最小二乘法擬合圓心

double64發表於2024-09-03
#include <map>
#include <vector>
#include <iostream>
#include <string>


void FitCenterByLeastSquares(std::map<int, std::vector<double>> mapPoint,
    std::vector<double> &centerP, double &radius)
{
    double sumX = 0, sumY = 0;
    double sumXX = 0, sumYY = 0, sumXY = 0;
    double sumXXX = 0, sumXXY = 0, sumXYY = 0, sumYYY = 0;

    for (std::map<int, std::vector<double>>::iterator it = mapPoint.begin(); it != mapPoint.end(); ++it)
    {
        std::vector<double> p = it->second;

        sumX += p[0];
        sumY += p[1];
        sumXX += p[0] * p[0];
        sumYY += p[1] * p[1];
        sumXY += p[0] * p[1];
        sumXXX += p[0] * p[0] * p[0];
        sumXXY += p[0] * p[0] * p[1];
        sumXYY += p[0] * p[1] * p[1];
        sumYYY += p[1] * p[1] * p[1];
    }

    int pCount = mapPoint.size();
    double M1 = pCount * sumXY - sumX * sumY;
    double M2 = pCount * sumXX - sumX * sumX;
    double M3 = pCount * (sumXXX + sumXYY) - sumX * (sumXX + sumYY);
    double M4 = pCount * sumYY - sumY * sumY;
    double M5 = pCount * (sumYYY + sumXXY) - sumY * (sumXX + sumYY);

    double a = (M1 * M5 - M3 * M4) / (M2*M4 - M1 * M1);
    double b = (M1 * M3 - M2 * M5) / (M2*M4 - M1 * M1);
    double c = -(a * sumX + b * sumY + sumXX + sumYY) / pCount;

    //圓心XY 半徑
    double xCenter = -0.5*a;
    double yCenter = -0.5*b;
    radius = 0.5 * sqrt(a * a + b * b - 4 * c);
    centerP[0] = xCenter;
    centerP[1] = yCenter;
}

int main()
{
    std::vector<double> p1{ 0, 10 };
    std::vector<double> p2{ 10, 0 };
    std::vector<double> p3{ 0, -10 };
    std::vector<double> p4{ -10, 0 };
    std::map<int, std::vector<double>> mapPoints
    {
        {0, p1},
        {1, p2},
        {2, p3},
        {3, p4},
    };
    std::vector<double> centerP{ 0,0 };
    double r = 0;
    FitCenterByLeastSquares(mapPoints, centerP, r);

    return 0;
}



測試程式碼來自：
https://blog.csdn.net/xinjiang666/article/details/103767319

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