Anybody help me with measure asymmetry an irregular object

What's the best method to sort the asymmetry?

basically, I need the result of the ratio between the perpendicular segment and the largest diagonal, where the score is quantified in (0 to 2 points): if the proportion is greater than 40.01%: 0 point, symmetric lesion. If the proportion is between 20.01% to 40%: 1 point, little asymmetric lesion. If the proportion is less than 20%: 2 points, very unbalanced lesion.

I need to define 'percent' and apply in the results

 %%Reads the original image and performs the conversion to binary.
A = imread('IRREGULAROBJECT.jpg'); % this image already is grayscale.
imshow(A);
BW = im2bw(A);
imshow(BW);

 %%Extract the Asymmetry Value of the object.

 % AREA VALUE
s = regionprops(BW,'area');
area = cat(1, s.Area)
imshow(area)
title('Value AREA image', 'FontSize', fontSize);

 %%CONDITIONS
 %If the proportion is greater than 40.01%: 0 point, symmetrical lesion.
if (percent > 40.01)
    asymmetry = 0;
end

 % If the proportion is between 20.01% to 40%: 1 point, little asymmetric lesion.
if (percent > 20.01 & percent < 40.00)
    asymmetry = 1;
end

 % If the proportion is less than 20%: 2 points, very unbalanced lesion.
if (percent < 20.00)
    asymmetry = 2;
end

[Image Analyst]

Unfortunately you delayed any answer by forgetting to attach 'IRREGULAROBJECT.jpg'. We'll check back later.

[Marlon Damaceno]

in fact it could be any irregular shape, but here is an example.

[Aishwarya Udhayakumar]

How can I define the percent value??can u help me with this

[Image Analyst]

I have no idea. Is it the solidity like you get from reginoprops(), which is the area divided by the area of the convex hull? Since it now appears that you did not write that code, why don't you ask whomever wrote the code?

Answer 1

[Image Analyst]

You might want to try the Dice-Sorensen coefficient. See attached demo.

Alternatively you can find the bounding rectangle by using the radon transform. No demo for that (yet). Simply find the widest width and the narrowest width. Or you could use John's function: https://www.mathworks.com/matlabcentral/fileexchange/34767-a-suite-of-minimal-bounding-objects

[Marlon Damaceno]

by regionprops () method would I be able to measure asymmetry? maybe majoraxislenght, minoraxislenght or orientation?

"I need the result of the ratio between the perpendicular segment and the largest diagonal"

[Image Analyst]

No.

[Marlon Damaceno]

OK but I Didn't understand how DiceSorensenSimilarityCoefficient.m gives me the asymmetry of my irregular shape.

[Image Analyst]

You basically flip the blob about its major axis of symmetry and count pixels that are common to both the flipped and unflipped blob, and count pixels that are different (exist only in one). The coefficient is a ratio of those and is a common metric for symmetry. Look it up in Wikipedia.

[Marlon Damaceno]

I have another idea maybe more simple for my problem. Image Analyst, if I divide the irregular shape in 2 areas and measure this 2 areas. For example,

if (area1 = area2)
disp(The object is SIMETRIC)
elseif (area1 > area2)
disp(The object is ASYMMETRIC)
elseif (area1 < area2)
disp(The object is ASYMMETRIC)
end

[Image Analyst]

You don't have to think too hard to imagine an object where that is not a good algorithm. What if area 1 is a long thin rectangle that just happens to have the same area as a disc area 2? You'd say they're symmetric but they'd be far, far from symmetric. Any reason why you're not using generally accepted metric like Dice Sorensen?

[JovanS]

Hello Image Analyst ,

As you said before in order to find the percentage of asymmetry we have to flip the blob about its major axis of symmetry and count pixels that are common to both the flipped and unflipped blob, and count pixels that are different. could you share matlab code with us ?

[Image Analyst]

I don't have code for that but you'd call regionprops to get orientation, then call imrotate to put the major axis vertical, then call dice.