Radius of a disk in a binary image - opencv

Question

I have binarized images like this one:
I need to determine the center and radius of the inner solid disk. As you can see, it is surrounded by a textured area which touches it, so that simple connected component detection doesn't work. Anyway, there is a void margin on a large part of the perimeter.
A possible cure could be by eroding until all the texture disappears or disconnects from the disk, but this can be time consuming and the number of iterations is unsure. (In addition, in some unlucky cases there are tiny holes in the disk, which will grow with erosion.)
Any better suggestion to address this problem in a robust and fast way ? (I tagged OpenCV, but this is not mandated, what matters is the approach.)

Answer 1

You can:
Invert the image
Find the largest axis-aligned rectangle containing only zeros, (I used my C++ code from this answer). The algorithm is pretty fast.
Get the center and radius of the circle from the rectangle
Code:
#include <opencv2\opencv.hpp>
using namespace std;
using namespace cv;
// https://stackoverflow.com/a/30418912/5008845
cv::Rect findMaxRect(const cv::Mat1b& src)
{
cv::Mat1f W(src.rows, src.cols, float(0));
cv::Mat1f H(src.rows, src.cols, float(0));
cv::Rect maxRect(0,0,0,0);
float maxArea = 0.f;
for (int r = 0; r < src.rows; ++r)
{
for (int c = 0; c < src.cols; ++c)
{
if (src(r, c) == 0)
{
H(r, c) = 1.f + ((r>0) ? H(r-1, c) : 0);
W(r, c) = 1.f + ((c>0) ? W(r, c-1) : 0);
}
float minw = W(r,c);
for (int h = 0; h < H(r, c); ++h)
{
minw = std::min(minw, W(r-h, c));
float area = (h+1) * minw;
if (area > maxArea)
{
maxArea = area;
maxRect = cv::Rect(cv::Point(c - minw + 1, r - h), cv::Point(c+1, r+1));
}
}
}
}
return maxRect;
}
int main()
{
cv::Mat1b img = cv::imread("path/to/img", cv::IMREAD_GRAYSCALE);
// Correct image
img = img > 127;
cv::Rect r = findMaxRect(~img);
cv::Point center ( std::round(r.x + r.width / 2.f), std::round(r.y + r.height / 2.f));
int radius = std::sqrt(r.width*r.width + r.height*r.height) / 2;
cv::Mat3b out;
cv::cvtColor(img, out, cv::COLOR_GRAY2BGR);
cv::rectangle(out, r, cv::Scalar(0, 255, 0));
cv::circle(out, center, radius, cv::Scalar(0, 0, 255));
return 0;
}

Answer 2

My method is to use morph-open, findcontours, and minEnclosingCircle as follow:
#!/usr/bin/python3
# 2018/11/29 20:03
import cv2
fname = "test.png"
img = cv2.imread(fname)
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
th, threshed = cv2.threshold(gray, 200, 255, cv2.THRESH_BINARY)
kernel = cv2.getStructuringElement(cv2.MORPH_ELLIPSE, (3,3))
morphed = cv2.morphologyEx(threshed, cv2.MORPH_OPEN, kernel, iterations = 3)
cnts = cv2.findContours(morphed, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)[-2]
cnt = max(cnts, key=cv2.contourArea)
pt, r = cv2.minEnclosingCircle(cnt)
pt = (int(pt[0]), int(pt[1]))
r = int(r)
print("center: {}\nradius: {}".format(pt, r))
The final result:
center: (184, 170)
radius: 103

Answer 3

My second attempt on this case. This time I am using morphological closing operation to weaken the noise and maintain the signal. This is followed by a simple threshold and a connectedcomponent analysis. I hope this code can run faster.
Using this method, i can find the centroid with subpixel accuracy
('center : ', (184.12244328746746, 170.59771290442544))
Radius is derived from the area of the circle.
('radius : ', 101.34704439389715)
Here is the full code
import cv2
import numpy as np
# load image in grayscale
image = cv2.imread('radius.png',0)
r,c = image.shape
# remove noise
blured = cv2.blur(image,(5,5))
# Morphological closing
morph = cv2.erode(blured,None,iterations = 3)
morph = cv2.dilate(morph,None,iterations = 3)
cv2.imshow("morph",morph)
cv2.waitKey(0)
# Get the strong signal
th, th_img = cv2.threshold(morph,200,255,cv2.THRESH_BINARY)
cv2.imshow("th_img",th_img)
cv2.waitKey(0)
# Get connected components
num_labels, labels, stats, centroids = cv2.connectedComponentsWithStats(th_img)
print(num_labels)
print(stats)
# displat labels
labels_disp = np.uint8(255*labels/np.max(labels))
cv2.imshow("labels",labels_disp)
cv2.waitKey(0)
# Find center label
cnt_label = labels[r/2,c/2]
# Find circle center and radius
# Radius calculated by averaging the height and width of bounding box
area = stats[cnt_label][4]
radius = np.sqrt(area / np.pi)#stats[cnt_label][2]/2 + stats[cnt_label][3]/2)/2
cnt_pt = ((centroids[cnt_label][0]),(centroids[cnt_label][1]))
print('center : ',cnt_pt)
print('radius : ',radius)
# Display final result
edges_color = cv2.cvtColor(image,cv2.COLOR_GRAY2BGR)
cv2.circle(edges_color,(int(cnt_pt[0]),int(cnt_pt[1])),int(radius),(0,0,255),1)
cv2.circle(edges_color,(int(cnt_pt[0]),int(cnt_pt[1])),5,(0,0,255),-1)
x1 = stats[cnt_label][0]
y1 = stats[cnt_label][1]
w1 = stats[cnt_label][2]
h1 = stats[cnt_label][3]
cv2.rectangle(edges_color,(x1,y1),(x1+w1,y1+h1),(0,255,0))
cv2.imshow("edges_color",edges_color)
cv2.waitKey(0)

Answer 4

Here is an example of using hough circle. It can work if you set the min and max radius to a proper range.
import cv2
import numpy as np
# load image in grayscale
image = cv2.imread('radius.png',0)
r , c = image.shape
# remove noise
dst = cv2.blur(image,(5,5))
# Morphological closing
dst = cv2.erode(dst,None,iterations = 3)
dst = cv2.dilate(dst,None,iterations = 3)
# Find Hough Circle
circles = cv2.HoughCircles(dst
,cv2.HOUGH_GRADIENT
,2
,minDist = 0.5* r
,param2 = 150
,minRadius = int(0.5 * r / 2.0)
,maxRadius = int(0.75 * r / 2.0)
)
# Display
edges_color = cv2.cvtColor(image,cv2.COLOR_GRAY2BGR)
for i in circles[0]:
print(i)
cv2.circle(edges_color,(i[0],i[1]),i[2],(0,0,255),1)
cv2.imshow("edges_color",edges_color)
cv2.waitKey(0)
Here is the result
[185. 167. 103.6]

Answer 5

Have you tried something along the lines of the Circle Hough Transform?
I see that OpenCv has its own implementation. Some preprocessing (median filtering?) might be necessary here, though.

Answer 6

Here is a simple approach:
Erode the image (using a large, circular SE), then find the centroid of the result. This should be really close to the centroid of the central disk.
Compute the mean as a function of the radius of the original image, using the computed centroid as the center.
The output looks like this:
From here, determining the radius is quite simple.
Here is the code, I'm using PyDIP (we don't yet have a binary distribution, you'll need to download and build form sources):
import matplotlib.pyplot as pp
import PyDIP as dip
import numpy as np
img = dip.Image(pp.imread('/home/cris/tmp/FDvQm.png')[:,:,0])
b = dip.Erosion(img, 30)
c = dip.CenterOfMass(b)
rmean = dip.RadialMean(img, center=c)
pp.plot(rmean)
r = np.argmax(rmean < 0.5)
Here, r is 102, as the radius in integer number of pixels, I'm sure it's possible to interpolate to improve precision. c is [184.02, 170.45].