最小二乘法擬合直線
概念：最小二乘法多項式直線擬合，根據給定的點，求出它的函數y=f(x)，當然求得准確的函數是不太可能的，但是我們能求出它的近似曲線y=φ(x)
原理
假設有點 , I = 1,2,3,……n,求近似曲線y=φ(x)，並且使得y=φ(x)與y=f(x)的平方偏差和最小，偏差
其中我們要找到一組最好的a b ，“最好的”就是要使選出的a b能使得所有的誤差達到最小化。
在此要注意以下，y=ax+b 這里面的未知量是什么，很自然的說法是x y,實際上並不是，我們不用去解這個x和y ，因為x和y已經是給定的值了，當我們在找這條直線的時候，我們實際上並不關心x的值有多好，我們要的就是a 和b這兩個變量，它們可以描述x和y之間的關系，我們就是在試圖找出那條最適合的直線所對應的a和b。
可以看到最小二乘法對各個變量求偏導，使得偏導值為0，即可得到最小值，因為e是關於a b的函數，導數為0的點必定是最小值，進入正題
分別對 a b求偏導可以得到：
Halcon最小二乘法擬合直線
1 首先隨機生成一組數據 2
3 Mx:=[100:10:500] 4
5 tuple_length(Mx,len) 6
7 tuple_gen_const(len,5,r) 8
9 Ma:=2
10
11 Mb:=40
12
13 tuple_rand(len , noise) 14
15 My:= Ma *Mx + Mb*noise 16
17 gen_circle(ContCircle, My, Mx, r)
1 接下來用矩陣進行最小二乘求解 2
3 * y = ax + b 4
5 * y1 = ax1 + b 6
7 * y2 = ax2 + b 8
9 * ... ....... 10
11 * yn = ax + b
1 create_matrix(len,1,My,y) 2
3 create_matrix(len,2,1,x) 4
5 set_value_matrix(x, [0:len-1], gen_tuple_const(len, 0),Mx) 6
7
8 * XT 代表X的轉置 inv(*)代表*的逆 9
10 * x beta = y 11
12 * xT x beta = xT y 13
14 * beta = inv( xT x) xT y 15
16 mult_matrix(x,x,'ATB',xtx) 17
18 mult_matrix(x,y,'ATB',xty) 19
20 invert_matrix(xtx,'general', 0, invxtx) 21
22 mult_matrix(invxtx,xty,'AB', beta) 23
24 get_full_matrix(beta, Values) 25
26 Newy:=Values[0] * [10,800] + Values[1] 27
28 gen_contour_polygon_xld(Contour, Newy, [10,800])
OpenCV最小二乘法擬合直線
1 #include<iostream>
2 #include<opencv2\opencv.hpp>
3 using namespace std; 4 using namespace cv; 5
6
7 int main() 8 { 9 vector<Point>points; 10 //(27 39) (8 5) (8 9) (16 22) (44 71) (35 44) (43 57) (19 24) (27 39) (37 52)
11
12 points.push_back(Point(27, 39)); 13 points.push_back(Point(8, 5)); 14 points.push_back(Point(8, 9)); 15 points.push_back(Point(16, 22)); 16 points.push_back(Point(44, 71)); 17 points.push_back(Point(35, 44)); 18 points.push_back(Point(43, 57)); 19 points.push_back(Point(19, 24)); 20 points.push_back(Point(27, 39)); 21 points.push_back(Point(37, 52)); 22 Mat src = Mat::zeros(400, 400, CV_8UC3); 23
24 for (int i = 0;i < points.size();i++) 25 { 26 //在原圖上畫出點
27 circle(src, points[i], 3, Scalar(0, 0, 255), 1, 8); 28 } 29 //構建A矩陣
30 int N = 2; 31 Mat A = Mat::zeros(N, N, CV_64FC1); 32
33 for (int row = 0;row < A.rows;row++) 34 { 35 for (int col = 0; col < A.cols;col++) 36 { 37 for (int k = 0;k < points.size();k++) 38 { 39 A.at<double>(row, col) = A.at<double>(row, col) + pow(points[k].x, row + col); 40 } 41 } 42 } 43 //構建B矩陣
44 Mat B = Mat::zeros(N, 1, CV_64FC1); 45 for (int row = 0;row < B.rows;row++) 46 { 47
48 for (int k = 0;k < points.size();k++) 49 { 50 B.at<double>(row, 0) = B.at<double>(row, 0) + pow(points[k].x, row)*points[k].y; 51 } 52 } 53 //A*X=B
54 Mat X; 55 //cout << A << endl << B << endl;
56 solve(A, B, X,DECOMP_LU); 57 cout << X << endl; 58 vector<Point>lines; 59 for (int x = 0;x < src.size().width;x++) 60 { // y = b + ax;
61 double y = X.at<double>(0, 0) + X.at<double>(1, 0)*x; 62 printf("(%d,%lf)\n", x, y); 63 lines.push_back(Point(x, y)); 64 } 65 polylines(src, lines, false, Scalar(255, 0, 0), 1, 8); 66 imshow("src", src); 67
68 //imshow("src", A);
69 waitKey(0); 70 return 0; 71 }
C++最小二乘法擬合直線
1 #include 2 #include 3 #include 4
5 using namespace std; 6
7 int main(int argc, char *argv[]) 8 { 9 int num = 0; 10
11 cout << " Input how many numbers you want to calculate:"; 12 cin >> num; 13
14 valarray data_x(num); 15 valarray data_y(num); 16
17 while( num ) 18 { 19 cout << "Input the "<< num <<" of x:"; 20 cin >> data_x[num-1]; 21 cout << "Input the "<< num <<" of y:"; 22 cin >> data_y[num-1]; 23 num--; 24 } 25
26 double A =0.0; 27 double B =0.0; 28 double C =0.0; 29 double D =0.0; 30
31 A = (data_x*data_x).sum(); 32 B = data_x.sum(); 33 C = (data_x*data_y).sum(); 34 D = data_y.sum(); 35
36 double k,b,tmp =0; 37 if(tmp=(A*data_x.size()-B*B)) 38 { 39 k = (C*data_x.size()-B*D)/tmp; 40 b = (A*D-C*B)/tmp; 41 } 42
43 else
44 { 45 k=1; 46 b=0; 47 } 48
49 cout <<"k="< cout <<"b="<</p>
50
51 return 0; 52 }