在C++／CLI中如何利用等高线逼近OpenCV寻找随机形状物体的长度和宽度

我使用C++/CLI应用程序中的find contours OpenCV函数选择不同的随机形状对象

我想找到图像中每个轮廓的最大长度。以及轮廓的宽度，每次的差异为40到50像素。你能帮帮我吗

findContours(K, contours, hierarchy, CV_RETR_LIST, CV_CHAIN_APPROX_SIMPLE); for(idx=0; idx<contours.size(); idx++)
{
    double cArea = contourArea(contours[idx]);

    // get only contours with area above specific threshold
    if (cArea > min_area && cArea < 40000)// && !k )
    {
        if (cArea > largest_area) {
            largest_area = cArea; 

            largest_contour_index = idx;

           // Find the bounding rectangle for biggest contour
            bounding_rect = boundingRect(contours[idx]);
        }

        for (unsigned int j = 0; j<contours[idx].size(); j++)      
        {
            circle(dst, Point(contours[idx][j].x, contours[idx][j].y), 3, Scalar(255, 255,0), FILLED, LINE_AA);

        }
   }

---

}

假设找到了轮廓，。。。。。是您正在使用的，其结果是

   std::vector<std::vector<cv::Point>> contours;

将旋转的矩形拟合到每个轮廓

 //Assuming right version of C++
 for(const auto& it: contours){
 Rect boundingRect = boundingRect(it);
 // boundRect.height, boundingRect.width will give you what you 
  //need
  }

 //Assuming C++ 11
 for(const auto& it: contours){
 RotatedRect rotRect = minAreaRect(it);
 // rotRect.size.height, rotRect.size.width will give you what you 
  //need
  }

 //Assuming C++ 11
 for(const auto& it: contours){
 Point2f center;
 float radius;
 minEnclosingCircle(it, center, radius);
 }

 //Assuming C++ 11
 for(const auto& it: contours){
 // note need atleast 5 points to fit an ellipse
 if(it.size()>5){
 RotatedRect boundingEllipse = minBoundingEllipse(it);
 }
 // boundingEllipse.size.height, boundingEllipse.size.width will 
 //give you what you need - in this case the major and minor axis
 }

为每个轮廓拟合一个封闭圆

 //Assuming right version of C++
 for(const auto& it: contours){
 Rect boundingRect = boundingRect(it);
 // boundRect.height, boundingRect.width will give you what you 
  //need
  }

 //Assuming C++ 11
 for(const auto& it: contours){
 RotatedRect rotRect = minAreaRect(it);
 // rotRect.size.height, rotRect.size.width will give you what you 
  //need
  }

 //Assuming C++ 11
 for(const auto& it: contours){
 Point2f center;
 float radius;
 minEnclosingCircle(it, center, radius);
 }

 //Assuming C++ 11
 for(const auto& it: contours){
 // note need atleast 5 points to fit an ellipse
 if(it.size()>5){
 RotatedRect boundingEllipse = minBoundingEllipse(it);
 }
 // boundingEllipse.size.height, boundingEllipse.size.width will 
 //give you what you need - in this case the major and minor axis
 }

将椭圆拟合到每个轮廓

 //Assuming right version of C++
 for(const auto& it: contours){
 Rect boundingRect = boundingRect(it);
 // boundRect.height, boundingRect.width will give you what you 
  //need
  }

 //Assuming C++ 11
 for(const auto& it: contours){
 RotatedRect rotRect = minAreaRect(it);
 // rotRect.size.height, rotRect.size.width will give you what you 
  //need
  }

 //Assuming C++ 11
 for(const auto& it: contours){
 Point2f center;
 float radius;
 minEnclosingCircle(it, center, radius);
 }

 //Assuming C++ 11
 for(const auto& it: contours){
 // note need atleast 5 points to fit an ellipse
 if(it.size()>5){
 RotatedRect boundingEllipse = minBoundingEllipse(it);
 }
 // boundingEllipse.size.height, boundingEllipse.size.width will 
 //give you what you need - in this case the major and minor axis
 }

4通常是最准确的。2和4还可以根据需要提供更多信息，如中心、角度

---

我们用这种方法发现的宽度和长度只根据轮廓面积计算，或者也包含不属于轮廓部分的日食面积。轮廓形状是完全随机的，在不同角度上宽度不同。那么它将为宽度和长度选择哪个区域。最大的还是最大的？我想得到长度和宽度的单位，比如毫米或英寸。我使用了你和我分享的以下方法：对于const auto&it:contours{//注意，如果椭圆的大小大于5，则至少需要5个点来拟合椭圆{//RotatedRect BoundingEllipseit=minBoundingEllipseit；RotatedRect BoundingEllipseit=fitEllipseit；浮动裂纹高度=boundingEllipse.size.height；浮动裂纹宽度=boundingEllipse.size.width；//浮动裂纹面积=boundingEllipse.size.area；ellipsetempimage，boundingEllipse，0,0,255,2}它给了我一个错误：float crack_area=boundingEllipse.size.area；@DavidHorn我很确定属性大小没有属性区域。如果你需要椭圆的面积，你可以自己计算pi*boundingEllipse.size.width*boundingEllipse.size.height，或者你可以直接用轮廓区域听起来不错。非常感谢。你能告诉我如何在一个轮廓内找到子轮廓或区域吗。我想将每个轮廓划分为不同的分包商或区域，然后计算每个子轮廓或区域的宽度。然后我想根据每个区域的宽度进行分类。再次感谢。