10

这可能已经得到回答,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;
}

我在这里想要完成的是创建一个基于原始图像中发现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,我只得到最大正方形的四个点,我想我可以从那里拿走。但如果我可以返回裁剪后的图像会更好。

4

3 回答 3

12

Took me a while to convert the C++ code to Java, but here it is :-)

Warning ! Raw code, totally not optimized and all.

I decline any liability in cases of injury or lethal accident

    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());
 }
于 2015-02-25T16:22:35.477 回答
12

精明之后

1-你需要用高斯模糊来减少噪音并找到所有的轮廓

2-查找并列出所有轮廓的区域

3-最大的轮廓将不过是绘画。

4-现在使用透视变换将您的形状转换为矩形。

检查数独求解器示例以查看类似的处理问题。(最大轮廓+透视)

于 2013-07-07T14:31:08.030 回答
2

SO中有一些相关的问题。去看一下:

OpenCV 还附带了一个示例:

获得矩形后,您可以通过计算矩形角的单应性并应用透视变换来对齐图片。

于 2013-07-07T14:26:26.590 回答