nlinfit with modelfun as an integral

Question

Armantas 2014-7-31

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/144262-nlinfit-with-modelfun-as-an-integral

评论： Star Strider 2014-8-3

Dear all,

I have discrete data A(x,y), which I want to fit by a specified function y=f(x). My function f(x) has the following restriction: df(x)/dlog(x) is equal to a sum of two Gaussians. df(x)/dlog(x) being the derivative of f(x) with respect to the argument log(x). Given this restriction f(x) can be expressed as an integral. This gives 6 fit parameters, namely: the heights of the Gaussians, their means and their standard deviations. I have gone this far:

par=[1e-9,1e-5,1,10000,1,1]; %initial Gaussian parameter guess
Integrand = @(x) ((par(3)/(par(5)*(2*pi)^0.5))*exp((-(log10(x)-log10(par(1))).^2)/(2*par(5)^2))+(par(4)/(par(6)*(2*pi)^0.5))*exp((-(log10(x)-log10(par(2))).^2)/(2*par(6)^2)))./(x*log(10)); %two gaussians
Integral = @(x) integral(Integrand,0,x);
nlinfit(A(:,1),A(:,2),Integral,par)

I however get the following error:

Error using nlinfit (line 142)
Error evaluating model function '@(x)integral(Integrand,0,x)'.
Caused by:
    Error using @(x)integral(Integrand,0,x)
    Too many input arguments.

How can I fix this? Thank you

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Star Strider 2014-8-1

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/144262-nlinfit-with-modelfun-as-an-integral#answer_147098

编辑：Star Strider 2014-8-1

在 MATLAB Online 中打开

There are a few problems I can see. There could be more, but this should get you started:

Add ‘par’ to the Integrand function arguments:

Integrand = @(par,x) ((par(3)/(par(5) ...

The Integral function then becomes:

Integral = @(par,x) integral(@(x) Integrand(par,x),0,x);

I suggest that you name your initial parameter estimates ‘par0’ (or something other than ‘par’) to avoid confusion:

par0 = [1e-9,1e-5,1,10000,1,1]; %initial Gaussian parameter guess;
est_par = nlinfit(A(:,1),A(:,2),Integral,par0)

The ‘est_par’ assignment are the parameters estimated by nlinfit.

I can’t test your code (so no guarantees), but these changes should at least allow it to run.

8 个评论
显示 6更早的评论隐藏 6更早的评论

Star Strider 2014-8-3

I don’t know what is throwing the error. ‘Undefined values’ I assume to be NaN or ±Inf, so you have to evaluate your function with your initial values of x and parameters to see what the problem is.

However, if you got it to fit well with nlinfit, I would just stop there. Just go with the parameters nlinfit estimated and be happy with them since they seem to provide good results. You can also get confidence intervals on them with nlparci and confidence intervals on the fitted data with nlpredci.

I suggested lsqcurvefit in case you wanted to constrain par(1) and par(2) to be greater than zero to avoid complex results. Since your model obviously converged nicely on positive numbers without constraints with nlinfit, you don’t need to use lsqcurvefit.

................................

With respect to lsqcurvefit though, you can’t set bounds on x — x is your independent variable. You have to have matching x and y values, so if you have an x that is out-of-bounds, you have to delete it and its corresponding y value from the data you give to lsqcurvefit.

You can constrain your parameters with lsqcurvefit. You have 6 parameters so you have to have vectors of upper and lower bounds that are the same dimensions as your parameter vector. So, for instance, if your parameter vector is a (6x1) column vector, your upper and lower bound vectors each have to be (6x1) column vectors as well. There have to be as many constraints as you have parameters. If you don’t want to constrain a particular parameter, set its bounds (in the respective vectors) as -Inf and +Inf. To set a parameter to be greater than zero, set its lower bound to eps.

Armantas 2014-8-3

编辑：Armantas 2014-8-3

Yes, it turns out NaN values in my A(x,y) where causing the problem.

Thank you for all the info, it was very helpful! Now the script is up and running :)

Star Strider 2014-8-3

My pleasure!

Yours is the most unusual curve-fitting design (with the integral) that I’ve thus far encountered, so I learned much from it. It is also interesting that The Statistics Toolbox function nlinfit could deal with the NaN values on its own.

请先登录，再进行评论。

nlinfit with modelfun as an integral

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

8 个评论
显示 6更早的评论隐藏 6更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

Community Treasure Hunt

nlinfit with modelfun as an integral

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

8 个评论 显示 6更早的评论隐藏 6更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

8 个评论
显示 6更早的评论隐藏 6更早的评论