imfindcircles doesn't work

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John
John am 29 Jun. 2025
Kommentiert: Mathieu NOE am 6 Jul. 2025
The image looks normal with range from 0 to 4096:
[x,y] = size(AA1); % output x and y both 512
Obviously radius range is [x/4 x/2].
However, both of below:
[centers, radii, metric] = imfindcircles(AA1,[fix(x/4) fix(x/2)],'ObjectPolarity','bright')
[centers, radii, metric] = imfindcircles(AA1,[fix(x/4) fix(x/2)])
output:
centers = []
radii = []
metric = []
changed to [x/6 x/2] doesn't help.
Thank you for finding the issue!
  3 Kommentare
John
John am 29 Jun. 2025
Even made it binary:
AAB = zeros(size(AA1));
AAB(AA1>max(AA1(:))/2) = 1;
figure;imshow(AAB,[])
it showed a great binary circle:
[centers, radii, metric] = imfindcircles(AAB,[for all ranges])
Output still is:
centers = []
radii = []
mtric = []
Image Analyst
Image Analyst am 30 Jun. 2025
If you have any more questions, then attach your image and code to read it in with the paperclip icon after you read this:

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Akzeptierte Antwort

Mathieu NOE
Mathieu NOE am 30 Jun. 2025
hello
let's try with your image
AA = imread('circle.png');
AA = rgb2gray(AA);
BW = imbinarize(AA);
[B,L,N] = bwboundaries(BW);
whos
Name Size Bytes Class Attributes AA 510x505 257550 uint8 B 1x1 17768 cell BW 510x505 257550 logical L 510x505 2060400 double N 1x1 8 double ans 1x34 68 char
figure;imshow(BW)
hold on
%Display object boundaries in red and hole boundaries in green.
N
N = 1
for k=1:length(B),
boundary = B{k};
if(k > N)
plot(boundary(:,2), boundary(:,1), 'g','LineWidth',2);
else
plot(boundary(:,2), boundary(:,1), 'r','LineWidth',2);
end
end
[m,n] =size(BW);
[centers,radii]=imfindcircles(BW,round([n/4,n/2]))
centers = 1×2
252.1513 250.1969
<mw-icon class=""></mw-icon>
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radii = 195.0940
  3 Kommentare
Image Analyst
Image Analyst am 4 Jul. 2025
I think @John's ignoring us now - he didn't answer my comment up above either.
@Mathieu NOE did you know there is a visboundaries function?
help visboundaries
visboundaries - Plot region boundaries This MATLAB function draws boundaries of regions in the binary image BW on the current axes. Syntax visboundaries(BW) visboundaries(B) visboundaries(ax,___) visboundaries(___,Name,Value) h = visboundaries(___) Input Arguments BW - Binary image logical array B - Boundary pixel locations cell array of Q-by-2 matrices ax - Image on which to draw boundaries current axes (default) | axes object Name-Value Arguments Color - Color of boundary 'red' (default) | RGB triplet | hexadecimal color code | color name | short color name LineStyle - Style of boundary line '-' (default) | '--' | ':' | '-.' | 'none' LineWidth - Width of the line used for the boundary 2 (default) | positive number EnhanceVisibility - Augment drawn boundary with contrasting features true or 1 (default) | false or 0 Output Arguments h - Boundary lines hggroup object Examples openExample('images/ComputeBoundariesAndPlotOnImageExample') openExample('images/VisualizeSegmentationResultExample') See also bwboundaries, bwperim, bwtraceboundary, viscircles Introduced in Image Processing Toolbox in R2015a Documentation for visboundaries doc visboundaries
Mathieu NOE
Mathieu NOE am 6 Jul. 2025
tx for the info, I'll try to remember
as you already know , I am not truly an image expert , that's not my every day job
I'm sometimes trying to improve my basic skills in that field
don't be surprised if my answers are still a bit "basic"

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Weitere Antworten (1)

Matt J
Matt J am 30 Jun. 2025
Verschoben: Image Analyst am 4 Jul. 2025
I can't reproduce the problem you claim to have with a binarized disk. That case is pretty well-handled by imfindcircles:
n=300;
[x,y]=deal(1:n);
mask=hypot(x-mean(x),y'-mean(y))<=n/3;
[centers,radii]=imfindcircles(mask,[n/4,n/2])
centers = 1×2
150.5011 150.5011
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
radii = 99.9962
imshow(mask,[])
viscircles(centers,radii);

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