清空筛选条件
清空筛选条件

circle centered on the object

4 次查看（过去 30 天）
Tomescu
Tomescu 2011-12-22
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
http://www.uploadup.com/di-5Y01.jpg
Object
http://www.uploadup.com/di-FNMJ.jpg
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.

请先登录，再回答此问题。

采纳的回答

Chandra Kurniawan
Chandra Kurniawan 2011-12-22
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:

更多回答（3 个）

Chandra Kurniawan
Chandra Kurniawan 2011-12-22
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
Chandra Kurniawan 2011-12-22
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 个评论
显示 1更早的评论
Chandra Kurniawan
Chandra Kurniawan 2011-12-22
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
Tomescu 2011-12-23
Hello!
I found this code on another forum.
The problem is that I think it's a bit complet.Can you look over it please?
Thank you.
%#read & convert the image
imgCol = imread('http://i.stack.imgur.com/tbnV9.jpg');
imgGray = rgb2gray(img);
obj = rgb2gray(imread('http://i.stack.imgur.com/GkYii.jpg'));
%# cross-correlate and find the offset
corr = normxcorr2(...);
[~,indx] = max(abs(corr(:))); %# Modify for multiple instances (generalize)
[yPeak, xPeak] = ind2sub(...);
corrOffset = [yPeak - ..., xPeak - ...];
%# create a mask
mask = zeros(size(...));
mask(...) = 1;
mask = imdilate(mask,ones(size(...)));
%# plot the above result
h1 = imshow(imgGray);
set(h1,'AlphaData',0.4)
hold on
h2 = imshow(imgCol);
set(h2,'AlphaData',mask)

请先登录，再进行评论。

Tomescu
Tomescu 2011-12-22
The ptoject say that the object must be identify in a circle, but i hope that it whont be a problem if is in a rectangle.
  1 个评论
Chandra Kurniawan
Chandra Kurniawan 2011-12-22
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 :)

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Feature Detection and Extraction 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by