problem in perimeter

1 Ansicht (letzte 30 Tage)
Mohammad Golam Kibria
Mohammad Golam Kibria am 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

Akzeptierte Antwort

Chandra Kurniawan
Chandra Kurniawan am 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

Weitere Antworten (3)

Chandra Kurniawan
Chandra Kurniawan am 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 Kommentar
Mohammad Golam Kibria
Mohammad Golam Kibria am 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 am 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

Image Analyst
Image Analyst am 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 Kommentar
Mohammad Golam Kibria
Mohammad Golam Kibria am 1 Dez. 2011
would u please mention the code, since the parameter of imdilate is not clear to me.

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