how to flip a segmented image along the major axis

I and my friends are doing project on skin cancer detection using image processing, We completed upto segmentation, now we have to find area of non overlapped region of the segmented image.For this we have to flip the image along its major axis andminor axis. so please help us in this matter.

Advance thanks.

[James Morris]

Hi, I have used (and slightly edited) your code to rotate and center my own skin lesions, but I am now struggling on being able to fold it along the X axis, and calculate the difference between the 2 overlapping areas. I need to show the overlapping regions in an image, furthermore I need to also be able sum the differences of the overlapping area, as this data needs to go into a classifier. I have linked my image of what I have so far, and also my code. Any guidance on how to solve my issue will be greatly appreciated. Thanks.

[Image Analyst]

See my answer below Just sum the images if you need a pixel count:

sum1 = sum(overlapped(:));
sum2 = sum(nonOverlapped(:));

Answer 1

[Image Analyst]

I gave you the initial part of the code already but it appears you need more help so here's some more. See attached m-file below the image. I don't know how you got a very high value for the orientation - I get 14 degrees. The only remaining part you asked for is " we have to flip the image along its major axis" and for that you can use flipud(rotatedImage) - very simple.

[Image Analyst]

narayana or anastasia - are either of you still there?

[anastasia]

There seems to be some error when I implement the code you gave. I slightly modified the threshold to using level=graythresh(image) and the lesion is not appearing in my bwImage Was modifying the threshold the error? Even when I used the same code you gave, it showed the same output image without the lesion.

I've attached the code below. Please help.

[anastasia]

I fixed the previous error but I need the final image to look something like this:

After following your method, I have two arrays, one for overllaped area and one for the non-overlapped area. However, I need a scalar value that gives me the asymmetry index according to the following equation. I have to feed this asymmetry index to a classifier and hence I need a scalar value.

Ax is the area difference about the x-axis(When the two halves are folded about the x-axis) and Ay is the area difference about the y-axis.

How to incorporate A1 and A2 values into the code you gave. Please help. Thank You.

[anastasia]

Image analyst, Please help with the above query. It's really important for my project. Thank You.

[Image Analyst]

Simply use xor() on the flipped and unflipped images:

diffImage = xor(flipped, notFlipped);

[anastasia]

I am getting this image when I use the above line of code.

flipped = fliplr(rotatedImage);
overlapped = flipped & rotatedImage;
nonOverlapped = xor(flipped, rotatedImage);
diffImage= xor( flipped, binaryImage);
figure();
imshow(diffImage);

In the image I attached in my previous comment, I just need the asymmetry part for one half of the image. Here my, notflipped part is the binaryImage right? Please suggest modification to my code.

[anastasia]

If I use

xor(flipped, overlapped)

I'm getting the below result. Which is the right one? I need the AREA DIFFERENCE between the two halves of the image.

[Image Analyst]

So did you compute it or not? I assume you simply got the areas of each half then subtracted like you said you needed to do.

topArea = sum(sum(binaryImage(1:rows/2, :)));
bottomArea = sum(sum(binaryImage(rows/2+1:end, :)));
areaDifference = abs(topArea - bottomArea);

Did this not give you the right answer? Please explain.

[anastasia]

I am getting a scalar value after computing the areas. But I have to find the area difference between the overlapped folds. For instance, if I am flipping the image about the x-axis, I need the area difference between the overlapping folds. Not the top area and the bottom area.

I am confused as to which two should I subtract. Overlapped and non-overlapped or top area and bottom area.

PLease explain the logic. Thank you so much.

[Image Analyst]

The binary image you have is an xor image, so that's already showing the pixels that are different between the unflipped and flipped image. So you just need to count them. You said you wanted a separate count for the top (above the mid-line) and bottom (below the mid-line), so that's what I gave you. If you just want the area of everything, just sum the original binary image, not the xor image, or use bwarea() to compute sums instead of the sum() function.

[anastasia]

It makes sense now! Thank you so much. Now, I have to do determine the color asymmetry index. The only variation being for color asymmetry, the * same procedure* is carried out except that, instead of area, pixel values were incorporated in the calculations: the sum of the absolute grey-level difference between the corresponding pixels in the two folds

I extrcated the color channels from my RGB image and followed the similar procedure above. But I am stuck at finding the sum of the absolute grey level difference between the folds. How do I proceed?

Thank you so much for the guidance! You've been of great help.

[Image Analyst]

To find the gray level differences between two images, you can do

diffImage = abs(double(image1) - double(image2));

This is an image the same size as your input images. You can then look at just parts of the image if you want (top half, bottom half, the xor mask, or whatever).

[waffle]

why i got error at this line ?

line([middlex, middlex], [1, rows], 'Color', 'm', 'LineWidth', 2);

[Image Analyst]

Because you removed the line above it that said:

middlex = columns/2;

That's my only guess, because you forgot to include the error message. Other possibilities include you redefining line() with your own m-file, or not defining rows.

[waffle]

I have try this your example to find the non overlap region but why i dont get the axis are not at the center of the image.Please help me..tq

[Image Analyst]

Why should it be? The blob could be absolutely anywhere in the image. It would be a miracle if the centroid of the blob happened by chance to be exactly at the center of the image.

[waffle]

How can I do to make the axis at the is at the center?

[Sadia Iffat]

Image Analyst.I run your code.But I didnot get correct output .

I think to get better overlapping result the lesion area should move into the middle .Thats why you move it to middle.I want to translate the lesion into the image middle but your code translate the image not the object.Am i misunderstood? or what should i do? i am a newbie

[Sadia Iffat]

If you see the comment please answer..i am waiting for your help..

[Image Analyst]

Your lesion is black instead of white. You reversed the sign during thresholding. It should be <, not >. Like

binaryImage = grayImage < 100; % or whatever number works.

so that you get dark stuff, not bright stuff.

[Sadia Iffat]

    clc;    % Clear the command window.
    close all;  % Close all figures (except those of imtool.)
    clear;  % Erase all existing variables. Or clearvars if you want.
    workspace;  % Make sure the workspace panel is showing.
    format long g;
    format compact;
    fontSize = 15;

    %===============================================================================
    % Get the name of the image the user wants to use.
    baseFileName = 'Figure9.jpg'; % Assign the one on the button that they clicked on.
    % Get the full filename, with path prepended.
    folder = []; % Determine where demo folder is (works with all versions).
    fullFileName = fullfile(folder, baseFileName);

    %===============================================================================
    % Read in a demo image.
    grayImage = imread(fullFileName);
    % Get the dimensions of the image.  
    % numberOfColorChannels should be = 1 for a gray scale image, and 3 for an RGB color image.
    [rows, columns, numberOfColorChannels] = size(grayImage);
    if numberOfColorChannels > 1
  % It's not really gray scale like we expected - it's color.
  % Use weighted sum of ALL channels to create a gray scale image.
  grayImage = rgb2gray(grayImage); 
  % ALTERNATE METHOD: Convert it to gray scale by taking only the green channel,
  % which in a typical snapshot will be the least noisy channel.
  % grayImage = grayImage(:, :, 2); % Take green channel. 
    end
    % Display the image.
    subplot(2, 2, 1);
    imshow(grayImage, []);
    axis on;
    caption = sprintf('Original Gray Scale Image');
    title(caption, 'FontSize', fontSize, 'Interpreter', 'None');
    drawnow;
    hp = impixelinfo();

    % Set up figure properties:
    % Enlarge figure to full screen.
    set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
    % Get rid of tool bar and pulldown menus that are along top of figure.
    set(gcf, 'Toolbar', 'none', 'Menu', 'none');
    % Give a name to the title bar.
    set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off') 
    drawnow;

    % Binarize the image by thresholding.
    binaryImage = grayImage<100;
    subplot(2, 2, 2);
    imshow(binaryImage);
    % Get rid of white surround.
    binaryImage = imclearborder(binaryImage);
    % Extract the largest blob only.
    binaryImage = bwareafilt(binaryImage, 1);

    axis on;
    title('Binary Image with centers found', 'fontSize', fontSize);
    drawnow;

    % Label the image
    labeledImage = bwlabel(binaryImage);
    % Make the measurements
    props = regionprops(labeledImage, 'Centroid', 'Orientation');
    xCentroid = props.Centroid(1)
    yCentroid = props.Centroid(2)
    %  Find the half way point of the image.
    middlex = columns/2;
    middley = rows/2;
    % Plot a + at the center of the blob and center of the image.
    hold on;
    line([middlex, middlex], [1, rows], 'Color', 'm', 'LineWidth', 2);
    line([1, columns], [middley, middley], 'Color', 'm', 'LineWidth', 2);
    plot(xCentroid, yCentroid, 'r+', 'MarkerSize', 30);
    plot(xCentroid, yCentroid, 'ro', 'MarkerSize', 30);
    % Translate the image
    deltax = middlex - xCentroid;
    deltay = middley - yCentroid;
    binaryImage = imtranslate(binaryImage,  [deltax, deltay]);
    % Display the image.
    subplot(2, 2, 3);
    imshow(binaryImage);
    axis on;
    title('Binary Image Translated to Middle', 'fontSize', fontSize);
    % Plot a + at the center of the blob and center of the image.
    hold on;
    line([middlex, middlex], [1, rows], 'Color', 'm', 'LineWidth', 2);
    line([1, columns], [middley, middley], 'Color', 'm', 'LineWidth', 2);
    plot(xCentroid, yCentroid, 'r+', 'MarkerSize', 30);
    plot(xCentroid, yCentroid, 'ro', 'MarkerSize', 30);
    drawnow;

    % Rotate the image
    angle = -props.Orientation
    rotatedImage = imrotate(binaryImage, angle, 'crop');
    %  Display the image.
    subplot(2, 2, 4);
    imshow(rotatedImage);
    axis on;
    caption = sprintf('Image Rotated by %f Degrees', angle);
    title(caption, 'fontSize', fontSize);
    % Plot a + at the center of the blob and center of the image.
    hold on;
    line([middlex, middlex], [1, rows], 'Color', 'm', 'LineWidth', 2);
    line([1, columns], [middley, middley], 'Color', 'm', 'LineWidth', 2);
     plot(xCentroid, yCentroid, 'r+', 'MarkerSize', 30);
    plot(xCentroid, yCentroid, 'ro', 'MarkerSize', 30);
    drawnow;

[Sadia Iffat]

Still..I get the result for the above code..

[Sadia Iffat]

I am waiting for your reply..

[Image Analyst]

Not sure what reply you're waiting for. Perhaps "Congratulations, you got it working!"?

[Sadia Iffat]

Thanks for your help.And sorry for annoying by asking silly question.What should i do now to flip along the minor axis compared to major axis?

[Sadia Iffat]

BY rotating 90 degree of image that was obtain from rotating for major axis?

[Image Analyst]

Once you've rotated the blob so that the major axis is along the horizontal axis, simply use flipud() to rotate it about the minor axis.

[Hannah Gabrielle Alagon]

Hello. can someone help on how to do this image below.

[Image Analyst]

I don't know what help you need beyond the code that I attached. Be more specific.

Answer 2

[Image Analyst]

First align the image so that the major axis is horizontal (or vertical) with imrotate(). Then use imtranslate() to put the center line at the center of the image. Then use flipud (or fliplr) and AND or XOR them together

rotatedBinary = imrotate(binaryImage, angle);
rotatedBinary  = imtranslate(rotatedBinary,.....
flipped = flipud(rotatedBinary);
overlapped = flipped & rotatedBinary;
nonOverlapped = xor(flipped, rotatedBinary);

I'm hoping you can find the angle yourself (it comes from Orientation in regionprops), and can find the amount of lines to translate it (it also comes from Centroid in regionprops)

[narayana reddy]

We have to fit an ellipse which coincides almost to the segmented image. For that, along major axis and minor axis, we have to flip the image.

This is the image after rotation by seeing the orientation value from regionprops. now we have to flip the image along its major axis

[Image Analyst]

OK.

[narayana reddy]

Orientation which is obtained by regionprops is very high, with this value the image is not looking straight. please help me

[anastasia]

how to find the major axis initially and then align it using imtranslate()? Please help with the initial part of the code.

Answer 3

[sandeep kumar]

Image Analyst I implemented your code testGrayimage.m but iam not able to get a proper segmented image iam attaching below the images, in fig.(1) the inner blob is suppose to be the lesion area but it is selecting the outerleft arc please help ..

[Image Analyst]

You didn't attach your original image but it looks like you do not have uniform exposure and that the edges of your image are darker than the middle, hence it's finding other dark blobs, not just the lesion. You need to do a background correction to compensate for your non-uniform illumination. Take a photo of a uniform gray sheet and divide your actual image by the white sheet image.

[sandeep kumar]

sir is ther any function to do that, iam new to image processing

[sandeep kumar]

sir images vary in sizes some are bigger some are smaller some has intense blob while in some images their is a very less difference between lesion and skin intensity. sir is there any way that out of all the regions in the binaryImage it could only select a region which is may be say circular or elliptical in shape like in above fig can we select only the inner blob and also in the fig attached

[Image Analyst]

Yes, you can do background division with the / symbol, but you'd need a snapshot of a totally uniform image, like I said. Divide the background image by the max value to get a "percentage image", then divide your sample image by the percentage image and rescale. I'm attaching a demo of a different background correction method.

Why don't you just call imclearborder to get rid of blobs touching the border? It just treats the symptom, not the cause, but I don't think you're skilled enough yet to do a proper background correction.

Answer 4

[James Morris]

Hi image analyst, from the matlab file you posted, you rotated and calculated the area difference along the x axis. I was just wondering if you could show and explain how to do it along the Y axis too?

[Image Analyst]

Use fliplr() instead of flipud().