2d gaussian function
243 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
lital
am 3 Aug. 2011
Kommentiert: Walter Roberson
am 17 Okt. 2020
I need to plot a 2d gaussian function, where x and y corresponds to the image pixels, my code uses a nested for loop which makes my program run extremely slow, is there a way to write this in a more faster way?
(right now it takes about 8-10 sec to run on 1920*1080 size matrix and i need to produce 173,340 2d gaussian functions images which is too much time...)
thanks in advance for any help.
this is my current code:
function mat = gauss2d(mat, sigma, center)
gsize = size(mat);
for r=1:gsize(1)
for c=1:gsize(2)
mat(r,c) = gaussC(r,c, sigma, center);
end
end
function val = gaussC(x, y, sigma, center)
xc = center(1);
yc = center(2);
exponent = ((x-xc).^2 + (y-yc).^2)./(2*sigma);
val = (exp(-exponent));
10 Kommentare
Walter Roberson
am 17 Okt. 2020
It is already Saturday in Atlantic Time and everywhere east of that. There is zero chance that anyone is going to write the code for you in the next few hours before you meet with your professor.
However, if you ask specific questions about MATLAB then there is a possibility that someone might answer in the next few hours.
Akzeptierte Antwort
Walter Roberson
am 3 Aug. 2011
function mat = gauss2d(mat, sigma, center)
gsize = size(mat);
[R,C] = ndgrid(1:gsize(1), 1:gsize(2));
mat = gaussC(R,C, sigma, center);
5 Kommentare
Andicha Zain
am 21 Sep. 2020
Bearbeitet: Walter Roberson
am 15 Okt. 2020
hello can anyone help me
how to write program The Influence of Coherence Length and Gaussian Beam on the Multibeam Interference by Matlab ?
please help me.
Weitere Antworten (5)
Image Analyst
am 26 Nov. 2011
Is that 8-10 seconds for all 173,340 Gaussians? That's not bad. I have a demo that randomly places Gaussians in a larger image using fspecial() and indexing, not one pixel at a time like you did. It takes about 7.4 seconds on an old computer for 173,340 randomly placed Gaussians in a 1920x1080 image. Here's the code, in case you're interested:
% Demo to randomly place Gaussians in an image.
% By ImageAnalyst
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
imtool close all; % Close all imtool figures.
clear; % Erase all existing variables.
workspace; % Make sure the workspace panel is showing.
tic;
% Set up some parameters.
fontSize = 20;
backgroundGrayLevel = 128;
windowSize = 50; % Could be random if you want.
sigma = 10; % Could be random if you want.
numberOfGaussians = 173340;
rows = 1080;
columns = 1920;
% Create one Gaussian.
g = fspecial('gaussian', windowSize, sigma);
grayImage = backgroundGrayLevel * ones(rows, columns);
% Create random signs so that the Gaussians are
% randomly brighter or darker than the background.
s = 2*randi(2, [1 numberOfGaussians])-3;
% Note: g and grayImage are floating point images, not uint8,
% though you could modify the program to have them be uint8 if you wanted.
% Get a list of random locations.
randomRow = randi(rows-windowSize+1, [1 numberOfGaussians]);
randomCol = randi(columns-windowSize+1, [1 numberOfGaussians]);
% Place the Gaussians on the image at those random locations.
for k = 1 : numberOfGaussians
grayImage(randomRow(k):randomRow(k)+windowSize-1, randomCol(k):randomCol(k)+windowSize-1) = ...
grayImage(randomRow(k):randomRow(k)+windowSize-1, randomCol(k):randomCol(k)+windowSize-1) + ...
s(k) * g;
end
toc;
% Display the final image.
imshow(grayImage, []);
caption = sprintf('%d Gaussians, Randomly Placed', numberOfGaussians);
title(caption, 'FontSize', fontSize);
axis on;
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1])
set(gcf,'name','Demo by ImageAnalyst','numbertitle','off')
2 Kommentare
Muhammed Sameed
am 24 Apr. 2020
This is a really good piece of code. But how would you do the reverse i.e. from a noisy image with N randomly placed Gaussians, extract the centroid location and sigma of each Gaussian? I am interested in N=2 for the moment, but a generalized function would be helpful.
Walter Roberson
am 24 Apr. 2020
Hmmm, would that be like finding the values of x(n), y(n), width(n), height(n) such that the sum of the gaussians generated by those parameters is everywhere less than (image + 1/2*EPS(image)), where EPS(image) is 1 for integer-valued images and eps(image) for floating point? Which is to say, that the sum of the gaussians can exceed the value of the image at any given location, but must not exceed it by enough that the total would become the next representable number ? For example on an integer image, if an image location was 42, and the sum of gaussians predicted 42.42 there, then that would be okay because uint8(42.42) would be 42, but predicting 42.52 would not be okay because uint8(42.52) would be 43.
Mahdi
am 22 Jul. 2014
I faced the same problem, just so others know you can use fspecial('gaussian', hsize, sigma) intrinsic function.
1 Kommentar
Image Analyst
am 22 Jul. 2014
Thanks for calling it out specially - it was kind of buried in my demo code and hard to notice.
Karbala'a Unvi. Science
am 28 Dez. 2014
Dear Sir, I am interested about the code that you wrote about the 2D Gaussian. I have a problem that I want to an image data to be distributed in another image ( image A is the Original, image B is the data one) so that when you see image A you find that there is a noise in it ( where that noise is image B)... I hope that is a good information to help me in building the code... Help will be appreciated.. Thank you in advance
1 Kommentar
Image Analyst
am 28 Dez. 2014
This is not an answer. Who are you talking to? You should have put it as a comment under their answer. Why don't you just try your best and then post your code as a new question? And explain it better there. I don't even understand what you want. I don't know if B is a "data" image or a "noise" image and if you can just add B to A.
rusgu fcf
am 6 Mai 2016
what does the variable 'center' signify in the code of gaussian2d by lital ?
3 Kommentare
Rym Benchaabane
am 29 Mär. 2020
is center a real number ie 0? or a two element array ie [0,0]?
Thank you
Image Analyst
am 29 Mär. 2020
It's a 2-element vector, as you can see by looking inside his gaussC() function.
I recommend you use my code or Walter's code since lital's had problems.
Siehe auch
Kategorien
Mehr zu Logical finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!