从 OpenCV 中的旋转图像旋转回点
我在轮换方面遇到了麻烦.我想做的是:
I’m having troubles with rotation. What I want to do is this:
- 旋转图像
- 检测旋转图像上的特征(点)
- 旋转点，这样我就可以得到与初始图像相对应的点坐标
我在第三步上有点卡住了.
I’m a bit stuck on the third step.
我设法使用以下代码旋转图像:
I manage to rotated the image with the following code:
cv::Mat M(2, 3, CV_32FC1);
cv::Point2f center((float)dst_img.rows / 2.0f, (float)dst_img.cols / 2.0f);
M = cv::getRotationMatrix2D(center, rotateAngle, 1.0);
cv::warpAffine(dst_img, rotated, M, cv::Size(rotated.cols, rotated.rows));
我尝试使用以下代码旋转点:
I try to rotate back the points with this code:
float xp = r.x * std::cos( PI * (-rotateAngle) / 180 ) - r.y * sin(PI * (rotateAngle) / 180);
float yp = r.x * sin(PI * (-rotateAngle) / 180) + r.y * cos(PI * (rotateAngle) / 180);
工作并不费劲，但图像上的点数不能很好地反映.有一个偏移量.
It is not to fare to be working but the points don’t go back well on the image. There is an offset.
感谢您的帮助
推荐答案
如果 M 是你从 cv::getRotationMatrix2D 得到的旋转矩阵，旋转一个 cv::Point p 使用这个矩阵你可以这样做:
If M is the rotation matrix you get from cv::getRotationMatrix2D, to rotate a cv::Point p with this matrix you can do this:
cv::Point result;
result.x = M.at<double>(0,0)*p.x + M.at<double>(0,1)*p.y + M.at<double>(0,2);
result.y = M.at<double>(1,0)*p.x + M.at<double>(1,1)*p.y + M.at<double>(1,2);
如果要旋转一个点，生成M的逆矩阵或使用cv::getRotationMatrix2D(center, -rotateAngle, scale)生成矩阵用于反向旋转.
If you want to rotate a point back, generate the inverse matrix of M or use cv::getRotationMatrix2D(center, -rotateAngle, scale) to generate a matrix for reverse rotation.
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