如何从OpenCV的fitEllipse函数中获取椭圆系数?

时间:2015-09-26 04:41:36

标签: opencv ellipse

我想从一张照片中提取红球并在图片中获取检测到的椭圆矩阵。

这是我的例子: enter image description here

我对图片进行阈值处理,使用findContour()函数找到红球的轮廓,并使用fitEllipse()来拟合椭圆。

但我想要的是得到这个椭圆的系数。因为fitEllipse()返回一个旋转矩形(RotatedRect),所以我需要重新编写这个函数。

一个椭圆可以表示为Ax ^ 2 + By ^ 2 + Cxy + Dx + Ey + F = 0;如果F为1(构造椭圆矩阵),我想得到u =(A,B,C,D,E,F)或u =(A,B,C,D,E)。

我读了fitEllipse()的源代码,共有三个SVD过程,我想我可以从那三个SVD过程的结果中得到上述系数。但我很困惑每个SVD过程的每个结果(变量cv :: Mat x)代表什么,为什么这里有三个SVD?

这是这个功能:

cv::RotatedRect cv::fitEllipse( InputArray _points )
{
   Mat points = _points.getMat();
   int i, n = points.checkVector(2);
   int depth = points.depth();
   CV_Assert( n >= 0 && (depth == CV_32F || depth == CV_32S));

   RotatedRect box;

   if( n < 5 )
        CV_Error( CV_StsBadSize, "There should be at least 5 points to fit the ellipse" );

    // New fitellipse algorithm, contributed by Dr. Daniel Weiss
    Point2f c(0,0);
    double gfp[5], rp[5], t;
    const double min_eps = 1e-8;
    bool is_float = depth == CV_32F;
    const Point* ptsi = points.ptr<Point>();
    const Point2f* ptsf = points.ptr<Point2f>();

    AutoBuffer<double> _Ad(n*5), _bd(n);
    double *Ad = _Ad, *bd = _bd;

    // first fit for parameters A - E
    Mat A( n, 5, CV_64F, Ad );
    Mat b( n, 1, CV_64F, bd );
    Mat x( 5, 1, CV_64F, gfp );

    for( i = 0; i < n; i++ )
    {
        Point2f p = is_float ? ptsf[i] : Point2f((float)ptsi[i].x, (float)ptsi[i].y);
        c += p;
    }
    c.x /= n;
    c.y /= n;

    for( i = 0; i < n; i++ )
    {
        Point2f p = is_float ? ptsf[i] : Point2f((float)ptsi[i].x, (float)ptsi[i].y);
        p -= c;

        bd[i] = 10000.0; // 1.0?
        Ad[i*5] = -(double)p.x * p.x; // A - C signs inverted as proposed by APP
        Ad[i*5 + 1] = -(double)p.y * p.y;
        Ad[i*5 + 2] = -(double)p.x * p.y;
        Ad[i*5 + 3] = p.x;
        Ad[i*5 + 4] = p.y;
    }

    solve(A, b, x, DECOMP_SVD);

    // now use general-form parameters A - E to find the ellipse center:
    // differentiate general form wrt x/y to get two equations for cx and cy
    A = Mat( 2, 2, CV_64F, Ad );
    b = Mat( 2, 1, CV_64F, bd );
    x = Mat( 2, 1, CV_64F, rp );
    Ad[0] = 2 * gfp[0];
    Ad[1] = Ad[2] = gfp[2];
    Ad[3] = 2 * gfp[1];
    bd[0] = gfp[3];
    bd[1] = gfp[4];
    solve( A, b, x, DECOMP_SVD );

    // re-fit for parameters A - C with those center coordinates
    A = Mat( n, 3, CV_64F, Ad );
    b = Mat( n, 1, CV_64F, bd );
    x = Mat( 3, 1, CV_64F, gfp );
    for( i = 0; i < n; i++ )
    {
        Point2f p = is_float ? ptsf[i] : Point2f((float)ptsi[i].x, (float)ptsi[i].y);
        p -= c;
        bd[i] = 1.0;
        Ad[i * 3] = (p.x - rp[0]) * (p.x - rp[0]);
        Ad[i * 3 + 1] = (p.y - rp[1]) * (p.y - rp[1]);
        Ad[i * 3 + 2] = (p.x - rp[0]) * (p.y - rp[1]);
    }
    solve(A, b, x, DECOMP_SVD);

    // store angle and radii
    rp[4] = -0.5 * atan2(gfp[2], gfp[1] - gfp[0]); // convert from APP angle usage
    if( fabs(gfp[2]) > min_eps )
        t = gfp[2]/sin(-2.0 * rp[4]);
    else // ellipse is rotated by an integer multiple of pi/2
        t = gfp[1] - gfp[0];
    rp[2] = fabs(gfp[0] + gfp[1] - t);
    if( rp[2] > min_eps )
        rp[2] = std::sqrt(2.0 / rp[2]);
    rp[3] = fabs(gfp[0] + gfp[1] + t);
    if( rp[3] > min_eps )
        rp[3] = std::sqrt(2.0 / rp[3]);

    box.center.x = (float)rp[0] + c.x;
    box.center.y = (float)rp[1] + c.y;
    box.size.width = (float)(rp[2]*2);
    box.size.height = (float)(rp[3]*2);
    if( box.size.width > box.size.height )
    {
        float tmp;
        CV_SWAP( box.size.width, box.size.height, tmp );
        box.angle = (float)(90 + rp[4]*180/CV_PI);
    }
    if( box.angle < -180 )
        box.angle += 360;
    if( box.angle > 360 )
        box.angle -= 360;

    return box;
}

源代码链接:https://github.com/Itseez/opencv/blob/master/modules/imgproc/src/shapedescr.cpp

2 个答案:

答案 0 :(得分:15)

函数fitEllipse返回包含椭圆所有参数的RotatedRect

椭圆由5个参数定义:

  • xc :中心的x坐标
  • yc :中心的y坐标
  • a :主要半轴
  • b :次要半轴
  • theta :旋转角度

您可以获取以下参数:

RotatedRect e = fitEllipse(points);

float xc    = e.center.x;
float yc    = e.center.y;
float a     = e.size.width  / 2;    // width >= height
float b     = e.size.height / 2;
float theta = e.angle;              // in degrees

您可以使用ellipse

绘制带有RotatedRect函数的椭圆
ellipse(image, e, Scalar(0,255,0)); 

或等效地使用椭圆参数:

ellipse(res, Point(xc, yc), Size(a, b), theta, 0.0, 360.0, Scalar(0,255,0));

如果您需要隐式方程的系数值,您可以这样做(来自Wikipedia):

enter image description here

因此,您可以从RotatedRect获取所需的参数,而不需要更改函数fitEllipsesolve函数用于解决线性系统或最小二乘问题。使用SVD分解方法可以过度定义系统和/或矩阵src1可以是单数。

有关算法的更多详细信息,您可以看到提出此拟合椭圆方法的paper of Fitzgibbon

答案 1 :(得分:0)

这是一些对我有用的代码,我基于此线程上的其他响应。

def getConicCoeffFromEllipse(e):
    # ellipse(Point(xc, yc),Size(a, b), theta)
    xc = e[0][0]
    yc = e[0][1]
    a = e[1][0]/2
    b = e[1][1]/2
    theta = math.radians(e[2])
    # See https://en.wikipedia.org/wiki/Ellipse
    # Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 is the equation
    A = a*a*math.pow(math.sin(theta),2) + b*b*math.pow(math.cos(theta),2)
    B = 2*(b*b - a*a)*math.sin(theta)*math.cos(theta)
    C = a*a*math.pow(math.cos(theta),2) + b*b*math.pow(math.sin(theta),2)
    D = -2*A*xc - B*yc
    E = -B*xc - 2*C*yc
    F = A*xc*xc + B*xc*yc + C*yc*yc - a*a*b*b
    coef = np.array([A,B,C,D,E,F]) / F
    return coef

def getConicMatrixFromCoeff(c):
    C = np.array([[c[0], c[1]/2, c[3]/2], # [ a, b/2, d/2 ]
                 [c[1]/2, c[2], c[4]/2],  # [b/2,  c, e/2 ]
                 [c[3]/2, c[4]/2, c[5]]])  # [d/2], e/2, f ]
    return C