Problem trying to access labeled portion of images (updated 7/5)
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unique(diff(unique(endpts_labeled)))
ans =
1
2
ORIGINAL QUESTION:
I am getting a rare error running the following code on an image of 1px thick skeleton segments:
V(V<=0) = 0; V(V>0) = 1; % Segments <=2 pixels are removed as well earlier
[K n] = bwlabel(V,8);
props = regionprops(K,'Perimeter');
tort = zeros(1,n);
endpts = bwmorph(K,'endpoints');
endpts_labeled = immultiply(K,endpts);
for m = 1:n
[rows,cols] = find(endpts_labeled==m);
p1 = [rows(1),cols(1)]; % Error happens here!!
p2 = [rows(2),cols(2)];
d = sqrt((p1(1)-p2(1))^2+(p1(2)-p2(2))^2);
c = props(m,1).Perimeter/2;
tort(m) = c/d;
end
The error happens when the loop tries to access rows(1) when the numel(rows) = 0. That case should never happen though. This error happened one in every 3 or 4 runs previously when I was using a non-labeled version of the image to find the endpoints, but now that I am using the labeled image to find the endpoints, it is only happening once in a blue moon and I cannot figure out why.
1 comentario
Ryan
el 5 de Jul. de 2012
Respuesta aceptada
Más respuestas (2)
Image Analyst
el 4 de Jul. de 2012
Editada: Image Analyst
el 4 de Jul. de 2012
I don't know how this could happen. If rows is null, then endpts_labeled==m is completely false. And if that's true, then n must be zero and so the loop never gets entered. That is a contradiction. Try "dbstop if error" and then when it errors, see what n is. Alternatively you can try this (untested)
endpts_labeled = int32(zeros(size(K)));
endpts_labeled(endpts) = K(endpts);
instead of this:
endpts_labeled = immultiply(K,endpts);
in case the immultiply is giving you slight mismatches as referred to in the FAQ http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F
3 comentarios
Image Analyst
el 6 de Jul. de 2012
immultiply may work if you cast it to int32 afterwards.
endpts_labeled = int32(immultiply(K,endpts));
just in case there are any really small fractional parts produced during the multiply because of quantization error.
Steve Eddins
el 6 de Jul. de 2012
K contains all integers, and endpts contains only 0s and 1s. There shouldn't be any quantization error going on here.
Ryan
el 6 de Jul. de 2012
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