Reputation: 123
I want to process an image in matlab
The image consists out of a solid back ground and two specimens (top and bottom side). I already have a code that separate the top and bottom and make it two images. But the part what I don't get working is to crop the image to the glued area only (red box in the image, I've only marked the top one). However, the cropped image should be a rectangle just like the red box (the yellow background, can be discarded afterwards).
I know this can be done with imcrop
, but this requires manual input from the user. The code needs to be automated such that it is possible to process more images without user input. All image will have the same colors (red for glue, black for material).
Can someone help me with this?
edit: Thanks for the help. I used the following code to solve the problem. However, I couldn't get rid of the black part right of the red box. This can be fix by taping that part off before making pictures. The code which I used looks a bit weird, but it succeeds in counting the black region in the picture and getting a percentage.
a=imread('testim0.png');
level = graythresh(a);
bw2=im2bw(a, level);
rgb2=bw2rgb(bw2);
IM2 = imclearborder(rgb2,4);
pic_negative = ait_imgneg(IM2);
%% figures
% figure()
% image(rgb2)
%
% figure()
% imshow(pic_negative)
%% Counting percentage
g=0;
for j=1:size(rgb2,2)
for i=1:size(rgb2,1)
if rgb2(i,j,1) <= 0 ...
& rgb2(i,j,2) <= 0 ...
& rgb2(i,j,3) <= 0
g=g+1;
end
end
end
h=0;
for j=1:size(pic_negative,2)
for i=1:size(pic_negative,1)
if pic_negative(i,j)== 0
h=h+1;
end
end
end
per=g/(g+h)
If anyone has some suggestions to improve the code, I'm happy to hear it.
Upvotes: 1
Views: 9937
Reputation: 21
since you've already been able to seperate the top and bottom, and also able to segment the region you want (including the small part of the right side that you don't want), I propose you just add a fix at the end of the code via the following.
after segmentation, sum each column Blue intensity value, so that you are compressing the image from 2d to 1d. so if original region is width=683 height=59, the new matrix/image will be simply width=683 height=1. now, you can apply a small threshold to determine where the edge should lie, and apply a crop to the image at that location. Now you get your stats.
Upvotes: 0
Reputation: 5014
For a straight-forward image segmentation into 2 regions (background, foreground) based on color (yellow, black are prominent in your case), an option can be clustering image color values using kmeans algorithm. For additional robustness you can transform the image from RGB to Lab* colorspace.
The example for your case follows the MATLAB Imape Processing example here.
% read and transform to L*a*b space
im_rgb = double(imread('testim0.png'))./256;
im_lab = applycform(im_rgb, makecform('srgb2lab'));
% keep only a,b-channels and form feature vector
ab = double(lab_I(:,:,2:3));
[nRows, nCols, ~] = size(ab);
ab = reshape(ab,nRows * nCols,2);
% apply k-means for 2 regions, repeat c times, e.g. c = 5
nRegions = 2;
[cluster_idx cluster_center] = kmeans(ab,nRegions, 'Replicates', 5);
% get foreground-background mask
im_regions = reshape(cluster_idx, nRows, nCols);
You can use the resulting binary image to index the regions of interest (or find the boundary) in the original reference image.
Upvotes: 2
Reputation: 2111
You can easily get the unknown bounded crop by
1) Contour plotting the image,
2) find()
on the result for the max/min X/ys,
3) use @slayton's method to perform the actual crop.
EDIT: Just looked at your actual image - it won't be so easy. But color enhance/threshold your image first, and the contours should work with reasonable accuracy. Needless to say, this requires tweaking to your specific situation.
Upvotes: 0
Reputation: 20319
Images are saved as matrices. If you know the bounds in pixels of the crop box want to crop you can execute the crop using indexing.
M = rand(100); % create a 100x100 matrix or load it from an image
left = 50;
right = 75;
top = 80;
bottom = 10;
croppedM = M(bottom:top, left:right);
%save croppedm
Upvotes: 0