opencv练习18椭圆、直线拟合

opencv练习18椭圆、直线拟合

使用的函数为 cv2.ellipse()，返回值其实就是旋转边界矩形的内切圆。

ellipse = cv2.fitEllipse(cnt)im = cv2.ellipse(im,ellipse,(0,255,0),2)

直线拟合

我们可以根据一组点拟合出一条直线，同样我们也可以为图像中的白色点拟合出一条直线。

rows,cols = img.shape[:2]#cv2.fitLine(points, distType, param, reps, aeps[, line ]) → line#points – Input vector of 2D or 3D points, stored in std::vector<> or Mat.#line#4 elements (likeVec4f) - (vx, vy, x0, y0), where (vx, vy) is a normalized#vector collinear to the line and (x0, y0) is a point on the line. In case of#3D fitting, it should be a vector of 6 elements (like Vec6f) - (vx, vy, vz,#x0, y0, z0), where (vx, vy, vz) is a normalized vector collinear to the line#and (x0, y0, z0) is a point on the line.#distType – Distance used by the M-estimator#distType=CV_DIST_L2#ρ(r) = r2 /2 (the simplest and the fastest least-squares method)#param – Numerical parameter ( C ) for some types of distances. If it is 0, an optimal value#is chosen.#reps – Sufficient accuracy for the radius (distance between the coordinate origin and the#line).#aeps – Sufficient accuracy for the angle. 0.01 would be a good default value for reps and#aeps.[vx,vy,x,y] = cv2.fitLine(cnt, cv2.DIST_L2,0,0.01,0.01)lefty = int((-x*vy/vx) + y)righty = int(((cols-x)*vy/vx)+y)img = cv2.line(img,(cols-1,righty),(0,lefty),(0,255,0),2)

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