Circularity of an image

I am trying to find circularity for my shape feature requirement in MRI image classification. I have used the matlab function REGIONPROPS to find area and perimeter and the circularity formula: 4*pi*area/(perimeter)^2.

My problem is i am getting perimeter as 0 value for many regions in an image. Then circularity will become infinite. I have attached a screenshot of perimeter values i am getting for a single image. Someone please suggest me how to calculate circularity in such case???

Answer 1

[Mohammad Abouali]

does it have to be circularity?

Can you replace circularity with the ratio between major/minor axis length?

[purti choudhary]

The research paper i am following considered area, perimeter, circularity as shape features.

Is major/minor axis length is also a shape feature??

[Mohammad Abouali]

Yes they are and they are calculated by regionprops. the ratio of that can tell you if the object is elongated or not. However this is not exactly the same as circularity. Because circularity is affected by how jaggedy the permitters are.

So if you have a very circular looking island but with a lot of Jaggedy permitter, actually the circularity goes down; since the length of permitter is increasing. Although it looks circular enough. In that case minor/major axis ratio is better.

Having said this, it means that they are not exactly the same. So that's why I asked if you can change circularity to minor/major ratio.

[purti choudhary]

Thank you...

Why this function is giving so many zeros for perimeter?? also the size of area and perimeter array is different for different images. This will create problem in creating train and test feature matrix for various images as the size will be different.(according to the area and perimeter array for the image)

[Mohammad Abouali]

The sizes are in pixels. So if your images don;t have the same spatial pixel size then 1pixel in first image could refer to 1meter, in the second image one pixel refers to 10meter (well the numbers are just examples).

So the sizes never come out exactly the same.

But why it is giving too many zeros. I don't know. I have to check the image and then perhaps I may find out.

[purti choudhary]

This is one of the image for which i am calculating perimeter.

And permiter(97x1) matrix is likewise

Answer 2

[Image Analyst]

What is the area of the blobs that have zero perimeter? Is it 1? Like 1 pixel? If so, maybe you just don't want to consider those. You can use bwareaopen() to filter out tiny things. By the way, I use the inverse of your equation.

allAreas = [measurements.Area];
allPerimeters = [measurements.Perimeter];
circularities = allPerimeters .^ 2 ./ (4 * pi * allAreas);

Of course zero perimeter is not a problem there.

[Image Analyst]

By the way, you shouldn't call regionprops() twice. Just get everything all in one shot:

measurements = regionprops(logical(BW), 'Area', 'Perimeter');

[purti choudhary]

yes u r right i am getting the same thing.

wherever perimeter is 0 its corresponding area values is 1.

[Image Analyst]

OK, so like I thought, the solution is to call bwareaopen(binaryImage, 1) to get rid of pixels that are of size 1 or less. Personally I'd use a min blob size of a few hundred.

[Image Analyst]

Attach your script if you want me to try it. Also, im2bw() is usually not good at picking thresholds in my experience. Anyway, you probably need to call imfill(binaryImage, 'holes') after you call im2bw().

[purti choudhary]

thank you so much..

I have used bwareaopen(binaryimage,2), now it is not showing all the regions whose area is 1 and perimeter is 0 earlier. This is right??

[Image Analyst]

Where's your script so I can try it?

[purti choudhary]

BW = im2bw(Im, map, 0.5);

BW2=bwareaopen(BW,2);

% Calculating area feature of the image

ar = regionprops(BW2, 'area');

area = cat(1, ar.Area);

area = area(1:18);

area=reshape(area,1,18);

% Calculating perimeter feature of the image

pm = regionprops(BW2, 'perimeter');

perimeter = cat(1, pm.Perimeter);

perimeter = perimeter(1:18);

perimeter = reshape(perimeter,1,18);

% Calculating circularity feature of the image

for i = 1:18

    circ(i) = ( 4 * 3.14 * area(i) / ( perimeter(i) ^2));

end

circ = reshape(circ,1,18);

Feature_Vec = [area,perimeter,circ];

[purti choudhary]

i have considered only 18 elements because different images are giving different sized area arrays and feature vector dimension should be same for all images.

[Image Analyst]

That's a poor decision.

And why didn't you take my advice on how to calculate things:

measurements = regionprops(labeledImage, 'Area', 'Perimeter');
allAreas = [measurements.Area];
allPerimeters = [measurements.Perimeter];
circularities = allPerimeters .^ 2 ./ (4 * pi * allAreas);

[purti choudhary]

Thank you very much... This labelled image is a grayscale image??

[Image Analyst]

It's an integer valued image. You can display it as a grayscale image if you want

imshow(labeledImage, []);

Or you can pseudocolor the blobs and show that:

% Assign labeled blobs rainbow colors.
coloredLabels = label2rgb (labeledImage, 'hsv', 'k', 'shuffle');
imshow(coloredLabels);