problem in perimeter

1 visualización (últimos 30 días)
Mohammad Golam Kibria
Mohammad Golam Kibria el 30 de Nov. de 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

Respuesta aceptada

Chandra Kurniawan
Chandra Kurniawan el 30 de Nov. de 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

Más respuestas (3)

Chandra Kurniawan
Chandra Kurniawan el 30 de Nov. de 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 comentario
Mohammad Golam Kibria
Mohammad Golam Kibria el 30 de Nov. de 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 el 30 de Nov. de 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

Image Analyst
Image Analyst el 30 de Nov. de 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 comentario
Mohammad Golam Kibria
Mohammad Golam Kibria el 1 de Dic. de 2011
would u please mention the code, since the parameter of imdilate is not clear to me.

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