Question

这是方检测例的输出我的问题是过滤此方块

第一个问题是它绘制的是相同区域的多条线;
第二个是我只需要检测对象而不是所有图像。

另一个问题是除了所有图像外，我必须采取最大的对象。

以下是检测代码：

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);
            }
        }
    }
}

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); } } } }

Answer 1

您需要查看findContours()的标记。您可以设置一个名为CV_RETR_EXTERNAL的标志，该标志仅返回最外面的轮廓（其中的所有轮廓都被丢弃）。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（）来完成此任务。我还建议删除你的扩张功能，因为它可能会导致一条线两侧出现重复的轮廓（如果你移除扩张，你甚至可能不需要边框）。这是我的输出： enter image description here

OpenCV正方形：过滤输出

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