Find the barycenter of luminance in an image

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I need to find the weighted center of luminosity of barycenter in an image. (the weighted L center in Lab a b)
  2 Comments
David Corwin
David Corwin on 29 Oct 2015
Edited: David Corwin on 29 Oct 2015
The image is an rgb image that is converted to an L*A*B* image. (I could just start with a gray scale image) For each pixel there is an L (value 1-100)multiplied by the distance to geometric center. The sum of all this should be the luminosity center or tonal center of the image or using tonal value as weight: the center of mass of the picture.

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Accepted Answer

Image Analyst
Image Analyst on 29 Oct 2015
Edited: Image Analyst on 29 Oct 2015
Since R2008a (or maybe b) regionprops has had a "WeightedCentroid" property that you can ask for. So pass in the L channel and ask for that if you want the centroid weighted by the L value.
If you want it weighted by the distance, that's the inertia or moment. See https://en.wikipedia.org/wiki/Image_moment I don't think there's anything built in for that but you can ask for the PixelList which is the location of every pixel in the blob and you can easily calculate it yourself.
  6 Comments
David Corwin
David Corwin on 30 Oct 2015
I will look at your demo. I am sorry to be confusing but it is partially due to terminology. I want to obtain the tonal center of gravity for a variety of images. I have a theory concerning how to achieve perfect pictorial composition and its effects on eye movements. i.e. why pictures can have a powerful effect as pictures. This is only the first half necessary to study eye movements.
Walter Roberson
Walter Roberson on 30 Oct 2015
No, L is not the label matrix here. L is the luminance channel, in accordance with IA's earlier "So pass in the L channel and ask for that if you want the centroid weighted by the L value."
A = imread('figure2.jpg');
LAB = rgb2lab(A);
L = LAB(:,:,1);
imageprops = regionprops(L, 'WeightedCentroid');
result = imageprops.WeightedCentroid;
If you want the "top half" then pass in the top half,
imageprops = regionprops(L(1:end/2,:), 'WeightedCentroid');
upperCent = imageprops.WeightedCentroid;
imageprops = regionprops(L(ceil(end/2):end,:), 'WeightedCentroid');
lowerCent = imageprops.WeightedCentroid;

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More Answers (1)

Image Analyst
Image Analyst on 30 Oct 2015
David:
I wrote a demo for you to compute the first 5x5 central spatial moments, as given by the Wikipedia page https://en.wikipedia.org/wiki/Image_moment. See attached m-file. The key code is this:
% Sum up the powers of the gray levels, weighted by the distance from the centroids.
% Ref: https://en.wikipedia.org/wiki/Image_moment
highestMomentNumber = 5;
mu = zeros(highestMomentNumber, highestMomentNumber); % Allocate up to 5 moment (probably way overkill).
for q = 1 : size(mu, 2) % in the column, x direction.
for p = 1 : size(mu, 1) % in the row, y direction
for location = 1 : length(allX)
% For every pixel in this blob....
% Sum up for moment "mu sub pq" - i.e. the pqth moment.
thisGrayLevel = grayImage(allY(location), allX(location));
mu(p, q) = mu(p, q) + ...
(allX(location) - xCenter).^q * ...
(allY(location) - yCenter).^p * ...
double(thisGrayLevel);
end
end
end
See file for a complete demo.
  6 Comments
David Corwin
David Corwin on 30 Oct 2015
Thank you very much for the solution. However, I can see the reason for the confusion.
take a rectangular image with each pixel having a tonal value. The picture is being suspended on the wall at its geometric center. In this situation light values exert a conceptual upward force while dark values exert a downward force
Now the upper left and right quadrants have a weight relative to the center which can cause the picture to spin.
Likewise, with the lower left and right quadrants
The center of gravity does not tell me the relative balance of the upper right and left quadrants. The center of gravity of each quadrant would place equal value on what would be interior pixels that should have relatively no value
It seems to me that I need an upper half center of balance relative to the vertical median such that values closer to the upper right and left edges have a greater weight relative to the geometric center
In a more sophisticated setup, this type of situation is found in airplanes
I hope I am explaining the problem clearly. Ultimately this will be part of a paper I am hoping to publish and I would be happy to give you credit.
David Corwin
David Corwin on 30 Oct 2015
Thanks for all your help. I figured out the solution to the above by simply placing 1 or 255 in what ever part of the image I do not want to be included.

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