Fit gaussian surface on 3D data
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Using the following script and attached image tif file, I am trying to fit a gaussian surface on the image data. Is there something that I skip? The result soes not seem to be gaussian. I also tried using the following but couldnt obtain a meaningful result.
https://www.mathworks.com/matlabcentral/fileexchange/41938-fit-2d-gaussian-with-optimization-toolbox
I appreciate any help/comment in advence.
pixelsize = 0.65512E-6; %pixel conversion [m]
folder = pwd;
baseFileName = 'imageprocessing.tif';
fullFileName = fullfile(folder, baseFileName);
% info = imfinfo(baseFileName);
info = imfinfo(fullFileName);
grayImage = imread(fullFileName);
Z = grayImage./max(grayImage);
xx = [1:size(grayImage,1)]*pixelsize;
yy = [1:size(grayImage,2)]*pixelsize;
[X,Y]=meshgrid(xx,yy);
surf(X,Y,Z');shading interp
xlabel('x [\mu m]')
ylabel('y [\mu m]')
zlabel('Signal')
hold on
xpdf = X(:); ypdf = Y(:); zpdf = Z(:);
gaussianfit = fit([xpdf, ypdf], zpdf, 'a*exp(-( ((x-b)/c)^2 + ((y-d)/e)^2 ) ) + f', 'StartPoint', [400, 3, 50, 0, 40, 450])
plot(gaussianfit, [xpdf, ypdf], zpdf)
3 Kommentare
Matt J
am 27 Mai 2023
Bearbeitet: Matt J
am 27 Mai 2023
a*exp(-( ((x-b)/c)^2 + ((y-d)/e)^2 ) )
It seems inadvisable to model the Gaussian term as separable in x and y, with only 5 parameters, instead of the more general 6-parameter model which has cross-terms,
Even if the object truly follows a separable Gaussian, a slight rotation of the imaging device would render the separable model invalid.
Akzeptierte Antwort
Matt J
am 27 Mai 2023
Bearbeitet: Matt J
am 27 Mai 2023
Using gaussfitn from,
load Image %cropped to include only the dark lobe
Z=double(grayImage);
Z=Z(1:2:end,1:2:end);
[X,Y]=ndgrid(1:size(Z,1),1:size(Z,2));
b=diag([inf,inf]);
p=gaussfitn([X(:),Y(:)],Z(:),[],...
{[],[],[],-b}, {[],[],[],+b}); %force covariance matrix to be diagonal
[D,A,mu,sig]=deal(p{:});
delta=([X(:),Y(:)]-mu')';
Zfit= D + A*exp( -0.5 * sum( (sig\delta).*delta,1) ); %gaussian fit
Zfit=reshape(Zfit,size(X));
%Display
f=@(q) reshape(q(1:10:end,1:10:end),[],1);
scatter3(f(X),f(Y),f(Z)); hold on
surf(X,Y,Zfit,'FaceColor','r','FaceAlpha',0.5,'EdgeColor','none'); hold off
xlabel X, ylabel Y; view(45,35)
legend Data Fit
4 Kommentare
Matt J
am 30 Mai 2023
Bearbeitet: Matt J
am 30 Mai 2023
The isocontours of a 2D Gaussian surface are ellipses. What do you consider to be the "width" of an ellipse? Is it the semi-major axis? The semi-minor axis? Something in between?
Weitere Antworten (1)
Image Analyst
am 27 Mai 2023
That's not 3-D data. That is 2-D data. How about you find the centroid with code I gave you before, then get the average radial profile, then fit a Guassian to that?
2 Kommentare
Image Analyst
am 27 Mai 2023
I don't think I recommended accumarray. What is the profile for any one point in time? Does it look like a Gaussian? If the profile is gaussian, then I think the histogram would also be Gaussian and you could also fit the histogram to a Gaussian. The surface shown in 'datasurface.png' looks slightly tilted so you might want to flatten/level it before, though I think my demo had a tilt allowed. If you had two different Gaussians, one for x and one for y, then you'd have to account for that with additional terms in the model.
I don't know what this dark thing represents but I think youi may be overthinking it by trying for more accuracy that the situation really needs. Let's say you were somehow able to get the two Gaussian parameter sets, one along each principal direction, then what would you do differently than if you just had the standard deviation of the thing assuming a single radially symmetric Gaussian like I suggested? Probably nothing different, so why make it more complicated than it needs to be?
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