Finding the width of a blob perpendicular to the major axis

Hi!

I've got a binarised blob here with a major axis plotted inside the blob. I am looking to find a series of width, perpendicular to the major axis plotted in the blob. Basically, I'm looking to find the distance of a line that has a starting coordinate from the left side of boundary, intersects with the coordinate of the major axis and ending with a coordinate from the right side of the boundary. The image below shows what I'm looking for. Apologies if my question isn't clear enough.

Basically I'm looking to measure the distance of the black arrows. Attached is the code I've used to draw the magenta coloured line (major axis) of the blob.

Really appreciate if someone can help me to solve this problem.

Cheers,

How

Answer 1

[DGM]

Unless you have some existing means to find the major axis, I'd still use feret properties to rotate the object. The rest of this really isn't much different than IA's answer, though.

% cropped screenshot with annotations removed
mask = imread('image.png');
mask = rgb2gray(mask)>128;

% pick the largest blob if desired
% otherwise, the script processes all blobs
mask = bwareafilt(mask,1);

% get properties
S = regionprops(mask,'minferetproperties','image');
th = vertcat(S.MinFeretAngle)-180;

% preallocate outputs
blobimgs = cell(numel(S),1);
blobwidth = cell(numel(S),1);
for k = 1:numel(S)
    % rotate blob image
    thisblob = imrotate(S(k).Image,th(k));
    
    % close-crop blob image
    yrange = any(thisblob,2);
    thisblob = thisblob(yrange,:);
    xrange = any(thisblob,1);
    thisblob = thisblob(:,xrange);
    
    % throw results into output arrays
    blobimgs{k} = thisblob;
    blobwidth{k} = sum(thisblob,2);
end

% view the blob and width profile
f = 1; % select a blob
subplot(2,1,1)
imshow(blobimgs{f})
subplot(2,1,2)
plot(blobwidth{f})

[Muhammad Syahmeer]

Thanks so much for the code, DGM. I wonder if it's possible to just get 90 minus the orientation angle to get the image vertical? It's just I'm using the R2018b version which is missing the 'minferetproperties' feature.

[Image Analyst]

@Muhammad Syahmeer, sorry it didn't work for you. To get that function, the best option is to upgrade. If you can't do that you can use the "orientation" option from regionprops like I showed you in my answer below.

[Muhammad Syahmeer]

Thanks for that @Image Analyst. Yup on my way to get the latest version installed.

I used the “orientation” option from regionprops like you suggested and subtracted it with 90 degrees to get the image vertical. Though wasn’t sure why 90 is the magic number but it seems to do the trick..

[Image Analyst]

@Muhammad Syahmeer With regionprops the Orientation value is in degrees, ranging from -90 degrees to 90 degrees. If you want a different range, you'd have to add the offset. For example to go from -90 to +90 to 0 to 180, you'd have to add 90 to the Orientation value.

Answer 2

[Image Analyst]

Use bwferet

[Muhammad Syahmeer]

Thanks for your swift reply Image Analyst. However I’m not looking for just the Min and Max Feret Diameters, but rather the distribution of the diameters perpendicular to the major axis if that make sense.

I have a dumb-bell shaped image here and would like to find the point when the width (perpendicular to the major axis) starts to increase ‘significantly’. Most of the width in the middle section of the shape should remain constant.

[Image Analyst]

Then use regionprops() to get the 'Image' and the 'Orientation'. Then rotate the blob so that it's vertical and then scan down line by line using find() or nnz() to count the white pixels on each line. I think you can do it. Something like

props = regionprops(mask, 'Image', 'Orientation');
for k = 1 : numel(props)
    thisImage = props(k).Image;
    thisImage = imrotate(thisImage, props(k).Orientation);
    [rows, columns] = size(thisImage);
    widths = nan(rows, 1), 1);
    for row = 1 : rows
        widths(row) = nnz(thisImage(row, :));
    end
end

If that's not right, I'm sure you can debug it.

[Muhammad Syahmeer]

Thanks so much for the code Image Analyst. That's really helpful!