problem in perimeter

조회 수: 1 (최근 30일)
Mohammad Golam Kibria
Mohammad Golam Kibria 2011년 11월 30일
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

채택된 답변

Chandra Kurniawan
Chandra Kurniawan 2011년 11월 30일
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

추가 답변 (3개)

Chandra Kurniawan
Chandra Kurniawan 2011년 11월 30일
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
  댓글 수: 1
Mohammad Golam Kibria
Mohammad Golam Kibria 2011년 11월 30일
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 2011년 11월 30일
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
  댓글 수: 1
Mohammad Golam Kibria
Mohammad Golam Kibria 2011년 12월 1일
It also works fine

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Image Analyst
Image Analyst 2011년 11월 30일
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.
  댓글 수: 1
Mohammad Golam Kibria
Mohammad Golam Kibria 2011년 12월 1일
would u please mention the code, since the parameter of imdilate is not clear to me.

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