problem in perimeter

1 vue (au cours des 30 derniers jours)
Mohammad Golam Kibria
Mohammad Golam Kibria le 30 Nov 2011
hi,
I have a matrix as follows:
I =
0 0 0 0 0 0
0 0 1 1 0 0
0 1 1 1 1 0
0 1 1 1 1 0
0 0 1 1 0 0
0 0 0 0 0 0
I want to have another matrix as follows
I =
0 2 2 2 2 0
2 2 1 1 2 2
2 1 1 1 1 2
2 1 1 1 1 2
2 2 1 1 2 2
0 2 2 2 2 0
i.e. replace every position with 2 within distance 1 from the perimeter of region 1. Thanks

Réponse acceptée

Chandra Kurniawan
Chandra Kurniawan le 30 Nov 2011
Sorry. Here I modified my code
clear all; clc
I =[0 0 1 1 0 0;
0 1 1 1 1 0;
0 1 1 1 1 0;
0 0 1 1 0 0;
0 0 0 0 0 0];
J = zeros(size(I,1)+2, size(I,2)+2);
K = zeros(size(I,1)+2, size(I,2)+2);
J(2:size(J,1)-1, 2:size(J,2)-1) = I;
for x = 2 : size(J,1)-1
for y = 2 : size(J,2)-1
neighbour = [J(x-1,y-1) J(x-1,y) J(x-1,y+1) ...
J(x,y-1) J(x,y+1) ...
J(x+1,y-1) J(x+1,y) J(x+1,y+1)];
if (find(neighbour))
K(x,y) = 2;
end
end
end
L = K - J;
L(1,:) = []; L(end,:) = [];
L(:,1) = []; L(:,end) = []
And the result :
L =
2 2 1 1 2 2
2 1 1 1 1 2
2 1 1 1 1 2
2 2 1 1 2 2
0 2 2 2 2 0
  1 commentaire
Mohammad Golam Kibria
Mohammad Golam Kibria le 30 Nov 2011
Thanks

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Plus de réponses (3)

Chandra Kurniawan
Chandra Kurniawan le 30 Nov 2011
clear all; clc;
I =[0 0 0 0 0 0;
0 0 1 1 0 0;
0 1 1 1 1 0;
0 1 1 1 1 0;
0 0 1 1 0 0;
0 0 0 0 0 0];
J = zeros(size(I,1)+2, size(I,2)+2);
K = zeros(size(I,1)+2, size(I,2)+2);
J(2:7,2:7) = I;
for x = 2 : 7
for y = 2 : 7
neighbour = [J(x-1,y-1) J(x-1,y) J(x-1,y+1) ...
J(x,y-1) J(x,y+1) ...
J(x+1,y-1) J(x+1,y) J(x+1,y+1)];
if (find(neighbour))
K(x,y) = 2;
end
end
end
L = K - J;
L(1,:) = []; L(end,:) = [];
L(:,1) = []; L(:,end) = []
And you will get L =
L =
0 2 2 2 2 0
2 2 1 1 2 2
2 1 1 1 1 2
2 1 1 1 1 2
2 2 1 1 2 2
0 2 2 2 2 0
  1 commentaire
Mohammad Golam Kibria
Mohammad Golam Kibria le 30 Nov 2011
but if my matrix is as follows:
I =[0 0 1 1 0 0;
0 1 1 1 1 0;
0 1 1 1 1 0;
0 0 1 1 0 0;
0 0 0 0 0 0];
It does not work, how to extend your idea for any matrix I give, I am not good in matlab. could u help me in this regard.

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Andrei Bobrov
Andrei Bobrov le 30 Nov 2011
variant use conv2 without Image Processing Toolbox
t = conv2(I,ones(3),'same')>0
out = t + 0
out(t>0&t~=I) = 2
or
out = 2*(conv2(I,ones(3),'same')>0+0)-I
variant use with function imdilate by Image Processing Toolbox
out = imdilate(I,ones(3))
out(out~=I) = 2
or
out = 2*imdilate(I,ones(3))-I
  1 commentaire
Mohammad Golam Kibria
Mohammad Golam Kibria le 1 Déc 2011
It also works fine

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Image Analyst
Image Analyst le 30 Nov 2011
If you have the Image Processing Toolbox you can call imdilate() and then bwperim() and then combine the perimeter image with the original by multiplying the perimeter image by 2 and adding to the original image.
  1 commentaire
Mohammad Golam Kibria
Mohammad Golam Kibria le 1 Déc 2011
would u please mention the code, since the parameter of imdilate is not clear to me.

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