cv :: RotatedRect中的非零像素数

问题描述

因为标题说我想在cv :: Mat的某个区域，即在RotatedRect中找到非零像素的数量。

对于一个常规的Rect，可以简单地在ROI上使用countNonZeroPixels。但是ROI只能是常规（非旋转）矩形。

另一个想法是绘制旋转后的矩形并将其用作掩码。但是openCV不支持旋转矩形的绘制，也不支持countNonZeroPixels接受一个掩码。

有没有人有解决方案如何优雅地解决这个问题？

谢谢！

解决方案

好吧，

这个想法是将图像旋转到矩形的旋转方向，然后在拉直的矩形上应用roi。

• 如果旋转的矩形不完全在图像内，则会破坏
。


• 你可以在旋转之前应用另一个roi来加快速度，整个图像...

    #include< highgui.h> 
   #include< cv.h> 
   
   
   //从http://stackoverflow.com/questions/2289690/opencv-how-to-rotate-iplimage 
   cv :: Mat rotateImage（const cv :: 
   {
   cv :: Mat rot_mat = cv :: getRotationMatrix2D（center，angle，1.0）; 
   cv :: Mat dst; 
   cv :: warpAffine（source，dst，rot_mat，source.size（））; 
   return dst; 
  } 
   
   int main（）
   {
   cv :: namedWindow（test1）; 
   
   //我们旋转的矩形
   int x = 300; 
   int y = 350; 
   int w = 200; 
   int h = 50; 
   float angle = 47; 
   cv :: RotatedRect rect = cv :: RotatedRect（cv :: Point2f（x，y），cv :: Size2f（w，h），angle）; 
   
   //空白图像
   cv :: Mat img = cv :: Mat（cv :: Size（640，480），CV_8UC3）; 
   
   //绘制旋转的矩形作为椭圆以获得一些视觉反馈
   cv :: ellipse（img，rect，cv :: Scalar（255,0,0），-1）; 
   
   //通过rect.angle旋转图像* -1 
   cv :: Mat rotimg = rotateImage（img，rect.center，-1 * rect.angle）; 
   
   //将roi设置为当前未旋转的矩形
   cv :: Rect roi; 
   roi.x = rect.center.x  - （rect.size.width / 2）; 
   roi.y = rect.center.y  - （rect.size.height / 2）; 
   roi.width = rect.size.width; 
   roi.height = rect.size.height; 
   
   cv :: imshow（test1，rotimg（roi））; 
   cv :: waitKey（0）; 
  } 
    

as the title says i'm trying to find the number of non-zero pixels in a certain area of a cv::Mat, namely within a RotatedRect.

For a regular Rect one could simply use countNonZeroPixels on a ROI. However ROIs can only be regular (non rotated) rectangles.

Another idea was to draw the rotated rectangle and use that as a mask. However openCV neither supports the drawing of rotated rectangles nor does countNonZeroPixels accept a mask.

Does anyone have a solution for how to elegantly solve this ?

Thank you !

解决方案

Ok, so here's my first take at it.

The idea is to rotate the image reverse to the rectangle's rotation and than apply a roi on the straightened rectangle.

• This will break if the rotated rectangle is not completely within the image

• You can probably speed this up by applying another roi before rotation to avoid having to rotate the whole image...

  #include <highgui.h>
  #include <cv.h>
  
  
  // From http://stackoverflow.com/questions/2289690/opencv-how-to-rotate-iplimage
  cv::Mat rotateImage(const cv::Mat& source, cv::Point2f center, double angle)
  {
    cv::Mat rot_mat = cv::getRotationMatrix2D(center, angle, 1.0);
    cv::Mat dst;
    cv::warpAffine(source, dst, rot_mat, source.size());
    return dst;
  }
  
  int main()
  {
    cv::namedWindow("test1");
  
    // Our rotated rect
    int x = 300;
    int y = 350;
    int w = 200;
    int h = 50;
    float angle = 47;
    cv::RotatedRect rect = cv::RotatedRect(cv::Point2f(x,y), cv::Size2f(w,h), angle);
  
    // An empty image
    cv::Mat img = cv::Mat(cv::Size(640, 480), CV_8UC3);
  
    // Draw rotated rect as an ellipse to get some visual feedback
    cv::ellipse(img, rect, cv::Scalar(255,0,0), -1);
  
    // Rotate the image by rect.angle * -1
    cv::Mat rotimg = rotateImage(img, rect.center, -1 * rect.angle);
  
    // Set roi to the now unrotated rectangle
    cv::Rect roi;
    roi.x = rect.center.x - (rect.size.width / 2);
    roi.y = rect.center.y - (rect.size.height / 2);
    roi.width = rect.size.width;
    roi.height = rect.size.height;
  
    cv::imshow("test1", rotimg(roi));
    cv::waitKey(0);
  }
  

这篇关于cv :: RotatedRect中的非零像素数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！