从 OpenCV 中的旋转图像旋转回点 [英] Rotating back points from a rotated image in OpenCV

问题描述

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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