How to speed up function approximation?

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Adrian T
Adrian T on 19 Aug 2019
Edited: Matt J on 19 Aug 2019
I have a code in MATLAB2016:
[f,r2]=fit(tr,is_t1(:,i),'1-b*exp(-((a*x)/(var_c)))','StartPoint',[0.8 0.8]);
T1(i)=-(var_c*log(-(0.6321 1)/f.b))/f.a;
Its execution time: ~0,031s
Number of approximation points in a given task: ~7000-10000
How to ,speed up this operation?
Thanks for any help

Accepted Answer

Matt J
Matt J on 19 Aug 2019
Edited: Matt J on 19 Aug 2019
There are a few inefficiencies that I see. Firstly, you shouldn't pick an arbitrary StartPoint like [0.8,0.8]. The problem can be log-transformed into a linear equation and solved algebraically for a much more informed initial guess,
z=[x(:), ones(size(xdata(:)))]\log(ydata);
In fact, depending on your needs and the noisiness of your data, the above analytical solution might be accurate enough already. Maybe you don't need to use an iterative routine like fit() after all.
But if the data noise is significant and you wish to further refine a0 and b0 with iterative nonlinear fitting, I would recommend fminspleas
which will let you iterate over a only,
modelfun=@(a) exp((-a/var_c)*x);
[a,b] = fminspleas({modelfun}, a0, xdata,ydata);
Notice also that we've optimized the implementation of your modelfun() a bit. Here, it only executes 2 vectorized operations whereas your original implementation had 5.

More Answers (2)

Chris on 19 Aug 2019
I dont have those toolboxes but often you can start an optimization problem with a randomly selected sub-set of data run for a short time to get a better initial guess before running the algorithm over all your data. YMMV. Take care with this and make sure you understand your data and how you are modeling it.
  1 Comment
Adrian T
Adrian T on 19 Aug 2019
Thank you for your response.
1. Numeric data are well matched - R2> 0.9. I know the nature of the phenomenon.
2. random data - no result.

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Yair Altman
Yair Altman on 19 Aug 2019
It may be useful to replace the string with a function handle.
Also, setting the convergence tolerances to smaller values than the defaults might converge faster without a meaningful degradation of the result.
Lastly, read where I discuss different ways of significantly speeding-up the fitdist function. It's not the same as your fit but similar ideas can possibly also be useful in your case.
  1 Comment
Adrian T
Adrian T on 19 Aug 2019
Thank you very much
I need to devote some time to this..

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