Question

我正在尝试理解使用OpenCV库来识别任何给定图像上的方块的代码。这条线朝向最后 - “gray = gray0＆gt; =（l + 1）* 255 / N;”。我从来没有见过＆gt; =以这种方式使用它看起来不像是在进行比较。那个运营商在那条线上做了什么？

static void findSquares(const Mat& image, vector<vector<Point> >& squares)
{
Mat pyr, timg, gray0(image.size(), CV_8U), gray;

pyrDown(image, pyr, Size(image.cols / 2, image.rows / 2));
pyrUp(pyr, timg, image.size());
vector<vector<Point> > contours;

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:
            //     tgray(x,y) = gray(x,y) < (l+1)*255/N ? 255 : 0
            gray = gray0 >= (l + 1) * 255 / N;
        }
 .....

更新

        // find contours and store them all as a list
        findContours(gray, contours, RETR_LIST, CHAIN_APPROX_SIMPLE);

        vector<Point> approx;

        // test each contour
        for (size_t i = 0; i < contours.size(); i++)
        {
            // approximate contour with accuracy proportional
            // to the contour perimeter
            approxPolyDP(Mat(contours[i]), approx, arcLength(Mat(contours[i]), 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.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 cosines of all angles are small
                // (all angles are ~90 degree) then write quandrange
                // vertices to resultant sequence
                if (maxCosine < 0.3)
                    squares.push_back(approx);
            }
        }
    }
}

Answer 1

基于OpenCV documentation on cv::Mat，此表达式返回一个通道掩码（cv::Mat），然后将其指定为灰色（另一个cv::Mat）：

比较：A cmpop B, A cmpop alpha, alpha cmpop A，其中cmpop 其中之一：>, >=, ==, !=, <=, <。比较结果是 8位单通道掩码，其元素设置为255（如果特定的话元素或元素对满足条件）或0。

＆gt; =在此上下文中的含义

1 个答案: