Remove background portion

Hi

May I know how to remove background by using original image's binary image? I found that I can use element wise multiplication with original mammogram to obtain the image I want, but how?

Please refer to the link below: http://www.mypicx.com/uploadimg/1811043486_10072011_1.png

Thank you.

[Walter Roberson]

My browser can't bring up that file... some problem with redirection for the site logo.

Answer 1

[Walter Roberson]

MaskedImage = MammogramImage .* repmat(BinaryImage,1,1,size(MammogramImage,3));

Anything set (1) in the binary image would have its corresponding position preserved, and anything clear (0) in the binary image would result in the correspond image location being zero'd (black).

Note: this won't work very well if the original image is a pseudocolor image ;-)

Answer 2

[Image Analyst]

I can't bring up the image either. I usually "dot multiply" like Walter did, but here's another method for masking that Sean uses:

% Mask the image.
maskedRgbImage = bsxfun(@times, rgbImage, cast(mask,class(rgbImage)));

It's a little more obtuse and cryptic than dot-multiplication but it gets the job done.

Answer 3

[ws]

Thanks for the reply. Here is another link: http://imageshack.us/photo/my-images/502/unled1lq.png/

my variables: # original image(unit8): I # binary image(unit8): I2 (only have 0 & 255)

LM = I .* repmat(I2,1,1,size(I,3));   % I cant get this works

LM = bsxfun(@times, I, cast(I, class(I2)));   % Cannot get the correct image.

Anything I'm wrong?

[Image Analyst]

How do we download the original images from that link?

[ws]

here is the link: http://peipa.essex.ac.uk/ipa/pix/mias/

[Walter Roberson]

LM = bsxfun(@times, I, cast(I, class(I2)));

should be

LM = bsxfun(@times, I, cast(I2, class(I)));

[ws]

The result is the same. Could you guide me through my code? Or any bugs in my code?

% image link: http://peipa.essex.ac.uk/ipa/pix/mias/
clear;

I = imread('mdb041.pgm');
imshow(I);

I = medfilt2(I, [1 5]);
% figure, imshow(I), title('Medfilt');
% Median filtering is a nonlinear operation often used in image processing to reduce "salt and pepper" noise. A median filter is more effective than convolution when the goal is to simultaneously reduce noise and preserve edges.
% Since all the mammograms are in high quality images, there is no need to
% perform median filtering

[x y] = size(I);

% find outer bw image
outer = I;
thres=3;

for i=1:x
    for j=1:y
        if (outer(i,j)<thres)
            outer(i,j)=0;   %black
        else             
            outer(i,j)=255; %white
        end
    end
end

%figure, imshow(outer), title('outer');

% find inner bw image
inner = I;
thres=12;

for i=1:x
    for j=1:y
        if (inner(i,j)<thres)
            inner(i,j)=0;
        else             
            inner(i,j)=255;
        end
    end
end

%figure, imshow(outer), title('inner');

result = outer - inner;
figure, imshow(result), title('outer - inner');

% RMLO process----------------------------------------
RMLO_result = result;
for i=1:x
    for j=1:y
        if (RMLO_result(i,j)==255)
            RMLO_result(i,j:y)=255;
        end
        j=1024;
    end
end

figure, imshow(RMLO_result), title('pre-RMLO');

RM = I - RMLO_result;
figure, imshow(RM), title('Done RMLO image');

% LMLO process-----------------------------------------
LMLO_result = result;
for i=1:x
    for j=y:-1:1
        if (LMLO_result(i,j)==255)
            LMLO_result(i,j:1024)=255;
            LMLO_result(i,1:j)=0;   
        end
    end
end

figure, imshow(LMLO_result), title('pre-LMLO');

%MaskedImage = MammogramImage .* repmat(BinaryImage,1,1,size(MammogramImage,3));
%LM = I .* repmat(LMLO_result,1,1,size(I,3));

%maskedRgbImage = bsxfun(@times, rgbImage, cast(mask,class(rgbImage)));
LM = bsxfun(@times, I, cast(LMLO_result, class(I)));

%%%%%LM = I - LMLO_result;
figure, imshow(LM), title('Done LMLO image');

%Identified LM or RM
mean_LM = mean(LM);
mean_RM = mean(RM);

if mean_LM > mean_RM
    view = LM;
    disp('Left')
else
    view = RM;
    disp('Right')
end    

%Remove pectoral muscle
% pect = view;
% thres=173;
% 
% for i=1:x
%     for j=1:y
%         if (pect(i,j)<thres)
%             pect(i,j)=0;   %black
%         end
%     end
% end
% 
% figure,imshow(I3),title('pre-pect');

% a) For every POLE ,it should not exceed no. of pixel travelled by previous 
%    row 
% b) Now if at all Rule a is violating for consecutive 5 times then by keeping
%    45o in mind decrease pole position for next row by 1 and replace all 
%    pixels with zero up to calculated POLE

[ws]

Hi,

I'm not sure my method is correct or not, although it's simple, but it works for me.

LM = abs(I - I2)

Thanks for those who try to help me :)