# Remove specific objects from an image

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Torkan on 18 Oct 2019
Edited: DGM on 23 Feb 2022
I want to remove some objects from a binary image based upon length to width ratio of the rectangular labeled each object, such that when a particular threshold is set the components satisfied then it removes t hose objects and show a new image.
Anybody has any idea?
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KALYAN ACHARJYA on 18 Oct 2019
Can you share the sample image?

Adam Danz on 18 Oct 2019
Edited: Adam Danz on 18 Oct 2019
Following similar logic to this answer, we'll use regionprops to get the length and width of the bounding box rectangles. Then we can compute the length:width ratio, set a threshold, and determine which boxes are greater than that threshold.
Then you can plot accepted objects with a red rectangle and rejected objects with a yellow rectangle.
See below for a second method.
% Read image as RGB
I = imread('bwimage.png'); %including the pull path is better
% Convert to binary
BW = imbinarize(rgb2gray(I));
% Get rid of white border
BWConv = bwconvhull(~BW);
BW = BW & BWConv;
% get region centers and major axis, and orientation
stats = regionprops('table',BW,'BoundingBox');
% Compute length/width ratio
stats.LenWdRatio = stats.BoundingBox(:,3) ./ stats.BoundingBox(:,4);
% Set threshold and determine which objects
% are greater than the threshold.
thresh = 2.0;
stats.isGreater = stats.LenWdRatio > thresh;
% Show bounding box for objects that met the threshold
figure()
h = imshow(BW);
axis on
hold on
% Plot red rectangles around accepted objects
arrayfun(@(i)rectangle('Position',stats.BoundingBox(i,:),'EdgeColor','r'), find(stats.isGreater));
% Plot yellow rectangles around rejected objects
arrayfun(@(i)rectangle('Position',stats.BoundingBox(i,:),'EdgeColor','y'), find(~stats.isGreater)); Second method: using major:minor axis length
Instead of using the horizontal and vertical extent of the bounding box, you could use the ratio of the MajorAxisLength to the MinorAxisLength. Here's an example why this method might be better. Suppose you have an long object that stretchs along a diagonal with slope=1. The width and height of the bounding box will be equal even though it's a very long and narrow object. If you used the ratio of the Major:Minor axis lengths, the object will be detected as long and narrow even through the bounding box is square.
To use this method, make the following changes.
% get region centers and major axis, and orientation
stats = regionprops('table',BW,'BoundingBox','MajorAxisLength','MinorAxisLength');
% Compute length/width ratio
stats.LenWdRatio = stats.MajorAxisLength ./ stats.MinorAxisLength;
[Update]
I forgot to show how to remove the objects.
When you call regionprop, add the SubarrayIdx property to the list.
stats = regionprops('table',BW,'BoundingBox','MajorAxisLength','MinorAxisLength','SubarrayIdx');
Then you can loop through each rejected object and replace its values in BW with false (black).
% Remove rejected objects (fill with black)
objRemoveIdx = find(~stats.isGreater);
for i = find(~stats.isGreater).'
BW(stats.SubarrayIdx{i,1},stats.SubarrayIdx{i,2}) = false;
end
% Show same figure with objects removed (yellow rectangles show where they were)
figure()
h = imshow(BW);
axis on
hold on
% Plot red rectangles around accepted objects
arrayfun(@(i)rectangle('Position',stats.BoundingBox(i,:),'EdgeColor','r'), find(stats.isGreater));
% Plot yellow rectangles around rejected objects
arrayfun(@(i)rectangle('Position',stats.BoundingBox(i,:),'EdgeColor','y'), find(~stats.isGreater));
DGM on 23 Feb 2022
Edited: DGM on 23 Feb 2022
Knowing what the rest of the error message says might also be helpful here.
I can run Adam's code out to 25 million objects before running into array size errors, and even then, they're not "maximum possible array size" errors; they're "maximum array size preferences" errors (I don't have much memory).
I question the likelihood that someone is ntentionally trying to segment images with tens of millions of objects in them. Any nontrivial images would need to be ridiculously large to even hold that many objects without them being connected.

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