[英]Getting NaN for image angle in emgucv with c#
我参考了以下链接来开发以下 csharp 代码来检测图像的角度。 https://stackoverflow.com/a/34285205/7805023
Image<Gray, byte> imgout = imgInput.Convert<Gray, byte>().Not().ThresholdBinary(new
Gray(50), new Gray(255));
VectorOfVectorOfPoint contours = new VectorOfVectorOfPoint();
Mat hier = new Mat();
CvInvoke.FindContours(imgout, contours, hier, Emgu.CV.CvEnum.RetrType.External, Emgu.CV.CvEnum.ChainApproxMethod.ChainApproxSimple);
for (int i = 0; i <= 1; i++)
{
Rectangle rect = CvInvoke.BoundingRectangle(contours[i]);
RotatedRect box = CvInvoke.MinAreaRect(contours[i]);
PointF[] Vertices = box.GetVertices();
PointF point = box.Center;
PointF edge1 = new PointF(Vertices[1].X - Vertices[0].X,
Vertices[1].Y - Vertices[0].Y);
PointF edge2 = new PointF(Vertices[2].X - Vertices[1].X,Vertices[2].Y - Vertices[1].Y);
double edge1Magnitude = Math.Sqrt(Math.Pow(edge1.X, 2) + Math.Pow(edge1.Y, 2));
double edge2Magnitude = Math.Sqrt(Math.Pow(edge2.X, 2) + Math.Pow(edge2.Y, 2));
PointF primaryEdge = edge1Magnitude > edge2Magnitude ? edge1 : edge2;
PointF reference = new PointF(Vertices[1].X, Vertices[0].Y);
double thetaRads = Math.Acos((primaryEdge.X * reference.X) + (primaryEdge.Y * reference.Y))/(edge1Magnitude* edge2Magnitude);
double thetaDeg = thetaRads * 180 / Math.PI;
}
我得到的 NaN 为 output 的角度。 请go通过代码让我知道我做错了什么?
我使用 emgucv 进行图像处理。
任何帮助将不胜感激。
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