从轮廓 OpenCV 检测卡片 MinArea Quadrilateral [英] Detect card MinArea Quadrilateral from contour OpenCV

问题描述

另一个关于检测图片中的卡片.我已经成功地隔离了图片中的卡片，我有一个很近的凸包，从这里我被卡住了.

Yet another about detecting card in a picture. I've managed to pretty much isolate the card in the picture, I have a convex hull that is close and from here I'm stuck.

对于上下文/约束，目标:

For the context/constraint, objective:

• 检测图片中的卡片
• 纯色背景(见示例)
• 固定在前面的卡片类型(意思是:我们有宽高比)
• 每张图片一个对象(至少目前是这样)

我使用的方法:

1. 缩小规模
2. 灰度
3. 轻度模糊
4. 精明
5. 寻找轮廓
6. 删除列表中少于 120 个点的所有轮廓(尝试/错误值)
7. 案例 1:我有 1 个轮廓:我的卡片的完美轮廓:第 9 步
8. 案例 2:我有多个轮廓
   • 凸包
   • 近似多边形?

步骤 1、3 和 6 主要是去除噪声和小伪影.

Step 1, 3 and 6 are mainly to remove noise and small artifacts.

所以我几乎卡在了第 9 步.我试过示例图片:

So I'm pretty much stuck at step 9. I've tried on a sample picture:

关于调试图片:

• 绿色:轮廓
• 红色:凸包
• 紫色/粉红色:使用了 approxPolyDp
• 黄色:minAreaRect

(结果图片是从minAreaRect中提取的)

(the result image is extracted from the minAreaRect)

所以轮廓是可以接受的，我可以通过调整 canny 或第一次模糊的参数来做得更好.但现在这是可以接受的，现在的问题是，我怎样才能得到将形成minarea 四边形"的 4 个点.如您所见，minAreaRect 给出了一个不完美的矩形，并且 approxPolyDp 丢失了太多卡片.

So the contour is acceptable, I can probably do a little better by tweaking the parameters from canny or the first blur. But for now this is acceptable, the issue now is, how can I get the the 4 points that will form the "minarea quadrilateral". As you can see, minAreaRect gives a rectangle which is not perfect, and the approxPolyDp is losing too much of the card.

有什么线索可以解决这个问题吗?我在使用 approxPolyDp 时尝试使用 epsilon 值(我使用了 arcLength*0.1)，但没有.

Any clue how I can approach this? I tried playing with the epsilon value when using approxPolyDp (I used arcLength*0.1), but nothing.

这种方法的另一个问题是在 canny 期间丢失了一个角(参见示例)，它不起作用(除非使用 minAreaRect 时).但这可能可以在之前(通过更好的预处理)或之后解决(因为我们知道宽/高比).

Another issue with this approach is that is a corner is lost during canny (see example) it'll not work (unless when using minAreaRect). But this can probably be resolved before (with a better pre-processing) or after (since we know the width/height ratio).

这里不要求代码，只是想办法解决这个问题，

Not asking for code here, just ideas how to approach this,

谢谢！

编辑:Yves Daoust 的解决方案:

Edit: Yves Daoust's solutions:

• 从凸包中获取与谓词匹配的 8 个点:(最大化 x, x+y, y, -x+y, -x, -x-y, -y, x-y)
• 从这个八边形，取4条最长的边，得到交点

结果:

编辑 2: 使用 Hough 变换(而不是 8 个极值点)可以为找到 4 个边的所有情况提供更好的结果.如果发现超过 4 行，可能我们有重复，所以使用一些数学来尝试过滤并保留 4 行.我使用行列式(如果平行则接近 0)和点线距离公式编写了一个草图)

Edit 2: Using Hough transform (instead of 8 extreme points) gives me better result for all cases where the 4 sides are found. If more than 4 lines are found, probably we have duplicates, so use some maths to try to filter and keep 4 lines. I coded a draft working by using the determinant (close to 0 if parallel) and the point-line distance formula)

推荐答案

这是我在您的输入图像上尝试的管道:

Here is the pipeline I tried on your input image:

• 模糊灰度输入并使用Canny过滤器
检测边缘

• Blur grayscale input and detect edges with Canny filter

• 计算轮廓
• 按长度对轮廓进行排序，只保留最大的
• 生成此轮廓的凸包
• 用凸包创建一个遮罩
• 使用HoughLinesP 找出你卡片的4 面
• 计算 4 条边的交点

• Compute the contours
• Sort the contours by length and only keep the largest one
• Generate the convex hull of this contour
• Create a mask out of the convex hull
• Use HoughLinesP to find the 4 sides of your cards
• Compute the intersections of the 4 sides

• 使用 findHomography 找到卡片的仿射变换(在 步骤 2 中找到 4 个交点)
• 变形使用计算单应矩阵的输入图像

• Use findHomography to find the affine transformation of your card (with the 4 intersection points found at Step 2)
• Warp the input image using the computed homography matrix

结果如下:

请注意，您必须找到一种方法来对 4 个交叉点进行排序，以便始终保持相同的顺序(否则 findHomography 将不起作用).

Note that you will have to find a way to sort the 4 intersection points so that there are always in the same order (otherwise findHomography won't work).

我知道你没有要求代码，但我必须测试我的管道，所以这里是... :)

I know you didn't ask for code, but I had to test my pipeline so here it is... :)

Vec3f calcParams(Point2f p1, Point2f p2) // line's equation Params computation
{
    float a, b, c;
    if (p2.y - p1.y == 0)
    {
        a = 0.0f;
        b = -1.0f;
    }
    else if (p2.x - p1.x == 0)
    {
        a = -1.0f;
        b = 0.0f;
    }
    else
    {
        a = (p2.y - p1.y) / (p2.x - p1.x);
        b = -1.0f;
    }

    c = (-a * p1.x) - b * p1.y;
    return(Vec3f(a, b, c));
}

Point findIntersection(Vec3f params1, Vec3f params2)
{
    float x = -1, y = -1;
    float det = params1[0] * params2[1] - params2[0] * params1[1];
    if (det < 0.5f && det > -0.5f) // lines are approximately parallel
    {
        return(Point(-1, -1));
    }
    else
    {
        x = (params2[1] * -params1[2] - params1[1] * -params2[2]) / det;
        y = (params1[0] * -params2[2] - params2[0] * -params1[2]) / det;
    }
    return(Point(x, y));
}

vector<Point> getQuadrilateral(Mat & grayscale, Mat& output) // returns that 4 intersection points of the card
{
    Mat convexHull_mask(grayscale.rows, grayscale.cols, CV_8UC1);
    convexHull_mask = Scalar(0);

    vector<vector<Point>> contours;
    findContours(grayscale, contours, RETR_EXTERNAL, CHAIN_APPROX_NONE);

    vector<int> indices(contours.size());
    iota(indices.begin(), indices.end(), 0);

    sort(indices.begin(), indices.end(), [&contours](int lhs, int rhs) {
        return contours[lhs].size() > contours[rhs].size();
    });

    /// Find the convex hull object
    vector<vector<Point> >hull(1);
    convexHull(Mat(contours[indices[0]]), hull[0], false);

    vector<Vec4i> lines;
    drawContours(convexHull_mask, hull, 0, Scalar(255));
    imshow("convexHull_mask", convexHull_mask);
    HoughLinesP(convexHull_mask, lines, 1, CV_PI / 200, 50, 50, 10);
    cout << "lines size:" << lines.size() << endl;

    if (lines.size() == 4) // we found the 4 sides
    {
        vector<Vec3f> params(4);
        for (int l = 0; l < 4; l++)
        {
            params.push_back(calcParams(Point(lines[l][0], lines[l][1]), Point(lines[l][2], lines[l][3])));
        }

        vector<Point> corners;
        for (int i = 0; i < params.size(); i++)
        {
            for (int j = i; j < params.size(); j++) // j starts at i so we don't have duplicated points
            {
                Point intersec = findIntersection(params[i], params[j]);
                if ((intersec.x > 0) && (intersec.y > 0) && (intersec.x < grayscale.cols) && (intersec.y < grayscale.rows))
                {
                    cout << "corner: " << intersec << endl;
                    corners.push_back(intersec);
                }
            }
        }

        for (int i = 0; i < corners.size(); i++)
        {
            circle(output, corners[i], 3, Scalar(0, 0, 255));
        }

        if (corners.size() == 4) // we have the 4 final corners
        {
            return(corners);
        }
    }
    
    return(vector<Point>());
}

int main(int argc, char** argv)
{
    Mat input = imread("playingcard_input.png");
    Mat input_grey;
    cvtColor(input, input_grey, CV_BGR2GRAY);
    Mat threshold1;
    Mat edges;
    blur(input_grey, input_grey, Size(3, 3));
    Canny(input_grey, edges, 30, 100);

    vector<Point> card_corners = getQuadrilateral(edges, input);
    Mat warpedCard(400, 300, CV_8UC3);
    if (card_corners.size() == 4)
    {
        Mat homography = findHomography(card_corners, vector<Point>{Point(warpedCard.cols, 0), Point(warpedCard.cols, warpedCard.rows), Point(0,0) , Point(0, warpedCard.rows)});
        warpPerspective(input, warpedCard, homography, Size(warpedCard.cols, warpedCard.rows));
    }

    imshow("warped card", warpedCard);
    imshow("edges", edges);
    imshow("input", input);
    waitKey(0);

    return 0;
}

我已经稍微调整了 Canny 和 HoughLinesP 函数的参数，以便更好地检测卡片(程序现在适用于两个输入样本).

I've have tweaked a little the parameters of Canny and HoughLinesP functions to have a better detection of the card (program now works on both input samples).

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