How to crop images in Matlab based on Intensity Profiles?

I have an image in Matlab which I want to crop based on the intensity profile. Below is my image.

I want to crop out the blue portion of the image based on the intensity profile. For example, based on the intensity profile, I want to crop the part of the image where the intensity of blue is higher than the other colors. Below is my sample code and example.

hFig = figure ; 
subplot(1, 2, 1 ); 
rgbImage = imread('Intensity1.jpg '); 
imshow(rgbImage ) 
subplot(1, 2, 2 ); 
croppedImage = rgbImage(200 : 400 , :, :); 
imshow(croppedImage);

In the image above, it can be seen that the blue portions still remain in the image which I don't want. Any help would be very much appreciated. Thank you.

[KALYAN ACHARJYA]

Crop or segment?

[Warid Islam]

Hi Kalyan,

I want to crop out the blue portion. That is I want an image with only the yellow and green color parts.

[KALYAN ACHARJYA]

What remain in the blue pixels on resultant image? What would be the size (rows x colm) of the output images, is it same as original images?

[Warid Islam]

Hi Kalyan,

The resultant image would somehow look below:

However, I used drawfreehand() above which should not be a correct process to crop the image above as this would cause loss in information. I want to crop out the blue parts of the image based on the intensity profile.

[KALYAN ACHARJYA]

If I am not wrong, 2-3 days before you asked the same question, right? I can help on segmentation only, this approach I have to try. Hope experts will give answer to your question.

Good Night!

[Warid Islam]

Hi Kalyan,

You are absolutely correct. I did ask similar question few days back. There, I wanted to crop using drawfreehand() approach. Now I want to crop using the intensity profile. Anyways, any help on segmentation would also be of a great help. Thank you very much for your cooperation.

[Image Analyst]

Can you post the original image? The grayscale image before you pseudocolored it? That's really what you need. If you don't have that, you'll have to "undo" the colormap, like I did in the attached example.

[Warid Islam]

Hi Image Analyst,

Please find the original image below. Thank you.

[Image Analyst]

CLoser, but that's not the original image. Can you attach the original image, in PNG, BMP, JPG or TIF format. I don't want the screenshot with axes numbers on it.

[Warid Islam]

Hi ,

My original image has been obtained from OCT machine which has axes numebrs on it. Unfortunately, I am unable to remove it from the image.

[Warid Islam]

Hi Image Analyst,

I tried to make changes to my code. But I still don't get any desired output.

clear all ; close all; clc
I = imread('Intensity5.png');
S = imcrop(I);

x = [215 215];
y = [0 size(I,1)];
c = improfile(I,x,y);

figure
subplot(2,1,1)
imshow(I);
hold on
plot(x,y,'r')
subplot(2,1,2)
plot(c(:,1,1),'r')
hold on
plot(c(:,1,2),'g')
plot(c(:,1,3),'b')


r= c(:,1,1);
g= c(:,1,2);
b= c(:,1,3);

if r > b
    I = imcrop(I);
elseif g > b
    I = imcrop(I);
else
    S = I.createMask();
    imshow(S)
end
     

%   find(c(:,1,3) > 250)==0;

[Image Analyst]

Looks like you've accepted the answer below. I am dealing with an OCT project in my company right now. I would not segment/threshold the RGB image like in the answer and like you've done it above. You have the original gray scale image and that's what you need to threshold. Do not do it on the RGB image.

brightStuff = grayImage > some Value;

[Warid Islam]

Hi Image Analyst,

Thank you for the suggestions. I would implement those in the final output.

Answer 1

[awezmm]

Try this where img is your original image:

% get red, green, blue channels
redC = img(:,:,1);
greenC = img(:,:,2);
blueC = img(:,:,3);

% mask where red channel is greater than blue channel and green channel greater than blue channel
mask = redC > blueC & greenC > blueC;

% overlay mask into original image
finalimg = bsxfun(@times, img, cast(mask,class(img)));

%plotting
figure
subplot(1,2,1)
imshow(img)
subplot(1,2,2)
imshow(finalimg)

[Warid Islam]

Hi Awezmm,

That works perfectly for me. Thank you very much for the help.

[KALYAN ACHARJYA]

+1 Great!