这是方块检测示例的输出我的问题是过滤这个方块

  • 的第一个问题是它在同一区域绘制一条以上的线;
  • 第二个是我只需要检测对象而不是所有图像。

  • 另一个问题是我必须只取除所有图像之外的最大对象。

    下面是检测代码:
    static void findSquares( const Mat& image, vector >& squares ){
    squares.clear();
    
    Mat pyr, timg, gray0(image.size(), CV_8U), gray;
    
    // down-scale and upscale the image to filter out the noise
    pyrDown(image, pyr, Size(image.cols/2, image.rows/2));
    pyrUp(pyr, timg, image.size());
    vector<vector<Point> > contours;
    
    // find squares in every color plane of the image
    for( int c = 0; c < 3; c++ )
    {
        int ch[] = {c, 0};
        mixChannels(&timg, 1, &gray0, 1, ch, 1);
    
        // try several threshold levels
        for( int l = 0; l < N; l++ )
        {
            // hack: use Canny instead of zero threshold level.
            // Canny helps to catch squares with gradient shading
            if( l == 0 )
            {
                // apply Canny. Take the upper threshold from slider
                // and set the lower to 0 (which forces edges merging)
                Canny(gray0, gray, 0, thresh, 5);
                // dilate canny output to remove potential
                // holes between edge segments
                dilate(gray, gray, Mat(), Point(-1,-1));
            }
            else
            {
                // apply threshold if l!=0:
                gray = gray0 >= (l+1)*255/N;
            }
    
            // find contours and store them all as a list
            findContours(gray, contours, CV_RETR_LIST, CV_CHAIN_APPROX_SIMPLE);
    
            vector<Point> approx;
    
            // test each contour
            for( size_t i = 0; i < contours.size(); i++ )
            {
                approxPolyDP(Mat(contours[i]), approx, arcLength(Mat(contours[i]), true)*0.02, true);
    
                if( approx.size() == 4 &&
                    fabs(contourArea(Mat(approx))) > 1000 &&
                    isContourConvex(Mat(approx)) )
                {
                    double maxCosine = 0;
    
                    for( int j = 2; j < 5; j++ )
                    {
                        // find the maximum cosine of the angle between joint edges
                        double cosine = fabs(angle(approx[j%4], approx[j-2], approx[j-1]));
                        maxCosine = MAX(maxCosine, cosine);
                    }
    
                    if( maxCosine < 0.3 )
                        squares.push_back(approx);
                }
            }
        }
    }
    
    }

    最佳答案

    您需要查看 findContours() 的标志。您可以设置一个名为 CV_RETR_EXTERNAL 的标志,它将仅返回最外层的轮廓(其中的所有轮廓都将被丢弃)。这可能会返回整个帧,因此您需要缩小搜索范围,以便它不会检查您的帧边界。使用函数 copyMakeBorder() 来完成此操作。我还建议删除您的 dilate 函数,因为它可能会导致线条两侧出现重复的轮廓(如果您删除了 dilate,您甚至可能不需要边框)。这是我的输出:

    关于OpenCV 平方 : filtering output,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/14975304/

    10-15 13:21