 0, sumYYY = 0;
     6  
     7     for (std::map<int, std::vector<double>>::iterator it = mapPoint.begin(); it != mapPoint.end(); ++it)
     8     {
     9         std::vector<double> p = it->second;
    10  
    11         sumX += p[0];
    12         sumY += p[1];
    13         sumXX += p[0] * p[0];
    14         sumYY += p[1] * p[1];
    15         sumXY += p[0] * p[1];
    16         sumXXX += p[0] * p[0] * p[0];
    17         sumXXY += p[0] * p[0] * p[1];
    18         sumXYY += p[0] * p[1] * p[1];
    19         sumYYY += p[1] * p[1] * p[1];
    20     }
    21  
    22     int pCount = mapPoint.size();
    23     double M1 = pCount * sumXY - sumX * sumY;
    24     double M2 = pCount * sumXX - sumX * sumX;
    25     double M3 = pCount * (sumXXX + sumXYY) - sumX * (sumXX + sumYY);
    26     double M4 = pCount * sumYY - sumY * sumY;
    27     double M5 = pCount * (sumYYY + sumXXY) - sumY * (sumXX + sumYY);
    28  
    29     double a = (M1 * M5 - M3 * M4) / (M2*M4 - M1 * M1);
    30     double b = (M1 * M3 - M2 * M5) / (M2*M4 - M1 * M1);
    31     double c = -(a * sumX + b * sumY + sumXX + sumYY) / pCount;
    32  
    33     //圆心XY 半径
    34     double xCenter = -0.5*a;
    35     double yCenter = -0.5*b;
    36     radius = 0.5 * sqrt(a * a + b * b - 4 * c);
    37     centerP[0] = xCenter;
    38     centerP[1] = yCenter;
    39 }
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  • 原文地址:https://www.cnblogs.com/ybqjymy/p/14201816.html
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