Identifying the circle that fits the outer white pixels in a binary image (or make external mask)

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Hi all,
I want to process a series of images similar to the attached one in order to measure their porosities: V_pores / V_total. For my case, in a 2D context, V_total is the surface of circle containing all pores. It is important to know that the radius and center of the circle changes, and also many circular shapes can be identified in between, however, I am only interested, at this moment, in measuring the general surface of the circle (Vt).
Alernatively, I would be interested in applying a mask to the image. But it seems to be more difficult.
Any help would be appreciated. Many thanks.

Accepted Answer

KSSV
KSSV on 27 Mar 2020
I = imread("SampleXRmorph.PNG");
[y,x] = find(I) ;
% Get center of cricle
C = [mean(x) mean(y)] ;
% GEt raidus
d = sqrt((C(1)-x).^2+(C(2)-y).^2) ;
R = max(d) ;
% plot circle
th = linspace(0,2*pi) ;
xc = C(1)+R*cos(th) ;
yc = C(2)+R*sin(th) ;
%
imshow(I)
hold on
plot(xc,yc,'r')
  9 Comments
Dyhia Bouhadjra
Dyhia Bouhadjra on 29 Mar 2020
Hi Again,
Following the previous question, and after identification of the boundary of the images, I've calculated the surface of the circle, however I don't know how to calculate the surface of black pixels inside the boundary circel.
To measure the surface of the white pixels I did:
A=imread('moreQuestion.png');
A=dounle(A);
[x,y]=find(A);
[idx Surf]=boundary(x,y); %this will allow drowing the boundary circle as mentioned above.
%Surf is the area of the circle.
Sw=sum(sum(A)); % surface of white pixels
Sb=sum(sum(~A)); %surface of the total black pixels.
Sb1= ? %The black pixels' surface incide the circle
%One may say: Sb1=Surf-Sw; but it's negative and shouldn't be!!
Do you please have an idea of how to do it?

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More Answers (1)

Image Analyst
Image Analyst on 29 Mar 2020
Edited: Image Analyst on 29 Mar 2020
Sorry I didn't see this earlier, but anyway...
How I'd solve it is
  1. Use bwconvhull() to get a convex hull of all the blobs.
  2. Uuse imerode() to shrink that a certain amount to get a smaller circular blob.
  3. Use the eroded image to erase every blob in the interior so you have only the outer blobs.
  4. If you need a perfect circle, call regionprops(mask, 'Centroid') to get a list of (x,y) coordinates.
  5. Use the FAQ to fit a circle through the centroids.
% Initialization steps.
clc; % Clear the command window.
fprintf('Beginning to run %s.m ...\n', mfilename);
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format short g;
format compact;
fontSize = 15;
mask = imread('morequestion.png');
imshow(mask);
if ndims(mask) > 2
mask = mask(:, :, 1); % Take red channel
end
mask2 = bwconvhull(mask, 'union');
se = strel('disk', 20, 0);
smallMask = imerode(mask2, se);
mask3 = mask & ~smallMask;
props = regionprops(mask3, 'Centroid');
xy = vertcat(props.Centroid);
x = xy(:, 1);
y = xy(:, 2);
[xc,yc,R,a] = circfit(x,y);
viscircles([xc, yc], R);
fprintf('Done running %s.m ...\n', mfilename);
function [xc,yc,R,a] = circfit(x,y)
%CIRCFIT Fits a circle in x,y plane
%
% [XC, YC, R, A] = CIRCFIT(X,Y)
% Result is center point (yc,xc) and radius R. A is an optional
% output describing the circle's equation:
%
% x^2+y^2+a(1)*x+a(2)*y+a(3)=0
% by Bucher izhak 25/oct/1991
n=length(x); xx=x.*x; yy=y.*y; xy=x.*y;
A=[sum(x) sum(y) n;sum(xy) sum(yy) sum(y);sum(xx) sum(xy) sum(x)];
B=[-sum(xx+yy) ; -sum(xx.*y+yy.*y) ; -sum(xx.*x+xy.*y)];
a=A\B;
xc = -.5*a(1);
yc = -.5*a(2);
R = sqrt((a(1)^2+a(2)^2)/4-a(3));
end
Adapt as needed.
  5 Comments
Image Analyst
Image Analyst on 31 Mar 2020
To get the area of the white in the mask, I'd do
numWhitePixels = nnz(mask)
% Now to get the area of the holes
holesImage = imclearborder(~mask);
imshow(holesImage);
numHolePixels = nnz(holesImage)
Dyhia Bouhadjra
Dyhia Bouhadjra on 1 Apr 2020
This works pretty well. Thabk you for the help, and keep in touch for future exchange.
Keep safe and happy confinement ^^

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