我在图像上运行cv :: warpPerspective()函数,以及在源图像中获取结果图像的某些点的位置,这里我走了多远:
int main (){
cv::Point2f srcQuad[4],dstQuad[4];
cv::Mat warpMatrix;
cv::Mat src, dst,src2;
src = cv::imread("card.jpg",1);
srcQuad[0].x = 0; //src Top left
srcQuad[0].y = 0;
srcQuad[1].x = src.cols - 1; //src Top right
srcQuad[1].y = 0;
srcQuad[2].x = 0; //src Bottom left
srcQuad[2].y = src.rows - 1;
srcQuad[3].x = src.cols -1; //src Bot right
srcQuad[3].y = src.rows - 1;
dstQuad[0].x = src.cols*0.05; //dst Top left
dstQuad[0].y = src.rows*0.33;
dstQuad[1].x = src.cols*0.9; //dst Top right
dstQuad[1].y = src.rows*0.25;
dstQuad[2].x = src.cols*0.2; //dst Bottom left
dstQuad[2].y = src.rows*0.7;
dstQuad[3].x = src.cols*0.8; //dst Bot right
dstQuad[3].y = src.rows*0.9;
warpMatrix =cv::getPerspectiveTransform(srcQuad,dstQuad);
cv::warpPerspective(src,dst,warpMatrix,src.size());
cv::imshow("source", src);
cv::imshow("destination", dst);
cv::warpPerspective(dst,src2,warpMatrix,dst.size(),CV_WARP_INVERSE_MAP);
cv::imshow("srouce 2 " , src2);
cv::waitKey();
return 0;
我的问题是,如果我从 dst 中选择一个点,怎么可以在** src或src2 **中得到它的坐标,因为 cv :: warpPerspective 函数没有'把cv :: Point作为参数?
答案 0 :(得分:6)
您需要perspectiveTransform(适用于Points矢量)而不是warpPerspective。 采取warpMatrix的反转;你可能需要调整最后一栏。
vector<Point2f> dstPoints, srcPoints;
dstPoints.push_back(Point2f(1,1));
cv::perspectiveTransform(dstPoints,srcPoints,warpMatrix.inv());
答案 1 :(得分:3)
透视变换以下列方式关联两点:
[x'] [m00 m01 m02] [x]
[y'] = [m10 m11 m12] [y]
[1] [m20 m21 m22] [1]
(x,y)
是原始的二维点坐标,(x', y')
是变换后的坐标。
在您的情况下,您知道(x', y')
,并想知道(x, y)
。这可以通过将已知点乘以变换矩阵的逆来实现:
cv::Matx33f warp = warpMatrix; // cv::Matx is much more useful for math
cv::Point2f warped_point = dstQuad[3]; // I just use dstQuad as an example
cv::Point3f homogeneous = warp.inv() * warped_point;
cv::Point2f result(homogeneous.x, homogeneous.y); // Drop the z=1 to get out of homogeneous coordinates
// now, result == srcQuad[3], which is what you wanted