近日考虑直线拟合相关的知识，大概有所了解，所以打算进行一些总结。

直线拟合常用的三种方法：

一、最小二乘法进行直线拟合

二、梯度下降法进行直线拟合

三、高斯牛顿，列-马算法进行直线拟合

 

一、使用最多的就是最小二乘法，这里我也对最小二乘法进行了一个总结。

1. 假设x是正确值，y存在误差。

根据上面两图的手推公式我们可以编写相关的代码了。此处我们借助opencv工具进行结果显示和分析。

void fitline3(){float b = 0.0f, k=0.0f;vector<Point>points;points.push_back(Point(27, 39));points.push_back(Point(8, 5));points.push_back(Point(8, 9));points.push_back(Point(16, 22));points.push_back(Point(44, 71));points.push_back(Point(35, 44));points.push_back(Point(43, 57));points.push_back(Point(19, 24));points.push_back(Point(27, 39));points.push_back(Point(37, 52));Mat src = Mat::zeros(400, 400, CV_8UC3);for (int i = 0; i < points.size(); i++){//在原图上画出点circle(src, points[i], 3, Scalar(0, 255, 0), 1, 8);}int n = points.size();double xx_sum = 0;double x_sum = 0;double y_sum = 0;double xy_sum = 0;for (int i = 0; i < n; i++){x_sum += points[i].x; //x的累加和y_sum += points[i].y; //y的累加和xx_sum += points[i].x * points[i].x; //x的平方累加和xy_sum += points[i].x * points[i].y; //x，y的累加和}k = (n*xy_sum - x_sum * y_sum) / (n*xx_sum - x_sum * x_sum); //根据公式求解kb = (-x_sum * xy_sum + xx_sum*y_sum) / (n*xx_sum - x_sum * x_sum);//根据公式求解bprintf("k = %f, b = %f\n", k, b); //k = 1.555569, b = -4.867031cv::Point first = { 5, int(k * 5 + b) }, second = { int((400 - b) / k), 400 };cv::line(src, first, second, cv::Scalar(0, 0, 255), 2);cv::imshow("name", src);cv::waitKey(0);
}

上面求解出来的结果是k = 1.555569, b = -4.867031。

图像显示结果：

 

 

下面我们使用opencv自带的函数求解k和b值。此处代码来自互联网。

void fitline1()
{vector<Point>points;//(27 39) (8 5) (8 9) (16 22) (44 71) (35 44) (43 57) (19 24) (27 39) (37 52)points.push_back(Point(27, 39));points.push_back(Point(8, 5));points.push_back(Point(8, 9));points.push_back(Point(16, 22));points.push_back(Point(44, 71));points.push_back(Point(35, 44));points.push_back(Point(43, 57));points.push_back(Point(19, 24));points.push_back(Point(27, 39));points.push_back(Point(37, 52));Mat src = Mat::zeros(400, 400, CV_8UC3);for (int i = 0; i < points.size(); i++){//在原图上画出点circle(src, points[i], 3, Scalar(0, 0, 255), 1, 8);}//构建A矩阵 int N = 2;Mat A = Mat::zeros(N, N, 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B矩阵Mat B = Mat::zeros(N, 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=BMat X;//cout << A << endl << B << endl;solve(A, B, X, DECOMP_LU);cout << X << endl;vector<Point>lines;for (int x = 0; x < src.size().width; x++){				// y = b + ax;double y = X.at<double>(0, 0) + X.at<double>(1, 0)*x; //b = -4.867031, k = 1.555569,printf("b = %f, k = %f, (%d,%lf)\n", X.at<double>(0, 0), X.at<double>(1, 0), x, y);lines.push_back(Point(x, y));}polylines(src, lines, false, Scalar(255, 0, 0), 1, 8);imshow("src", src);//imshow("src", A);waitKey(0);
}

求解得出对应的k = 1.555569, b = -4.867031。

这两个结果值是一样的。说明我们的推导结果是正确的。

 

2. 上面的公式推导是假设了x是正确值，y存在误差，是普通的一元线性回归。

此处将离散的点拟合为 ax + by +c = 0型直线。

即假设每个点的x，y坐标的误差都是 符合0均值的正态分布的。

有如下代码：

void fitline2(){float a = 0.0f, b = 0.0f, c = 0.0f;vector<Point>points;points.push_back(Point(27, 39));points.push_back(Point(8, 5));points.push_back(Point(8, 9));points.push_back(Point(16, 22));points.push_back(Point(44, 71));points.push_back(Point(35, 44));points.push_back(Point(43, 57));points.push_back(Point(19, 24));points.push_back(Point(27, 39));points.push_back(Point(37, 52));Mat src = Mat::zeros(400, 400, CV_8UC3);for (int i = 0; i < points.size(); i++){//在原图上画出点circle(src, points[i], 3, Scalar(0, 255, 0), 1, 8);}int size = points.size();double x_mean = 0;double y_mean = 0;for (int i = 0; i < size; i++){x_mean += points[i].x;y_mean += points[i].y;}x_mean /= size;y_mean /= size; //至此，计算出了 x y 的均值double Dxx = 0, Dxy = 0, Dyy = 0;for (int i = 0; i < size; i++){Dxx += (points[i].x - x_mean) * (points[i].x - x_mean);Dxy += (points[i].x - x_mean) * (points[i].y - y_mean);Dyy += (points[i].y - y_mean) * (points[i].y - y_mean);}double lambda = ((Dxx + Dyy) - sqrt((Dxx - Dyy) * (Dxx - Dyy) + 4 * Dxy * Dxy)) / 2.0;double den = sqrt(Dxy * Dxy + (lambda - Dxx) * (lambda - Dxx));a = Dxy / den;b = (lambda - Dxx) / den;c = -a * x_mean - b * y_mean;printf("a = %f, b = %f, c = %f\n", a, b, c);printf("k = %f, b = %f\n", - a/b, -c/b); //k = 1.593043, b = -5.856340float k = -a / b;b = -c / b;cv::Point first = { 5, int(k * 5 + b) }, second = { int((400 - b) / k), 400 };cv::line(src, first, second, cv::Scalar(0, 0, 255), 2);cv::imshow("name", src);cv::waitKey(0);}

求解结果为k = 1.593043, b = -5.856340。（对比发现此方法和方法一结果还是存在差异的）

显示结果：

 

二、梯度下降法进行直线拟合

下图是我的手写推导公式：

推导公式中alpha为学习率，theta0和theta1分别是斜率k和截距b。m是为了求平均损失，2是为了求偏导方便。

根据推导公式写出如下代码：

//梯度下降法进行线性拟合
// y = theta0 * x + theta1
void fitline4(){const int m = 10;double Train_set_x[m] = { 27, 8, 8, 16, 44, 35, 43, 19, 27, 37 };double Train_set_y[m] = { 39, 5, 9, 22, 71, 44, 57, 24, 39, 52 };double theta0 = 0.0, theta1 = 0.0, alpha = 0.002, error = 1e-8; //alpha是学习率，error是结束误差，theta0就是k，theta1就是b。double tmp_theta0 = theta0, tmp_theta1 = theta1;double sum_theta0 = 0.0, sum_theta1 = 0.0;while (1){sum_theta0 = 0.0, sum_theta1 = 0.0;for (size_t i = 0; i < m; i++) {sum_theta0 += (theta0 * Train_set_x[i] + theta1 - Train_set_y[i])*Train_set_x[i]; //累加和sum_theta1 += (theta0 * Train_set_x[i] + theta1 - Train_set_y[i]);  //累加和}theta0 = theta0 - alpha * (1.0 / m) * sum_theta0; //k更新公式theta1 = theta1 - alpha * (1.0 / m) * sum_theta1; //b更新公式printf("k=%lf, b=%lf\n", theta0, theta1);if (abs(theta0 - tmp_theta0) < error && abs(theta1 - tmp_theta1) < error){ //break;}tmp_theta0 = theta0;tmp_theta1 = theta1;}
}

求解结果为k=1.555569, b=-4.867004。与最小二乘法的结果极为相似。

 

 

三、高斯牛顿，列-马算法