What is the best way (in c/c++) to rotate an IplImage by 90 degrees? I would assume that there must be something better than transforming it using a matrix, but I can't seem to find anything other than that in the API and online.
+2
A:
Update for transposition:
You should use cvTranspose() or cv::transpose() because (as you rightly pointed out) it's more efficient. Again, I recommend upgrading to OpenCV2.0 since most of the cvXXX functions just convert IplImage* structures to Mat objects (no deep copies). If you stored the image in a Mat object, Mat.t() would return the transpose.
Any rotation:
You should use cvWarpAffine by defining the rotation matrix in the general framework of the transformation matrix. I would highly recommend upgrading to OpenCV2.0 which has several features as well as a Mat class which encapsulates matrices and images. With 2.0 you can use warpAffine to the above.
Jacob
2010-02-14 00:39:52
Thanks! Are you sure that's not less efficient than simply transposing the pixel array? The thing is, I wasn't sure if OpenCV even internally represents the image as an array of pixels, i.e. something like a 32-bit int*, and if it did, how to get a pointer to that array.
Ben Herila
2010-02-15 16:20:02
I'm sorry, I didn't read your question properly, I've updated my answer.
Jacob
2010-02-15 18:30:57
+1
A:
cvFlip (img, img, XorYaxe) - is the best way (in c/c++) to rotate an IplImage by 90 degrees.
Your Friend
2010-04-19 17:50:10
A:
Well I was looking for some details and didn't find any example. So I post here a transposeImage function which, I hope, will help others who are looking for a direct way to rotate 90° without loosing data.
IplImage* transposeImage(IplImage* image) {
IplImage *rotated = cvCreateImage(cvSize(image->height,image->width), IPL_DEPTH_8U,image->nChannels);
CvPoint2D32f center;
float center_val = (float)((image->width)-1) / 2;
center.x = center_val;
center.y = center_val;
CvMat *mapMatrix = cvCreateMat( 2, 3, CV_32FC1 );
cv2DRotationMatrix(center, 90, 1.0, mapMatrix);
cvWarpAffine(image, rotated, mapMatrix, CV_INTER_LINEAR + CV_WARP_FILL_OUTLIERS, cvScalarAll(0));
cvReleaseMat(&mapMatrix);
return rotated;
}
Question : Why
float center_val = (float)((image->width)-1) / 2;
?
Answer : Because it works :) The only center I found that doesn't translate image. Though if somebody had some explanation I would be interested.
Artem
2010-07-01 15:48:45
The reason for your math is probably due to integer division. What you want to do is: float center_val = (float)image->width / 2.0f.
Ben Herila
2010-07-02 22:22:24