Issue
I want to retrieve the rgb of a pixel in the image. But the location is not integer location but real values (x,y). I want a bilinear interpolated value. How could I do it opencv?
Thanks a lot
Solution
There is no simple function for subpixel access but I can suggest you few options:
Use getRectSubPix and extract 1 pixel region:
cv::Vec3b getColorSubpix(const cv::Mat& img, cv::Point2f pt) { cv::Mat patch; cv::getRectSubPix(img, cv::Size(1,1), pt, patch); return patch.at<cv::Vec3b>(0,0); }
Use more flexible but less precise remap with one-pixel map:
cv::Vec3b getColorSubpix(const cv::Mat& img, cv::Point2f pt) { cv::Mat patch; cv::remap(img, patch, cv::Mat(1, 1, CV_32FC2, &pt), cv::noArray(), cv::INTER_LINEAR, cv::BORDER_REFLECT_101); return patch.at<cv::Vec3b>(0,0); }
Implement bilinear interpolation yourself, as it is not a rocket science:
cv::Vec3b getColorSubpix(const cv::Mat& img, cv::Point2f pt) { assert(!img.empty()); assert(img.channels() == 3); int x = (int)pt.x; int y = (int)pt.y; int x0 = cv::borderInterpolate(x, img.cols, cv::BORDER_REFLECT_101); int x1 = cv::borderInterpolate(x+1, img.cols, cv::BORDER_REFLECT_101); int y0 = cv::borderInterpolate(y, img.rows, cv::BORDER_REFLECT_101); int y1 = cv::borderInterpolate(y+1, img.rows, cv::BORDER_REFLECT_101); float a = pt.x - (float)x; float c = pt.y - (float)y; uchar b = (uchar)cvRound((img.at<cv::Vec3b>(y0, x0)[0] * (1.f - a) + img.at<cv::Vec3b>(y0, x1)[0] * a) * (1.f - c) + (img.at<cv::Vec3b>(y1, x0)[0] * (1.f - a) + img.at<cv::Vec3b>(y1, x1)[0] * a) * c); uchar g = (uchar)cvRound((img.at<cv::Vec3b>(y0, x0)[1] * (1.f - a) + img.at<cv::Vec3b>(y0, x1)[1] * a) * (1.f - c) + (img.at<cv::Vec3b>(y1, x0)[1] * (1.f - a) + img.at<cv::Vec3b>(y1, x1)[1] * a) * c); uchar r = (uchar)cvRound((img.at<cv::Vec3b>(y0, x0)[2] * (1.f - a) + img.at<cv::Vec3b>(y0, x1)[2] * a) * (1.f - c) + (img.at<cv::Vec3b>(y1, x0)[2] * (1.f - a) + img.at<cv::Vec3b>(y1, x1)[2] * a) * c); return cv::Vec3b(b, g, r); }
Answered By - Andrey Kamaev Answer Checked By - Willingham (PHPFixing Volunteer)
0 Comments:
Post a Comment
Note: Only a member of this blog may post a comment.