finding region

2 vues (au cours des 30 derniers jours)
Mohammad Golam Kibria
Mohammad Golam Kibria le 29 Juin 2011
Hi I have a matrix as follows:
I=
1 1 0 0 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0 0
0 0 0 0 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
Here 1 has divided the matrix in two parts I need to have the indices of these two separate parts. say idx1 will be upper zeros and idx2 contains lower zero indices.Can any one help.

Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 29 Juin 2011
BW2 = bwmorph(I,'diag');
L = bwlabel(~BW2);
Id1 = bwdist(I,'chessboard') == 1;
for i1 = 1:max(L(:))
L(bwdist(L == i1,'chessboard')==1 & Id1) = i1;
end
idx1 = find(max(L(1,:))==L);
idx2 = find(max(L(end,:))==L);
EDIT as at Sean
L = bwlabel(~I,4);
idx1 = find(max(L(1,:))==L);
idx2 = find(max(L(end,:))==L);
  1 commentaire
Mohammad Golam Kibria
Mohammad Golam Kibria le 30 Juin 2011
Thanks

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Plus de réponses (2)

Sean de Wolski
Sean de Wolski le 29 Juin 2011
CC = bwconncomp(~I,4);
CC.PixelIdxList will contain the linear indices. To get the sub indices either use regionprops with the 'pixellist' option or ind2sub
  1 commentaire
Andrei Bobrov
Andrei Bobrov le 29 Juin 2011
+1
Hi Sean! I stupid, forgotten about the variable 4-connectivity. Thanks

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Paulo Silva
Paulo Silva le 29 Juin 2011
Just for fun
I=[1 1 0 0 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0 0
0 0 0 0 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0];
up=[];down=[];
for c=1:size(I,2)
d=0;u=1;
for r=1:size(I,1)
if(d==1)
down=[down sub2ind(size(I),r,c)];
end
if(I(r,c)==1)
d=1;u=0;
end
if u==1
up=[up sub2ind(size(I),r,c)];
end
end
end
disp('index of zeros bellow the 1')
down
disp('index of zeros above the 1')
up
%for big arrays it's better to pre-allocate the up and down vectors, much faster
  1 commentaire
Mohammad Golam Kibria
Mohammad Golam Kibria le 30 Juin 2011
Thanks. Its also ok.How to pre-allocate the up and down vectors?

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