problem in perimeter

1 次查看（过去 30 天）
Mohammad Golam Kibria
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

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采纳的回答

Chandra Kurniawan
Chandra Kurniawan 2011-11-30
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

更多回答（3 个）

Chandra Kurniawan
Chandra Kurniawan 2011-11-30
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 个评论
Mohammad Golam Kibria
Mohammad Golam Kibria 2011-11-30
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 2011-11-30
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 2011-11-30
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 个评论
Mohammad Golam Kibria
Mohammad Golam Kibria 2011-12-1
would u please mention the code, since the parameter of imdilate is not clear to me.

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