c++ - 可靠的卡检测/永久性更正OpenCV

Question

我目前有一种方法可以检测图像中的卡片，并且大部分情况下，当光线相当一致且背景非常平静时，它就可以工作。

这是我用来执行此操作的代码:

    Mat img = inImg.clone();
    outImg = Mat(inImg.size(), CV_8UC1);
    inImg.copyTo(outImg);

    Mat img_fullRes = img.clone();

    pyrDown(img, img);

    Mat imgGray;
    cvtColor(img, imgGray, CV_RGB2GRAY);
    outImg_gray = imgGray.clone();
    // Find Edges //

    Mat detectedEdges = imgGray.clone();

    bilateralFilter(imgGray, detectedEdges, 0, 185, 3, 0);
    Canny( detectedEdges, detectedEdges, 20, 65, 3 );

    dilate(detectedEdges, detectedEdges, Mat::ones(3,3,CV_8UC1));
    Mat cdst = img.clone();

    vector<Vec4i> lines;
    HoughLinesP(detectedEdges, lines, 1, CV_PI/180, 60, 50, 3 );
    for( size_t i = 0; i < lines.size(); i++ )
    {
        Vec4i l = lines[i];
        // For debug
        //line( cdst, cv::Point(l[0], l[1]), cv::Point(l[2], l[3]), Scalar(0,0,255), 1);
    }
    //cdst.copyTo(inImg);

//    // Find points of intersection //

    cv::Rect imgROI;
    int ext = 10;
    imgROI.x = ext;
    imgROI.y = ext;
    imgROI.width = img.size().width - ext;
    imgROI.height = img.size().height - ext;

    int N = lines.size();
    // Creating N amount of points // N == lines.size()
    cv::Point** poi = new cv::Point*[N];
    for( int i = 0; i < N; i++ )
        poi[i] = new cv::Point[N];
    vector<cv::Point> poiList;
    for( int i = 0; i < N; i++ )
    {
        poi[i][i] = cv::Point(-1,-1);
        Vec4i line1 = lines[i];
        for( int j = i + 1; j < N; j++ )
        {
            Vec4i line2 = lines[j];
            cv::Point p = computeIntersect(line1, line2, imgROI);

            if( p.x != -1 )
            {
                //line(cdst, p-cv::Point(2,0), p+cv::Point(2,0), Scalar(0,255,0));
                //line(cdst, p-cv::Point(0,2), p+cv::Point(0,2), Scalar(0,255,0));

                poiList.push_back(p);
            }

            poi[i][j] = p;
            poi[j][i] = p;
        }
    }

    cdst.copyTo(inImg);

    if(poiList.size()==0)
    {
        outImg = inImg.clone();
        //circle(outImg, cv::Point(100,100), 50, Scalar(255,0,0), -1);
        return;
    }

    convexHull(poiList, poiList, false, true);

    for( int i=0; i<poiList.size(); i++ )
    {
        cv::Point p = poiList[i];
        //circle(cdst, p, 3, Scalar(255,0,0), 2);
    }
     //Evaluate all possible quadrilaterals

    cv::Point cardCorners[4];
    float metric_max = 0;
    int Npoi = poiList.size();
    for( int p1=0; p1<Npoi; p1++ )
    {
        cv::Point pts[4];
        pts[0] = poiList[p1];

        for( int p2=p1+1; p2<Npoi; p2++ )
        {
            pts[1] = poiList[p2];
            if( isCloseBy(pts[1],pts[0]) )
                continue;

            for( int p3=p2+1; p3<Npoi; p3++ )
            {
                pts[2] = poiList[p3];
                if( isCloseBy(pts[2],pts[1]) || isCloseBy(pts[2],pts[0]) )
                    continue;


                for( int p4=p3+1; p4<Npoi; p4++ )
                {
                    pts[3] = poiList[p4];
                    if( isCloseBy(pts[3],pts[0]) || isCloseBy(pts[3],pts[1])
                       || isCloseBy(pts[3],pts[2]) )
                        continue;


                    // get the metrics
                    float area = getArea(pts);

                    cv::Point a = pts[0]-pts[1];
                    cv::Point b = pts[1]-pts[2];
                    cv::Point c = pts[2]-pts[3];
                    cv::Point d = pts[3]-pts[0];
                    float oppLenDiff = abs(a.dot(a)-c.dot(c)) + abs(b.dot(b)-d.dot(d));

                    float metric = area - 0.35*oppLenDiff;
                    if( metric > metric_max )
                    {
                        metric_max = metric;
                        cardCorners[0] = pts[0];
                        cardCorners[1] = pts[1];
                        cardCorners[2] = pts[2];
                        cardCorners[3] = pts[3];
                    }

                }
            }
        }
    }

    // find the corners corresponding to the 4 corners of the physical card
    sortPointsClockwise(cardCorners);

    // Calculate Homography //

    vector<Point2f> srcPts(4);
    srcPts[0] = cardCorners[0]*2;
    srcPts[1] = cardCorners[1]*2;
    srcPts[2] = cardCorners[2]*2;
    srcPts[3] = cardCorners[3]*2;


    vector<Point2f> dstPts(4);
    cv::Size outImgSize(1400,800);

    dstPts[0] = Point2f(0,0);
    dstPts[1] = Point2f(outImgSize.width-1,0);
    dstPts[2] = Point2f(outImgSize.width-1,outImgSize.height-1);
    dstPts[3] = Point2f(0,outImgSize.height-1);

    Mat Homography = findHomography(srcPts, dstPts);

    // Apply Homography
    warpPerspective( img_fullRes, outImg, Homography, outImgSize, INTER_CUBIC );
    outImg.copyTo(inImg);

其中 computeIntersect定义为:

cv::Point computeIntersect(cv::Vec4i a, cv::Vec4i b, cv::Rect ROI)
{
    int x1 = a[0], y1 = a[1], x2 = a[2], y2 = a[3];
    int x3 = b[0], y3 = b[1], x4 = b[2], y4 = b[3];

    cv::Point  p1 = cv::Point (x1,y1);
    cv::Point  p2 = cv::Point (x2,y2);
    cv::Point  p3 = cv::Point (x3,y3);
    cv::Point  p4 = cv::Point (x4,y4);
    // Check to make sure all points are within the image boundrys, if not reject them.
    if( !ROI.contains(p1) || !ROI.contains(p2)
       || !ROI.contains(p3) || !ROI.contains(p4) )
        return cv::Point (-1,-1);

    cv::Point  vec1 = p1-p2;
    cv::Point  vec2 = p3-p4;

    float vec1_norm2 = vec1.x*vec1.x + vec1.y*vec1.y;
    float vec2_norm2 = vec2.x*vec2.x + vec2.y*vec2.y;
    float cosTheta = (vec1.dot(vec2))/sqrt(vec1_norm2*vec2_norm2);

    float den = ((float)(x1-x2) * (y3-y4)) - ((y1-y2) * (x3-x4));
    if(den != 0)
    {
        cv::Point2f pt;
        pt.x = ((x1*y2 - y1*x2) * (x3-x4) - (x1-x2) * (x3*y4 - y3*x4)) / den;
        pt.y = ((x1*y2 - y1*x2) * (y3-y4) - (y1-y2) * (x3*y4 - y3*x4)) / den;

        if( !ROI.contains(pt) )
            return cv::Point (-1,-1);

        // no-confidence metric
        float d1 = MIN( dist2(p1,pt), dist2(p2,pt) )/vec1_norm2;
        float d2 = MIN( dist2(p3,pt), dist2(p4,pt) )/vec2_norm2;

        float no_confidence_metric = MAX(sqrt(d1),sqrt(d2));

        // If end point ratios are greater than .5 reject
        if( no_confidence_metric < 0.5 && cosTheta < 0.707 )
            return cv::Point (int(pt.x+0.5), int(pt.y+0.5));
    }

    return cv::Point(-1, -1);
}

sortPointsClockWise定义为:

void sortPointsClockwise(cv::Point a[])
{
    cv::Point b[4];

    cv::Point ctr = (a[0]+a[1]+a[2]+a[3]);
    ctr.x /= 4;
    ctr.y /= 4;
    b[0] = a[0]-ctr;
    b[1] = a[1]-ctr;
    b[2] = a[2]-ctr;
    b[3] = a[3]-ctr;

    for( int i=0; i<4; i++ )
    {
        if( b[i].x < 0 )
        {
            if( b[i].y < 0 )
                a[0] = b[i]+ctr;
            else
                a[3] = b[i]+ctr;
        }
        else
        {
            if( b[i].y < 0 )
                a[1] = b[i]+ctr;
            else
                a[2] = b[i]+ctr;
        }
    }

}

getArea定义为:

float getArea(cv::Point arr[])
{
    cv::Point  diag1 = arr[0]-arr[2];
    cv::Point  diag2 = arr[1]-arr[3];

    return 0.5*(diag1.cross(diag2));
}

isCloseBy定义为:

bool isCloseBy( cv::Point p1, cv::Point p2 )
{
    int D = 10;
    // Checking that X values are within 10, same for Y values.
    return ( abs(p1.x-p2.x)<=D && abs(p1.y-p2.y)<=D );
}

最后 dist2:

float dist2( cv::Point p1, cv::Point p2 )
{
    return float((p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y));
}

以下是一些测试图像及其结果:





很抱歉，篇幅太长了，但是我希望有人可以提出一种方法，使我从图像中提取卡的方法更加可靠。一种可以更好地处理破坏性背景以及光线不一致的情况。

当将卡片放置在具有良好照明效果的对比背景上时，我的方法几乎可以在90％的时间内使用。但是很明显，我需要一种更强大的方法。

有没有人有什么建议？

谢谢。

dhanushka解决方案的ATTEMPT

Mat gray, bw;     pyrDown(inImg, inImg);
cvtColor(inImg, gray, CV_RGB2GRAY);
int morph_size = 3;
Mat element = getStructuringElement( MORPH_ELLIPSE, cv::Size( 4*morph_size + 1, 2*morph_size+1 ), cv::Point( morph_size, morph_size ) );

morphologyEx(gray, gray, 2, element);
threshold(gray, bw, 160, 255, CV_THRESH_BINARY);

vector<vector<cv::Point> > contours;
vector<Vec4i> hierarchy;
findContours( bw, contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE, cv::Point(0, 0) );

int largest_area=0;
int largest_contour_index=0;
cv::Rect bounding_rect;

for( int i = 0; i< contours.size(); i++ )
{
    double a=contourArea( contours[i],false);  //  Find the area of contour
    if(a>largest_area){
        largest_area=a;
        largest_contour_index=i;                //Store the index of largest contour
        bounding_rect=boundingRect(contours[i]);
    }
}

//Scalar color( 255,255,255);
rectangle(inImg, bounding_rect,  Scalar(0,255,0),1, 8,0);
Mat biggestRect = inImg(bounding_rect);
Mat card1 = biggestRect.clone();

Answer 1

更通用的方法肯定是像Rutger Nijlunsing在他的答案中建议的那样。但是，就您而言，至少对于所提供的样本图像而言，非常简单的方法(例如形态学打开，阈值处理，轮廓处理和凸包)将产生所需的结果。使用缩小版本的图像进行处理，这样就不必使用大型内核进行形态学操作。下面是用这种方法处理过的图像。

    pyrDown(large, rgb0);
    pyrDown(rgb0, rgb0);
    pyrDown(rgb0, rgb0);

    Mat small;
    cvtColor(rgb0, small, CV_BGR2GRAY);

    Mat morph;
    Mat kernel = getStructuringElement(MORPH_ELLIPSE, Size(11, 11));
    morphologyEx(small, morph, MORPH_OPEN, kernel);

    Mat bw;
    threshold(morph, bw, 0, 255.0, CV_THRESH_BINARY | CV_THRESH_OTSU);

    Mat bdry;
    kernel = getStructuringElement(MORPH_ELLIPSE, Size(3, 3));
    erode(bw, bdry, kernel);
    subtract(bw, bdry, bdry);

    // do contour processing on bdry

这种方法通常不会起作用，因此我强烈建议使用Rutger建议的方法。