如何使用圆方程实现圆检测

Question

我正在尝试通过方程实现圆形霍夫变换，r = sqrt((x-h)^2-(y-k)^2)用于从图像中检测圆。

我应用了像高斯模糊这样的步骤列表，精明。之后，如果半径和边界点可用，我不知道如何实现上述等式。实现后，我将获得包含检测到的圆的半径和中心的累加器空间。我想在没有opencv的HoughCircle功能的情况下实现。有什么想法可以帮助我吗？这需要很多时间。

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

此代码中仍然需要一些升级以使累加器空间的进程更快。

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