如何利用圆方程实现圆检测？

Question

我试图用方程实现圆hough变换，r = sqrt((x-h)^2-(y-k)^2)从图像中检测圆。

我申请了一系列的步骤，比如高斯模糊，精明。在此之后，如果半径和边界点是可用的，我就不知道如何实现上述方程。实现后，我将得到累加器空间，其中包含半径和中心的检测圆。我想用opencv的HoughCircle函数来实现。有什么能帮到我的吗?花了这么多时间。

import numpy as np
import cv2
import math

image = cv2.imread(imagepath)

h, w = image.shape[:2]
print h, w

grayimg = cv2.cvtColor(image,cv2.COLOR_BGR2GRAY)

bimg = cv2.bilateralFilter(grayimg, 5, 175, 175)

cann = cv2.Canny(bimg,100,200)

pixel = np.argwhere(cann == 255)


accum = []
ct = 0
for r in range(10,21):
    for h in range(0,20):
        for k in range(0,20):
            for p in pixel:
                print r,h,k,p
                xpart = (h - p[0])**2
                ypart = (k - p[1])**2
                rhs = xpart + ypart
                lhs = r * r
                if(lhs == rhs):
                    accum.append((r,(h,k),p))

print len(accum)
cv2.waitKey(0)

Answer 1

但是，在此代码中仍然需要对累加器空间进行快速处理。

accum = [[[0 for r in range(10,21)]for h in range(0,30)]for k in range(0,30)]
print accum
ct = 0
for r in range(10,21):
    for h in range(0,30):
        for k in range(0,30):
            for p in pixel:
                #print r,h,k,p
                xpart = (h - p[0])**2
                ypart = (k - p[1])**2
                rhs = xpart + ypart
                lhs = r * r
                if(lhs == rhs):
                    accum[k][h][r-10] += 1