Azzera filtri
Azzera filtri

Radius and coordinates of the center for a big circle in a bidimensional image

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I have a bidimensional (N x N) matrix of pixels like the one shown in the attached figure. Black areas contain zero values, while other areas contain random non-zero values. How could I determine the radius (in pixels) and coordinates of the center for the big circle? Consider that its center doesn't correspond to the center of the matrix, like in the figure.

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Image Analyst
Image Analyst il 11 Feb 2016
Modificato: Image Analyst il 11 Feb 2016
Get a binary image by thresholding. Then fill holes, label, and call regionprops().
binaryImage = grayImage > 10; % or whatever value works.
% Fill holes
binaryImage = imfill(binaryImage, 'holes');
% Label
labeledImage = bwlabel(binaryImage);
% Measure
measurements = regionprops(labeledImage, 'EquivDiameter');
% Extract
theRadius = measurements.EquivDiameter / 2;
By the way, instead of the very verbose and overly complicated "bidimensional (N x N) matrix of pixels", most people would simply just say "2D image".
  1 Commento
matnewbie il 11 Feb 2016
Thanks for your answer. I modified the original post, since I forgot to say that I am also interested in determining the coordinates of the center for the big circle. Is it possible to obtain the same results without using functions from the Image Processing Toolbox? Since I have a matrix of values, is it possible to find the coordinates of the center by minimizing the number of black pixels (zero values) from the edges of the black square?

Accedi per commentare.

Più risposte (1)

Anand il 11 Feb 2016
You can use the awesome imfindcircles function to do this as well.
Here's some code that should help:
% Read the image
I = imread('');
% Find circles in the image
radiusRange = [20 30]; % You need an approximate range of what you expect the radius to be.
[centers,radii] = imfindcircles(I, radiusRange, 'ObjectPolarity', 'dark','Sensitivity',.9);
% Display found circles
The variables centers and radii hold what you are looking for.

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