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。
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
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:
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
:
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):
因此,您可以从RotatedRect
获取所需的参数,而不需要更改函数fitEllipse
。
solve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