circle centered on the object
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I have a reference image that contains 10 items. I have separate objects in small images (cut from the reference image). Code sequence must find an object that you selected (if the write code sequence "object 1" must make a circle around the "object 1" If it says "Objective 2" to make the circle around the "item 2" Can you help me please? reference image
Object
clear; clc;
I = imread('di-5Y01.jpg');
object = imread('di-FNMJ.jpg');
c = normxcorr2(object(:,:,1),I(:,:,1));
[max_c, imax] = max(abs(c(:)));
[ypeak, xpeak] = ind2sub(size(c),imax(1));
corr_offset = [(xpeak-size(object,2)) (ypeak-size(object,1))];
figure, imshow(I); hold on;
rectangle('position',[corr_offset(1) corr_offset(2) 30 60],...
'curvature',[1,1],'edgecolor','g','linewidth',2);
Code sequence works but not centered the circle on the object.
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Respuesta aceptada
Chandra Kurniawan
el 22 de Dic. de 2011
Hello, Why don't you use rectangle??
Why must circle??
I have my code detect the exact area if you consider to use rectangle??
Would you use rectangle? Then I'll give u the code
Here the results :
Object 1:
Object 2:
Object 3:
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Más respuestas (3)
Chandra Kurniawan
el 22 de Dic. de 2011
Here the code if you consider to use rectangle
I = imread('di-5Y01.jpg');
object = imread('obiect1.jpg');
[m n o] = size(object);
c = normxcorr2(object(:,:,1),I(:,:,1));
[max_c, imax] = max(abs(c(:)));
[ypeak, xpeak] = ind2sub(size(c),imax(1));
corr_offset = [(xpeak-size(object,2)) (ypeak-size(object,1))];
figure, imshow(I); hold on;
rectangle('position',[corr_offset(1) corr_offset(2) n m],...
'edgecolor','g','linewidth',2)
I hope this helps you :)
Chandra Kurniawan
el 22 de Dic. de 2011
Hello, It's quite simple.
You just need to increase or decrease the 'corr_offset' values with small random value.
I = imread('di-5Y01.jpg');
object = imread('di-FNMJ.jpg');
c = normxcorr2(object(:,:,1),I(:,:,1));
[max_c, imax] = max(abs(c(:)));
[ypeak, xpeak] = ind2sub(size(c),imax(1));
corr_offset = [(xpeak-size(object,2)) (ypeak-size(object,1))];
figure, imshow(I); hold on;
rectangle('position',[corr_offset(1)-9 corr_offset(2)+5 45 45],...
'curvature',[1,1],'edgecolor','g','linewidth',2);
The result :
3 comentarios
Chandra Kurniawan
el 22 de Dic. de 2011
Hello, Why don't you use rectangle??
Why must circle??
I have my code detect the exact area if you consider to use rectangle??
Would you use rectangle? Then I'll give u the code
Tomescu
el 22 de Dic. de 2011
1 comentario
Chandra Kurniawan
el 22 de Dic. de 2011
So, how about your own decision?
You'll stay use circle with non exact area or
You'll use rectangle with exact area.
You can ask your professor about it :)
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