在点上运行cv :: warpPerspective [英] Running cv::warpPerspective on points

问题描述

我正在图像上运行cv :: warpPerspective()函数，以及在源图像中获得结果图像某些点的位置的方法，这是我走了多远:

I'm running the cv::warpPerspective() function on a image and what to get the position of the some points of result image the I get in the source image, here how far I came :

 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，那么由于 cv :: warpPerspective 函数不会在** src或src2 **中获取其坐标， t以cv :: Point作为参数??

my problem is that if I select a point from dst how can get its coordinates in ** src or src2 ** since the cv::warpPerspective function doesn't take cv::Point as parameter ??

推荐答案

透视变换通过以下方式关联了两个点:

A perspective transform relates two points in the following manner:

[x']   [m00 m01 m02] [x]
[y'] = [m10 m11 m12] [y]
[1]    [m20 m21 m22] [1]

其中(x,y)是原始2D点坐标，而(x', y')是变换后的坐标.

Where (x,y) are the original 2D point coordinates, and (x', y') are the transformed coordinates.

在您的情况下，您知道(x', y')，并且想知道(x, y).这可以通过将已知点乘以变换矩阵的逆来实现:

In your case, you know (x', y'), and want to know (x, y). This can be achieved by multiplying the known point by the inverse of the transformation matrix:

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

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