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How to plot the radial profile of a 2D image
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My goal is to plot the image values (interpolated but not averaged) along an arbitrary diameter (arbitrary azimuth angle). I have attached my attempt to do that but it only works for 1/2 the diameter and only for square (nxn) images. Furthermore, it fails on some test images, even if represented by a squared array, printing the following error that I cannot interpret:
>> sampled_radial_slice = interp2(X,Y,img,x,y); Error using griddedInterpolant Sample values must be a single or double array. Error in interp2>makegriddedinterp (line 228) F = griddedInterpolant(varargin{:}); Error in interp2 (line 136) F = makegriddedinterp(X, Y, V, method,extrap);
I would like to extend this script to rectangular 2D images and to plot the profile along a diameter not just a radius for arbitrary azimuthal angle. The diameter should pass through the image center. Any suggestion and helpis welcome. Thank you in advance. maura
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Risposte (4)
Image Analyst
il 31 Mar 2016
Modificato: Image Analyst
il 25 Gen 2019
See attached demo to get the average radial profile.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/173899/image.jpeg)
8 Commenti
Image Analyst
il 29 Gen 2019
Yes, you can oversample like I, and you, said. In other words, just have a larger matrix and you'll have a lot more samples and higher precision.
Hugo Trentesaux
il 7 Feb 2019
The proposed function seems very complicated to me. Here is my version :
function profile = radialAverage(IMG, cx, cy, w)
% computes the radial average of the image IMG around the cx,cy point
% w is the vector of radii starting from zero
[a,b] = size(IMG);
[X, Y] = meshgrid( (1:a)-cx, (1:b)-cy);
R = sqrt(X.^2 + Y.^2);
profile = [];
for i = w % radius of the circle
mask = (i-1<R & R<i+1); % smooth 1 px around the radius
values = (1-abs(R(mask)-i)) .* double(IMG(mask)); % smooth based on distance to ring
% values = IMG(mask); % without smooth
profile(end+1) = mean( values(:) );
end
end
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Maura Monville
il 5 Apr 2016
2 Commenti
Image Analyst
il 5 Apr 2016
1. When linear indexing, it goes down rows first, then over column-by-column.
2. The left and top pixels are 1. The middle would be (columns/2, rows/2). To get from the middle of the picture to pixel coordinates, you need to add half the width or height in pixels, taking special care for whether the dimension has an odd or even number of pixels in that direction.
Sergey Loginov
il 5 Nov 2021
Accumarray function allows to build both radial profiler (like this one: https://www.mathworks.com/matlabcentral/fileexchange/101480-very-fast-radial-profile) or an azimutal profiler as well.
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