使用houghcircle函数opencv python检测不规则形状

Question

我目前正在对图像进行圆检测看起来像这样,但是一些墨滴合并并形成一些不规则的形状(原始图像中的红色标记).我在opencv中使用houghcircle函数来检测圆圈.对于那些不规则的形状,该功能只能将它们检测为几个小圆圈,但我真的希望程序将不规则形状视为一个完整的大形状并得到一个像我在输出图像中绘制的大圆圈.

我的代码将检测所有圆圈并获得它​​们的直径.

这是我的代码:

def circles(filename, p1, p2, minR, maxR):
# print(filename)
img  = cv2.imread(filename, 0)
img = img[0:1000, 0:1360]
l = len(img)
w = len(img[1])

cimg = cv2.cvtColor(img,cv2.COLOR_GRAY2BGR)

circles = cv2.HoughCircles(img, cv2.HOUGH_GRADIENT, 1, 25,
                            param1 = int(p1) ,param2 = int(p2), minRadius = int(minR), maxRadius = int(maxR))

diameter = open(filename[:-4] + "_diamater.txt", "w")
diameter.write("Diameters(um)\n")
for i in circles[0,:]:
    diameter.write(str(i[2] * 1.29 * 2) + "\n")

count = 0
d = []
area = []
for i in circles[0,:]:
    cv2.circle(cimg,(i[0],i[1]),i[2],(0,255,0),2)
    cv2.circle(cimg,(i[0],i[1]),2,(0,0,255),3)
    count += 1
    d += [i[2]*2]
    area += [i[2]*i[2]*pi*1.286*1.286]

f = filename.split("/")[-1]
cv2.imwrite(filename[:-4] + "_circle.jpg", cimg)

# cv2.imwrite("test3/edge.jpg", edges)
print "Number of Circles is %d" % count

diaM = []
for i in d:
    diaM += [i*1.286]

bWidth = range(int(min(diaM)) - 10, int(max(diaM)) + 10, 2)

txt = '''
Sample name: %s 
Average diameter(um): %f     std: %f
Drop counts: %d
Average coverage per drop(um^2): %f     std: %f
''' % (f, np.mean(diaM), np.std(diaM), count, np.mean(area), np.std(area))

fig = plt.figure()
fig.suptitle('Histogram of Diameters', fontsize=14, fontweight='bold')
ax1 = fig.add_axes((.1,.4,.8,.5))
ax1.hist(diaM, bins = bWidth)
ax1.set_xlabel('Diameter(um)')
ax1.set_ylabel('Frequency')
fig.text(.1,.1,txt)
plt.savefig(filename[:-4] + '_histogram.jpg')
plt.clf()

print "Total area is %d" % (w*l)
print "Total covered area is %d" % (np.sum(area))

rt = "Number of Circles is " + str(count) + "\n" + "Coverage percent is " + str(np.divide(np.sum(area), (w*l))) + "\n"
return rt

Answer 1

如果您仍然想使用 HoughCircles 函数，您可以查看两个圆是否重叠并用它们创建一个新圆。