C++:最小二乘法拟合直线

#include <opencv2/opencv.hpp>   
#include <opencv2/highgui/highgui.hpp>

using namespace std;
using namespace cv;
/*
* @brief 得到最小二乘法拟合直线的系数矩阵X    A*X = B
* param[in]  points 单个虚线轮廓内的所有中点
* param[out] 最小二乘法拟合出的直线的系数  直线 y= ax + b,返回结果即为(a,b)
*/
Mat getParameterMatrix(vector<Point2i> points)
{
	// 最小二乘拟合
	// 构建矩阵A
	Mat A = Mat::zeros(2, 2, CV_64FC1);
	for (int row = 0; row < A.rows; row++)
	{
		for (int col = 0; col < A.cols; col++)
		{
			for (int k = 0; k < points.size(); k++)
			{
				A.at<double>(row, col) = A.at<double>(row, col) + pow(points[k].x, row + col);
			}
		}
	}
	//构建矩阵B
	Mat B = Mat::zeros(2, 1, CV_64FC1);
	for (int row = 0; row < B.rows; row++)
	{
		for (int k = 0; k < points.size(); k++)
		{
			B.at<double>(row, 0) = B.at<double>(row, 0) + pow(points[k].x, row) * points[k].y;
		}
	}
	// A*X = B, 其中X为要求解的系数矩阵
	Mat X;
	solve(A, B, X, DECOMP_LU);
	return X;
}

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