在opencv中检测半圆-Java 学习之路

我试图检测图像中的完整圆圈和半圆 .
enter image description here

我遵循下面提到的过程：过程图像（包括Canny边缘检测）查找轮廓并在空图像上绘制它们，以便我可以消除不需要的组件 . （处理后的图像正是我想要的 . ）使用HoughCircles检测圆圈 . 这就是我得到的 .

enter image description here

我尝试改变HoughCircles中的参数，但结果不一致，因为它根据光照和图像中圆圈的位置而变化 . 我根据它的大小接受或拒绝一个圆圈 . 所以结果是不可接受的 . 我还有一长串“可接受的”圈子，所以我需要在HoughCircle参数中留出一些余量 . 至于完整的圆圈，很容易 - 我可以简单地找到轮廓的“圆度” . 问题是半圈！

请在hough变换前找到编辑过的图像
enter image description here

5 回答

直接在图像上使用 houghCircle ，不要先提取边缘 . 然后测试每个检测到的圆圈，图像中确实存在多少百分比：

int main()
{
    cv::Mat color = cv::imread("../houghCircles.png");
    cv::namedWindow("input"); cv::imshow("input", color);

    cv::Mat canny;

    cv::Mat gray;
    /// Convert it to gray
    cv::cvtColor( color, gray, CV_BGR2GRAY );

    // compute canny (don't blur with that image quality!!)
    cv::Canny(gray, canny, 200,20);
    cv::namedWindow("canny2"); cv::imshow("canny2", canny>0);

    std::vector<cv::Vec3f> circles;

    /// Apply the Hough Transform to find the circles
    cv::HoughCircles( gray, circles, CV_HOUGH_GRADIENT, 1, 60, 200, 20, 0, 0 );

    /// Draw the circles detected
    for( size_t i = 0; i < circles.size(); i++ ) 
    {
        Point center(cvRound(circles[i][0]), cvRound(circles[i][1]));
        int radius = cvRound(circles[i][2]);
        cv::circle( color, center, 3, Scalar(0,255,255), -1);
        cv::circle( color, center, radius, Scalar(0,0,255), 1 );
    }

    //compute distance transform:
    cv::Mat dt;
    cv::distanceTransform(255-(canny>0), dt, CV_DIST_L2 ,3);
    cv::namedWindow("distance transform"); cv::imshow("distance transform", dt/255.0f);

    // test for semi-circles:
    float minInlierDist = 2.0f;
    for( size_t i = 0; i < circles.size(); i++ ) 
    {
        // test inlier percentage:
        // sample the circle and check for distance to the next edge
        unsigned int counter = 0;
        unsigned int inlier = 0;

        cv::Point2f center((circles[i][0]), (circles[i][1]));
        float radius = (circles[i][2]);

        // maximal distance of inlier might depend on the size of the circle
        float maxInlierDist = radius/25.0f;
        if(maxInlierDist<minInlierDist) maxInlierDist = minInlierDist;

        //TODO: maybe paramter incrementation might depend on circle size!
        for(float t =0; t<2*3.14159265359f; t+= 0.1f) 
        {
            counter++;
            float cX = radius*cos(t) + circles[i][0];
            float cY = radius*sin(t) + circles[i][1];

            if(dt.at<float>(cY,cX) < maxInlierDist) 
            {
                inlier++;
                cv::circle(color, cv::Point2i(cX,cY),3, cv::Scalar(0,255,0));
            } 
           else
                cv::circle(color, cv::Point2i(cX,cY),3, cv::Scalar(255,0,0));
        }
        std::cout << 100.0f*(float)inlier/(float)counter << " % of a circle with radius " << radius << " detected" << std::endl;
    }

    cv::namedWindow("output"); cv::imshow("output", color);
    cv::imwrite("houghLinesComputed.png", color);

    cv::waitKey(-1);
    return 0;
}

对于此输入：

enter image description here

它给出了这个输出：

enter image description here

红圈是霍夫的结果 .

圆上的绿色采样点是内点 .

蓝点是异常值 .

控制台输出：

100 % of a circle with radius 27.5045 detected
100 % of a circle with radius 25.3476 detected
58.7302 % of a circle with radius 194.639 detected
50.7937 % of a circle with radius 23.1625 detected
79.3651 % of a circle with radius 7.64853 detected

如果你想测试RANSAC而不是Hough，请看this .

回复于 2024-05-03T20:30:02+08:00

我知道它有点晚了，但我使用了更简单的方法 . 从 cv2.HoughCircles(...) ，你得到圆的中心和直径（x，y，r） . 因此，我只需浏览圆圈的所有中心点，然后检查它们是否远离图像边缘而不是直径 .

这是我的代码：

height, width = img.shape[:2]

        #test top edge
        up = (circles[0, :, 0] - circles[0, :, 2]) >= 0

        #test left edge
        left = (circles[0, :, 1] - circles[0, :, 2]) >= 0

        #test right edge
        right = (circles[0, :, 0] + circles[0, :, 2]) <= width

        #test bottom edge
        down = (circles[0, :, 1] + circles[0, :, 2]) <= height

        circles = circles[:, (up & down & right & left), :]

回复于 2024-05-03T20:30:02+08:00

这是另一种方法，一个简单的RANSAC版本（要提高速度的大量优化），适用于Edge Image .

该方法循环这些步骤，直到它被取消

随机选择3个边缘像素
从它们估计圆圈（3个点足以识别圆圈）
验证或伪造它实际上是一个圆圈：计算给定边缘表示圆圈的百分比
如果验证了圆圈，请从输入/ egdes中删除圆圈

int main()
{
//RANSAC

//load edge image
cv::Mat color = cv::imread("../circleDetectionEdges.png");

// convert to grayscale
cv::Mat gray;
cv::cvtColor(color, gray, CV_RGB2GRAY);

// get binary image
cv::Mat mask = gray > 0;
//erode the edges to obtain sharp/thin edges (undo the blur?)
cv::erode(mask, mask, cv::Mat());

std::vector<cv::Point2f> edgePositions;
edgePositions = getPointPositions(mask);

// create distance transform to efficiently evaluate distance to nearest edge
cv::Mat dt;
cv::distanceTransform(255-mask, dt,CV_DIST_L1, 3);

//TODO: maybe seed random variable for real random numbers.

unsigned int nIterations = 0;

char quitKey = 'q';
std::cout << "press " << quitKey << " to stop" << std::endl;
while(cv::waitKey(-1) != quitKey)
{
    //RANSAC: randomly choose 3 point and create a circle:
    //TODO: choose randomly but more intelligent, 
    //so that it is more likely to choose three points of a circle. 
    //For example if there are many small circles, it is unlikely to randomly choose 3 points of the same circle.
    unsigned int idx1 = rand()%edgePositions.size();
    unsigned int idx2 = rand()%edgePositions.size();
    unsigned int idx3 = rand()%edgePositions.size();

    // we need 3 different samples:
    if(idx1 == idx2) continue;
    if(idx1 == idx3) continue;
    if(idx3 == idx2) continue;

    // create circle from 3 points:
    cv::Point2f center; float radius;
    getCircle(edgePositions[idx1],edgePositions[idx2],edgePositions[idx3],center,radius);

    float minCirclePercentage = 0.4f;

    // inlier set unused at the moment but could be used to approximate a (more robust) circle from alle inlier
    std::vector<cv::Point2f> inlierSet;

    //verify or falsify the circle by inlier counting:
    float cPerc = verifyCircle(dt,center,radius, inlierSet);

    if(cPerc >= minCirclePercentage)
    {
        std::cout << "accepted circle with " << cPerc*100.0f << " % inlier" << std::endl;
        // first step would be to approximate the circle iteratively from ALL INLIER to obtain a better circle center
        // but that's a TODO

        std::cout << "circle: " << "center: " << center << " radius: " << radius << std::endl;
        cv::circle(color, center,radius, cv::Scalar(255,255,0),1);

        // accept circle => remove it from the edge list
        cv::circle(mask,center,radius,cv::Scalar(0),10);

        //update edge positions and distance transform
        edgePositions = getPointPositions(mask);
        cv::distanceTransform(255-mask, dt,CV_DIST_L1, 3);
    }

    cv::Mat tmp;
    mask.copyTo(tmp);

    // prevent cases where no fircle could be extracted (because three points collinear or sth.)
    // filter NaN values
    if((center.x == center.x)&&(center.y == center.y)&&(radius == radius))
    {
        cv::circle(tmp,center,radius,cv::Scalar(255));
    }
    else
    {
        std::cout << "circle illegal" << std::endl;
    }

    ++nIterations;
    cv::namedWindow("RANSAC"); cv::imshow("RANSAC", tmp);
}

std::cout << nIterations <<  " iterations performed" << std::endl;


cv::namedWindow("edges"); cv::imshow("edges", mask);
cv::namedWindow("color"); cv::imshow("color", color);

cv::imwrite("detectedCircles.png", color);
cv::waitKey(-1);
return 0;
}


float verifyCircle(cv::Mat dt, cv::Point2f center, float radius, std::vector<cv::Point2f> & inlierSet)
{
 unsigned int counter = 0;
 unsigned int inlier = 0;
 float minInlierDist = 2.0f;
 float maxInlierDistMax = 100.0f;
 float maxInlierDist = radius/25.0f;
 if(maxInlierDist<minInlierDist) maxInlierDist = minInlierDist;
 if(maxInlierDist>maxInlierDistMax) maxInlierDist = maxInlierDistMax;

 // choose samples along the circle and count inlier percentage
 for(float t =0; t<2*3.14159265359f; t+= 0.05f)
 {
     counter++;
     float cX = radius*cos(t) + center.x;
     float cY = radius*sin(t) + center.y;

     if(cX < dt.cols)
     if(cX >= 0)
     if(cY < dt.rows)
     if(cY >= 0)
     if(dt.at<float>(cY,cX) < maxInlierDist)
     {
        inlier++;
        inlierSet.push_back(cv::Point2f(cX,cY));
     }
 }

 return (float)inlier/float(counter);
}


inline void getCircle(cv::Point2f& p1,cv::Point2f& p2,cv::Point2f& p3, cv::Point2f& center, float& radius)
{
  float x1 = p1.x;
  float x2 = p2.x;
  float x3 = p3.x;

  float y1 = p1.y;
  float y2 = p2.y;
  float y3 = p3.y;

  // PLEASE CHECK FOR TYPOS IN THE FORMULA :)
  center.x = (x1*x1+y1*y1)*(y2-y3) + (x2*x2+y2*y2)*(y3-y1) + (x3*x3+y3*y3)*(y1-y2);
  center.x /= ( 2*(x1*(y2-y3) - y1*(x2-x3) + x2*y3 - x3*y2) );

  center.y = (x1*x1 + y1*y1)*(x3-x2) + (x2*x2+y2*y2)*(x1-x3) + (x3*x3 + y3*y3)*(x2-x1);
  center.y /= ( 2*(x1*(y2-y3) - y1*(x2-x3) + x2*y3 - x3*y2) );

  radius = sqrt((center.x-x1)*(center.x-x1) + (center.y-y1)*(center.y-y1));
}



std::vector<cv::Point2f> getPointPositions(cv::Mat binaryImage)
{
 std::vector<cv::Point2f> pointPositions;

 for(unsigned int y=0; y<binaryImage.rows; ++y)
 {
     //unsigned char* rowPtr = binaryImage.ptr<unsigned char>(y);
     for(unsigned int x=0; x<binaryImage.cols; ++x)
     {
         //if(rowPtr[x] > 0) pointPositions.push_back(cv::Point2i(x,y));
         if(binaryImage.at<unsigned char>(y,x) > 0) pointPositions.push_back(cv::Point2f(x,y));
     }
 }

 return pointPositions;
}

输入：

enter image description here

输出：

enter image description here

控制台输出：

press q to stop
    accepted circle with 50 % inlier
    circle: center: [358.511, 211.163] radius: 193.849
    accepted circle with 85.7143 % inlier
    circle: center: [45.2273, 171.591] radius: 24.6215
    accepted circle with 100 % inlier
    circle: center: [257.066, 197.066] radius: 27.819
    circle illegal
    30 iterations performed`

优化应包括：

使用所有inlier来适应更好的圆圈
在每个检测到的圆圈之后不计算距离变换（它非常昂贵） . 直接从点/边集设置inlier并从该列表中删除inlier边 .
如果图像中有许多小圆圈（和/或大量噪声），则不太可能随机打3个边缘像素或圆圈 . =>首先尝试轮廓检测并检测每个轮廓的圆圈 . 之后尝试检测图像中剩余的所有“其他”圆圈 .
很多其他的东西

回复于 2024-05-03T20:30:02+08:00

0

霍夫算法检测到的半圆最可能是正确的 . 这里的问题可能是，除非您严格控制场景的几何形状，即相机相对于目标的精确位置，以便图像轴垂直于目标平面，您将获得省略号而不是投影在图像上的圆圈平面 . 更不用说由光学系统引起的扭曲，这进一步使几何图形退化 . 如果你依靠精确度，我会推荐camera calibration .

回复于 2024-05-03T20:30:02+08:00
0
你最好尝试不同的 kernel for gaussian blur . 这对你有帮助
```
GaussianBlur( src_gray, src_gray, Size(11, 11), 5,5);
```
所以改变 size(i,i),j,j)
回复于 2024-05-03T20:30:02+08:00

在opencv中检测半圆

5 回答

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