Fit x and y data

Question

Thor 2021-2-22

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/753029-fit-x-and-y-data

编辑： Thor 2021-5-14

Test1.xls

Hello everyone,

I got this data and I want to create a script that plots a gaussian curve that fits the major peak only. I have seen many examples.However, they involve superimposing normal distribution funcitons or are for multiple peaks , so the fitting i get isnt correct.

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Answer 1

Star Strider 2021-2-22

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https://ww2.mathworks.cn/matlabcentral/answers/753029-fit-x-and-y-data#answer_630474

在 MATLAB Online 中打开

Try this:

D1 = readmatrix('Test1.xls');
x = D1(:,1);
y = D1(:,2);
gausfcn = @(b,x) b(1).*exp(-(x-b(2)).^2/b(3));
[maxy,idx] = max(y);
Lv = y >= 1.5;                                                                  % Restrict Region Of Fit To Region Of Symmetry In ‘y’
B = fminsearch(@(b)norm(y(Lv) - gausfcn(b,x(Lv))), [maxy, x(idx), 1/12.5]);
figure
plot(x, y)
hold on
plot(x, gausfcn(B,x), '-r')
hold off
grid
text(x(idx), 0.4, sprintf('$y = %5.2f \\cdot e^{\\frac{-(x-%7.2f)^2}{%7.2f}}$', B), 'Interpreter','latex', 'HorizontalAlignment','center', 'FontSize',12)

producing:

.

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Star Strider 2021-3-2

在 MATLAB Online 中打开

There are two ways to combine them, adding them or multiplying them.

I let the ga (genetic algorithm) function have a go at this, since the parameters are difficult to estimate otherwise.

For the additive function:

gausfcn2a = @(b,x) b(1).*exp(-(x-b(2)).^2/b(3)) + b(4).*exp(-(x-b(5)).^2/b(6));

two parameter sets were:

		B(1) =  1.82832
		B(2) = 532.81617
		B(3) = 1865.81457
		B(4) =  1.57228
		B(5) = 592.02143
		B(6) = 1739.77333

and:

		B(1) =  1.44994
		B(2) = 592.98443
		B(3) = 2069.30812
		B(4) =  1.74877
		B(5) = 533.39050
		B(6) = 2223.33610

For the multiplicative function:

gausfcn2m = @(b,x) b(1).*exp(-(x-b(2)).^2/b(3)) .* b(4).*exp(-(x-b(5)).^2/b(6));

two parameter sets were:

		B(1) = 42.76331
		B(2) = 468.93401
		B(3) = 2709.89032
		B(4) = 25.28253
		B(5) = 649.40457
		B(6) = 2746.58184

and:

		B(1) = 42.41765
		B(2) = 468.55038
		B(3) = 2697.30963
		B(4) = 27.28298
		B(5) = 649.53395
		B(6) = 2731.39634

The additive version (‘gausfcn2a’) appears to provide the best results. You can use the parameter sets as initial estimates if you want to use other functions to improve on the fit. If you have the Global Optimization Toolbox, I will post the code I used for this so you can use the ga function to experiment with it. It converged relatively quickly, in about two minutes on my Ryzen 9 3900 machine.

Thor 2021-3-16

Hi @Star Strider, I was referring to this

'' If you have the Global Optimization Toolbox, I will post the code I used for this so you can use the ga function to experiment with it ''

Star Strider 2021-3-16

在 MATLAB Online 中打开

Sure!

I ran it again to be certain that it works.

D1 = readmatrix('Test1.xls');
x = D1(:,1);
y = D1(:,2);
[maxy,idx] = max(y);
Lv = y >= 1.08;                                                                  % Restrict Region Of Fit To Region Of Symmetry In ‘y’
gausfcn2a = @(b,x) b(1).*exp(-(x-b(2)).^2/b(3)) + b(4).*exp(-(x-b(5)).^2/b(6));
% gausfcn2m = @(b,x) b(1).*exp(-(x-b(2)).^2/b(3)) .* b(4).*exp(-(x-b(5)).^2/b(6));
ftns = @(b)norm(y(Lv) - gausfcn2a(b,x(Lv)))
PopSz = 50;
Parms = 6;
opts = optimoptions('ga', 'PopulationSize',PopSz, 'InitialPopulationMatrix',randi(1E+4,PopSz,Parms)*1E-3+[maxy, x(idx), 1/12.5, maxy, x(idx), rand], 'MaxGenerations',2E5, 'PlotFcn',@gaplotbestf, 'PlotInterval',1);
t0 = clock;
fprintf('\nStart Time: %4d-%02d-%02d %02d:%02d:%07.4f\n', t0)
[B,fval,exitflag,output] = ga(ftns, Parms, [],[],[],[],zeros(Parms,1),Inf(Parms,1),[],[],opts)
t1 = clock;
fprintf('\nStop Time: %4d-%02d-%02d %02d:%02d:%07.4f\n', t1)
GA_Time = etime(t1,t0)
DT_GA_Time = datetime([0 0 0 0 0 GA_Time], 'Format','HH:mm:ss.SSS');
fprintf('\nElapsed Time: %23.15E\t\t%s\n\n', GA_Time, DT_GA_Time)
fprintf(1,'\tRate Constants:\n')
for k1 = 1:length(B)
    fprintf(1, '\t\tB(%d) = %8.5f\n', k1, B(k1))
end
figure
plot(x, y)
hold on
plot(x, gausfcn2a(B,x), '-r')
hold off
grid
% text(x(idx), 0.4, sprintf('$y = %5.2f \\cdot e^{\\frac{-(x-%7.2f)^2}{%7.2f}}$', B), 'Interpreter','latex', 'HorizontalAlignment','center', 'FontSize',12)
% text(x(idx), 0.2, sprintf('$y = %5.2f \\cdot e^{(\\frac{-(x-%7.2f)}{%7.2f})^2}$', B(1:2),sqrt(B(3))), 'Interpreter','latex', 'HorizontalAlignment','center', 'FontSize',12)

Experiment with it to get the result you want. It may be necessary to tweak the tolerances to produce a slightly better result than this code produces. You can use any of the objective function variations in the ‘ftns’ function, or create your own, to get a suitable result.

Adding a bit of documentation —

PopSz = population size (rows of the 'InitialPopulationMatrix')

Parms = Number of Parameters in the objective function (columns of the 'InitialPopulationMatrix')

This vector: ‘[maxy, x(idx), 1/12.5, maxy, x(idx), rand]’ scales the columns of the 'InitialPopulationMatrix' to different magnitudes. That makes it easier for the ga algorithm to converge. Change that (or eliminate it entirely) as necessary.

If you have other questions, post them and I will do my best to provide appropriate replies.

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Fit x and y data

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