Question

这可能已经得到了回答，但我迫切需要一个答案。我想在Android中使用OpenCV找到图像中最大的正方形或矩形。我找到的所有解决方案都是C ++，我尝试转换它但它不起作用，我不知道我错在哪里。

private Mat findLargestRectangle(Mat original_image) {
    Mat imgSource = original_image;

    Imgproc.cvtColor(imgSource, imgSource, Imgproc.COLOR_BGR2GRAY);
    Imgproc.Canny(imgSource, imgSource, 100, 100);

    //I don't know what to do in here

    return imgSource;
}

我在这里要完成的是创建一个基于原始图像中找到的最大正方形的新图像（返回值Mat图像）。

这就是我想要发生的事情：

1 http://img14.imageshack.us/img14/7855/s7zr.jpg

我也可以获得最大广场的四个点，我想我可以从那里拿走它。但如果我能够返回裁剪后的图像会更好。

Answer 1

在精灵之后

1-您需要使用gaussian blur和find all the contours

降低噪音

2-查找并列出所有contours' areas。

3-最大的轮廓将只是绘画。

4-现在使用perpective transformation将您的形状转换为矩形。

检查sudoku solver examples以查看类似的处理问题。（最大轮廓+透视）

Answer 2

我花了一些时间将C ++代码转换为Java，但这里是： - ）

警告！原始代码，完全没有优化，全部。

如果受伤或致命事故，我不承担任何责任

    List<MatOfPoint> squares = new ArrayList<MatOfPoint>();

    public Mat onCameraFrame(CvCameraViewFrame inputFrame) {

        if (Math.random()>0.80) {

            findSquares(inputFrame.rgba().clone(),squares);

        }

        Mat image = inputFrame.rgba();

        Imgproc.drawContours(image, squares, -1, new Scalar(0,0,255));

        return image;
    }

    int thresh = 50, N = 11;

 // helper function:
 // finds a cosine of angle between vectors
 // from pt0->pt1 and from pt0->pt2
    double angle( Point pt1, Point pt2, Point pt0 ) {
            double dx1 = pt1.x - pt0.x;
            double dy1 = pt1.y - pt0.y;
            double dx2 = pt2.x - pt0.x;
            double dy2 = pt2.y - pt0.y;
            return (dx1*dx2 + dy1*dy2)/Math.sqrt((dx1*dx1 + dy1*dy1)*(dx2*dx2 + dy2*dy2) + 1e-10);
    }

 // returns sequence of squares detected on the image.
 // the sequence is stored in the specified memory storage
 void findSquares( Mat image, List<MatOfPoint> squares )
 {

     squares.clear();

     Mat smallerImg=new Mat(new Size(image.width()/2, image.height()/2),image.type());

     Mat gray=new Mat(image.size(),image.type());

     Mat gray0=new Mat(image.size(),CvType.CV_8U);

     // down-scale and upscale the image to filter out the noise
     Imgproc.pyrDown(image, smallerImg, smallerImg.size());
     Imgproc.pyrUp(smallerImg, image, image.size());

     // find squares in every color plane of the image
     for( int c = 0; c < 3; c++ )
     {

         extractChannel(image, gray, c);

         // try several threshold levels
         for( int l = 1; l < N; l++ )
         {
             //Cany removed... Didn't work so well


             Imgproc.threshold(gray, gray0, (l+1)*255/N, 255, Imgproc.THRESH_BINARY);


             List<MatOfPoint> contours=new ArrayList<MatOfPoint>();

             // find contours and store them all as a list
             Imgproc.findContours(gray0, contours, new Mat(), Imgproc.RETR_LIST, Imgproc.CHAIN_APPROX_SIMPLE);

             MatOfPoint approx=new MatOfPoint();

             // test each contour
             for( int i = 0; i < contours.size(); i++ )
             {

                 // approximate contour with accuracy proportional
                 // to the contour perimeter
                 approx = approxPolyDP(contours.get(i),  Imgproc.arcLength(new MatOfPoint2f(contours.get(i).toArray()), true)*0.02, true);


                 // square contours should have 4 vertices after approximation
                 // relatively large area (to filter out noisy contours)
                 // and be convex.
                 // Note: absolute value of an area is used because
                 // area may be positive or negative - in accordance with the
                 // contour orientation

                 if( approx.toArray().length == 4 &&
                     Math.abs(Imgproc.contourArea(approx)) > 1000 &&
                     Imgproc.isContourConvex(approx) )
                 {
                     double maxCosine = 0;

                     for( int j = 2; j < 5; j++ )
                     {
                         // find the maximum cosine of the angle between joint edges
                         double cosine = Math.abs(angle(approx.toArray()[j%4], approx.toArray()[j-2], approx.toArray()[j-1]));
                         maxCosine = Math.max(maxCosine, cosine);
                     }

                     // if cosines of all angles are small
                     // (all angles are ~90 degree) then write quandrange
                     // vertices to resultant sequence
                     if( maxCosine < 0.3 )
                         squares.add(approx);
                 }
             }
         }
     }
 }

 void extractChannel(Mat source, Mat out, int channelNum) {
     List<Mat> sourceChannels=new ArrayList<Mat>();
     List<Mat> outChannel=new ArrayList<Mat>();

     Core.split(source, sourceChannels);

     outChannel.add(new Mat(sourceChannels.get(0).size(),sourceChannels.get(0).type()));

     Core.mixChannels(sourceChannels, outChannel, new MatOfInt(channelNum,0));

     Core.merge(outChannel, out);
 }

 MatOfPoint approxPolyDP(MatOfPoint curve, double epsilon, boolean closed) {
     MatOfPoint2f tempMat=new MatOfPoint2f();

     Imgproc.approxPolyDP(new MatOfPoint2f(curve.toArray()), tempMat, epsilon, closed);

     return new MatOfPoint(tempMat.toArray());
 }

Answer 3

SO中有一些相关的问题。检查出来：

OpenCV附带了一个示例：

https://code.ros.org/trac/opencv/browser/trunk/opencv/samples/cpp/squares.cpp?rev=4079

获得矩形后，您可以通过计算单应性与矩形角并应用透视变换来对齐图片。

Android OpenCV查找最大的方形或矩形

3 个答案: