OpenCV 3.0 LineIterator

大佬教程收集整理的这篇文章主要介绍了OpenCV 3.0 LineIterator，大佬教程大佬觉得挺不错的，现在分享给大家，也给大家做个参考。

如何解决OpenCV 3.0 LineIterator？

开发过程中遇到OpenCV 3.0 LineIterator的问题如何解决？下面主要结合日常开发的经验，给出你关于OpenCV 3.0 LineIterator的解决方法建议，希望对你解决OpenCV 3.0 LineIterator有所启发或帮助；

我已经解决了自己的问题。行迭代器似乎在cv2库中不可用。因此，我做了自己的行迭代器。没有使用循环，因此它应该非常快。如果有人需要，下面是代码：

def createlineIterator(P1, P2, img):
    """
    Produces and array that consists of the coordinates and intensitIEs of each pixel in a line between two points

    Parameters:
        -P1: a numpy array that consists of the coordinate of the first point (x,y)
        -P2: a numpy array that consists of the coordinate of the second point (x,y)
        -img: the image being processed

    Returns:
        -it: a numpy array that consists of the coordinates and intensitIEs of each pixel in the radii (shape: [numPixels, 3], row = [x,y,intensity])     
    """
   #define local variables for readability
   imageH = img.shape[0]
   imageW = img.shape[1]
   P1X = P1[0]
   P1Y = P1[1]
   P2X = P2[0]
   P2Y = P2[1]

   #difference and absolute difference between points
   #used to calculate slope and relative LOCATIOn between points
   dX = P2X - P1X
   dY = P2Y - P1Y
   dXa = np.abs(dX)
   dYa = np.abs(dY)

   #predefine numpy array for output based on distance between points
   itbuffer = np.empty(shape=(np.maximum(dYa,dXa),3),dtype=np.float32)
   itbuffer.fill(np.nan)

   #Obtain coordinates along the line using a form of Bresenham's algorithm
   negY = P1Y > P2Y
   negX = P1X > P2X
   if P1X == P2X: #vertical line segment
       itbuffer[:,0] = P1X
       if negY:
           itbuffer[:,1] = np.arange(P1Y - 1,P1Y - dYa - 1,-1)
       else:
           itbuffer[:,1] = np.arange(P1Y+1,P1Y+dYa+1)              
   elif P1Y == P2Y: #horizontal line segment
       itbuffer[:,1] = P1Y
       if negX:
           itbuffer[:,0] = np.arange(P1X-1,P1X-dXa-1,-1)
       else:
           itbuffer[:,0] = np.arange(P1X+1,P1X+dXa+1)
   else: #diagonal line segment
       steepSlope = dYa > dXa
       if steepSlope:
           slope = dX.astype(np.float32)/dY.astype(np.float32)
           if negY:
               itbuffer[:,1] = np.arange(P1Y-1,P1Y-dYa-1,-1)
           else:
               itbuffer[:,1] = np.arange(P1Y+1,P1Y+dYa+1)
           itbuffer[:,0] = (slope*(itbuffer[:,1]-P1Y)).astype(np.int) + P1X
       else:
           slope = dY.astype(np.float32)/dX.astype(np.float32)
           if negX:
               itbuffer[:,0] = np.arange(P1X-1,P1X-dXa-1,-1)
           else:
               itbuffer[:,0] = np.arange(P1X+1,P1X+dXa+1)
           itbuffer[:,1] = (slope*(itbuffer[:,0]-P1X)).astype(np.int) + P1Y

   #Remove points outsIDe of image
   colX = itbuffer[:,0]
   colY = itbuffer[:,1]
   itbuffer = itbuffer[(colX >= 0) & (colY >=0) & (colX<imageW) & (colY<imageH)]

   #Get intensitIEs from img ndarray
   itbuffer[:,2] = img[itbuffer[:,1].astype(np.uint),itbuffer[:,0].astype(np.uint)]

   return itbuffer

解决方法

大佬总结

以上是大佬教程为你收集整理的OpenCV 3.0 LineIterator全部内容，希望文章能够帮你解决OpenCV 3.0 LineIterator所遇到的程序开发问题。

如果觉得大佬教程网站内容还不错，欢迎将大佬教程推荐给程序员好友。

本图文内容来源于网友网络收集整理提供，作为学习参考使用，版权属于原作者。
如您有任何意见或建议可联系处理。小编QQ：384754419，请注明来意。