HoughLines在opencv中转换 [英] HoughLines transform in opencv

问题描述

我正在使用opencv和Eclipse进行图片处理。

  vector< Vec2f>线; 
 HoughLines（dst，lines，1，CV_PI / 180，100，0，0）; 
 
 for（size_t i = 0; i< lines.size（）; i ++）
 {
 float rho = lines [i] [0] i] [1]; 
 Point pt1，pt2; 
 double a = cos（theta），b = sin（theta）; 
 double x0 = a * rho，y0 = b * rho; 
 pt1.x = cvRound（x0 + 1000 *（ -  b））; 
 pt1.y = cvRound（y0 + 1000 *（a））; 
 pt2.x = cvRound（x0  -  1000 *（ -  b））; 
 pt2.y = cvRound（y0  -  1000 *（a））; 
 line（cdst，pt1，pt2，Scalar（0,0,255），3，CV_AA）; 
} 
  

任何人都可以解释这些代码如何定义点。我们使用

  y =（ -  cos（theta）/ sin（theta））x + r /（sin ）
 rho = xo * cos（theta）+ yo * sin（theta）
  

无法理解为什么在行中执行1000的乘法

  pt1.x = cvRound（x0 + 1000 * （-b））; 
  

请尝试用简单的语句解释这个问题。
提前感谢

解决方案

问题已经得到解答。但是，由于我花了最后十五分钟绘制这个图，我可能会发布它无论如何。可能会有帮助：

/ p>

所以你有一个Point p0 =（x0，y0）
然后，计算每个方向上距离 p0 1000个单位的其他两个点。

I am working on image processing using opencv and Eclipse.

  vector<Vec2f> lines;  
  HoughLines(dst, lines, 1, CV_PI/180, 100, 0, 0 );

  for( size_t i = 0; i < lines.size(); i++ )  
  {  
     float rho = lines[i][0], theta = lines[i][1];  
     Point pt1, pt2;  
     double a = cos(theta), b = sin(theta);  
     double x0 = a*rho, y0 = b*rho;  
     pt1.x = cvRound(x0 + 1000*(-b));  
     pt1.y = cvRound(y0 + 1000*(a));  
     pt2.x = cvRound(x0 - 1000*(-b));  
     pt2.y = cvRound(y0 - 1000*(a));  
     line( cdst, pt1, pt2, Scalar(0,0,255), 3, CV_AA);  
  }  

Can anyone explain that how the points are being defined by this code. We are using

y=(-cos(theta)/sin(theta))x + r/(sin(theta))
rho=xo*cos(theta) + yo*sin(theta)

I am not able to understand why the multiplication of 1000 is being done in the line

pt1.x = cvRound(x0 + 1000*(-b));  

please try to explain this in simple terms. Thanks in advance

解决方案

The question has already been answered. But since I spend the last fifteen minutes drawing this diagram I might as well post it anyway. Maybe it helps:

So what you have is a Point p0 = (x0,y0) which is on the line. You then compute two other points on the line which are 1000 units away from p0 in each direction.

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