使用Homography链接两个图片像素 [英] Link two pictures pixels using Homography

问题描述

我有两张相同的图片，但第二张失真并且尺寸不同.我希望能够在第一张图片上选择一个点，并在第二张图片上找到完全相同的点. 图片的同步将由每张图片上的4点组成.

I have two pictures that are the same but the second is distorted and have different dimensions. I want to be able to select a point on the first picture and find the exact same point on the second picture. The synchronisation of the pictures will be made by 4 points on each picture.

示例:
普通图片

Example:
Normal picture

图片失真

结果将是一个函数，该函数从第一张图片获取X，Y点，并返回与第二张图片上完全相同的Point相对应的X'，Y'.

The result will be a function getting X,Y Point from first picture and returning X',Y' corresponding to the exact same Point on the second picture.

我支持OpenCV的findHomography()函数，该函数返回Homography矩阵，但在所有示例中，该矩阵都用于通过warpPerspective()扭曲第二张图像，这不是我想要的.我只想要两张图片之间的相应坐标/像素.

I foud OpenCV's findHomography() function which return a Homography matrix but in all the examples this matrix is used to distort the second image with warpPerspective() which is not what I want. I just want the corresponding coordinates / pixels between the two pictures.

有关如何在C#中执行此操作的任何提示? (是否使用OpenCV)

Any tips on how to do that in C# ? (With OpenCV or not)

要点示例:

普通图片:[250，25]，[250，287]，[399，287]，[400，28]

Normal picture : [250, 25] , [250, 287] , [399, 287] , [400, 28]

目标图片:[0，0]，[2098，29538]，[19127，28164]，[17097，-1200]

Target picture : [0, 0] , [2098, 29538] , [19127, 28164] , [17097, -1200]

使用OPEN CV

OpenCvSharp.Mat hCv = OpenCvSharp.Cv2.FindHomography(srcPoints, dstPoints);
OpenCvSharp.Point2d[] resultCv = OpenCvSharp.Cv2.PerspectiveTransform(targetCv, hCv);

推荐答案

感谢Alexander Reynolds的评论，我终于使用了 OpenCvSharp 3 库. 您可以使用以下方法获得单应性矩阵:

Thanks to Alexander Reynolds comment, I finally used the OpenCvSharp 3 library. You get the homography matrix using:

homographyMatrix = OpenCvSharp.Cv2.FindHomography(originalPicturePoints, targetPicturePoints);

，然后从原始图片中获取目标图片上的任何点:

and then to get any point on the target picture from the original picture:

targetPoint = OpenCvSharp.Cv2.PerspectiveTransform(orignalPoint, homographyMatrix);

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