How to get the barcode region to red color??

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Kim
Kim on 26 Jan 2012
How can I write the code to get the red region works on vertical barcode image shown on figure 1(a&b).
Figure 2(a&b) is what I want.
I also attach my coding for your reference.
----
Figure 1a (Original image)
Figure 1b (Processed Image)
Figure 2a (Original Image)
Figure 2b (Processed Image)
Coding
rgb = imread('barcode12.jpg');
% Resize Image
rgb = imresize(rgb,0.33);
figure(),imshow(rgb);
% Convert from RGB to Gray
Igray = rgb2gray(rgb);
Igray = double(Igray);
% Calculate the Gradients
[dIx, dIy] = gradient(Igray);
B = abs(dIx) - abs(dIy);
% Low-Pass Filtering
H = fspecial('gaussian', 20, 10);
C = imfilter(B, H);
figure(),imagesc(C); colorbar;

Accepted Answer

Chandra Kurniawan
Chandra Kurniawan on 26 Jan 2012
When use the first image (vertical barcode image),
I just simply try to change the position of abs(dIx) and abs(dIy) into :
B = abs(dIy) - abs(dIx);
But when I use the second image (horizontal barcode image),
I have the same problem.
Then, I decide to use this command :
B = imabsdiff(abs(dIx),abs(dIy));
  3 Comments
Kim
Kim on 27 Jan 2012
After I did a 'imfilter(B, H)' , how can I convert the image to the colorbar image so that I can do a bwareaopen. Thereafter, draw a box or a line across to determine the barcode. I don't know whether my idea is it good or not.
Anyway, thanks for the code, really appreciate it! Just that not the thing that I wanted.

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More Answers (1)

Chandra Kurniawan
Chandra Kurniawan on 27 Jan 2012
Hi, Kim
I agree with you. To determine the barcode just use boundingbox from regionprops :)
Irgb = imread('15wmqe9.jpg');
Irgb = imread('av672a.jpg');
Iresize = imresize(Irgb,0.33);
Igray = double(rgb2gray(Iresize));
[dIx dIy] = gradient(Igray);
B = imabsdiff(abs(dIx),abs(dIy));
H = fspecial('gaussian', 20, 10);
C = imfilter(B, H);
Th = max(C(:));
D = C < Th-10;
stat = regionprops(~D,'Area','BoundingBox');
for i = 1 : numel(stat),
Iarea(i) = stat(i).Area;
end
[C I] = max(Iarea);
bb = stat(I).BoundingBox;
imshow(Iresize); hold on
rectangle('position',bb,'edgecolor','r');
  2 Comments
Image Analyst
Image Analyst on 17 Oct 2012
This is not a robust algorithm. It's fine tuned for these particular images. It may need to be tweaked for your images. Reply to the thread you started with the URL to your image.

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