set parameters of nlinfit function

Question

Mani Ahmadian 2014-9-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/154707-set-parameters-of-nlinfit-function

评论： Star Strider 2014-10-15

nlinfit_Data.txt

Hi

I’m beginner in MATLAB. I have some experimental points and I used nlinfit function to solve my problem and find the best LS curve fitting result. When I use the nlinfit, it pass MSE= 1.5873e+07 and the result is not very good. I attached my data, please check output of nlinfit function to see the output.

Note: I want to find the best curve fitting based on spherical model, soI use: fun = @(p,X) p(1)*(1.5*X/p(2)-0.5*(X/p(2)).^3);

How I should set function’s arguments to reach the best results? Please help me to solve my problem.

Best regards

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 2014-9-14

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/154707-set-parameters-of-nlinfit-function#answer_151634

在 MATLAB Online 中打开

One problem might be that you did not completely vectorise your objective function code (notably the divisions and at least one multiplication. I did here, so that could help:

fun = @(p,X) p(1).*(1.5*X./p(2)-0.5*(X./p(2)).^3);

See if running it makes a difference. If it solves your problem, let me know.

I’d actually like to know a bit more about what you did. It would help to know your starting parameter estimates.

Past midnight here, so I’ll run your code with your data later in the morning and see what results I get.

10 个评论
显示 8更早的评论隐藏 8更早的评论

Star Strider 2014-9-14

在 MATLAB Online 中打开

My pleasure!

‘I think spherical model is the best one based on diagrams in plots. Is it true?’

Don’t depend on how the plots look. This is a statistical exploration.

Note that nonlinear parameter estimation routines are sensitive to the initial parameter estimates.

Use:

pguess = [2E+4; 1E+3];

for your initial estimates for every model, and you may conclude that other models may offer closer fits to your data than your earlier initial estimates provided. (The exponential fit starts looking better with those initial estimates.)

I also suggest you return to your original syntax in your call to nlinfit:

[p,R,J,CovB,MSE,ErrorModelInfo] = nlinfit(xdata,ydata,fun,pguess)

because you need those outputs to input to nlparci and nlpredci and other calculations to help you determine the best fit. Remember to name them slightly differently in each call to nlinfit so you don’t overwrite earlier results in your subsequent calls to nlinfit.

As to the best fit, that is entirely up to you to determine. (The norm of the residuals is a good place to start.) They are your data and regressions and statistics.

Star Strider 2014-9-15

My pleasure!

‘Guessing the initial values’ involves knowing something about your data and the function you are fitting. In your application here, the x- and y-data are on the order of 1E+4, so it makes sense to initialise the parameters to a similar magnitude, considering that your parameters are scaled by the x-data in all your functions. (For instance, if you multiplied your data by your parameters instead of dividing your data by the parameters, the appropriate initial parameter estimates magnitudes would be on the order of 1E-4 instead.) The important considerations in choosing initial parameter estimates for a particular problem are understanding the maths of your regression objective functions and the data you are using in your nonlinear parameter estimations.

There are a number of good textbooks on linear and nonlinear parameter estimation and regression that use MATLAB. See: MATLAB and Simulink Based Books. There are a number of online resources available as well. It is probably best that you search for those yourself, since you have a much better understanding of your background and applications than I do, and while having an encyclopaedic knowledge of these techniques is desirable, it is not usually practical.

Another source I use to guide me to good references are university courses on the subjects I’m interested in. If the course looks as though it covers the topics that interest me, I search for the syllabus. The professors usually mention the textbooks for the course in the syllabus for the course.

Star Strider 2014-9-15

My pleasure!

If you haven’t already done so, I suggest you start with a good introductory Statistics course. Then depending on what you want to do, take specialty courses, for instance in linear and nonlinear regression if you want to continue with what you are doing now.

The Spherical model seems to fit its parameters well even if given an initial estimate that is far from ideal. (You would have to calculate its Jacobian to see the reason.) I used various initial estimates for it, and it always converged on the same values. Assume it is already optimised as much as it can be.

I would use the norm of the residuals or analogously ‘MSE’ (mean squared error) as a way to determine the best model, so the ‘best’ model would have the least MSE. (If you want, you can estimate the statistical significance of the difference between two models with the Likelihood Ratio Test. For your purposes, it is safe to assume the the MSE is the likelihood, since all your models have the same number of parameters. This is the only way I know to compare two nonlinear models with different objective functions.)

Did we solve your problem or do you still have questions?

Mani Ahmadian 2014-9-16

Of course... you helped me to solve the most parts of my problem. But condition of spherical model doesn't satisfied me. I think it's diagram is out of tune and its mse is a huge value. I guess I have a big calculation or mathematical mistake about it, but I'm not sure where probably it happens.

I downloaded some textbooks about nonlinear curve fitting and I'll study them as soon as possible to have better understand.

Your view points and guidance was very helpful for me. Thank you very much, my friend.

Have a good time. Mani

Star Strider 2014-9-16

My pleasure!

Compare the MSE values for the various models. If any of the models adequately explain the process that produced your data, those should take precedence. If they all do, then the one with the lowest MSE is likely the correct one. If with the Likelihood Ratio Test they are not significantly different from the one with the lowest MSE, they are all valid. Developing and estimating other models that could explain your data are appropriate and acceptable. If you can find a more appropriate model, fit it as well and compare it to the others. That is the fun — and frustration — of nonlinear parameter estimation and mathematical modeling.

You have fun as well! I wish you well in your research. We’re here if you need us for other MATLAB projects.

请先登录，再进行评论。

Answer 2

Mani Ahmadian 2014-10-2

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/154707-set-parameters-of-nlinfit-function#answer_153577

编辑：Mani Ahmadian 2014-10-2

Hi again

I'm trying to solve the problem with spherical model fit function, this model has a mathematical formula as below:

I think the un-fit problem is due to two criteria function of above model. I don't know how I should define this type of function to nlinfit.

Would you please help me to solve the problem?

Thanks

Mani

12 个评论
显示 10更早的评论隐藏 10更早的评论

Star Strider 2014-10-14

编辑：Star Strider 2014-10-14

在 MATLAB Online 中打开

Truncating the x and y values so that x<=a0 for successive values of ‘a0’, both parameter values converge to essentially stable values, although those values are a function of ‘a0’.

The following code (for the spherical model) loops, calculating ‘a0’ at each step and truncating the (x,y) arrays to accommodate x<=a0 in the subsequent step, estimating the parameters with the new data set, and repeating for 25 iterations:

fidi = fopen('nlinfit_Data.txt');
D = textscan(fidi, '%f %f', 'HeaderLines',1);
x = D{1};
y = D{2};
% b(1) = c0,  b(2) = a0
gam2 = @(b,h) b(1).*(1.5*h./b(2) - 0.5*(h./b(2)).^3);
B(:,1) = [1E+4; x(length(x))];
Lxy(1) = length(x);
for k1 = 2:25
    B(:,k1) = nlinfit(x(x<=(B(2,k1-1))),y(x<=(B(2,k1-1))),gam2, B(:,k1-1));
    Lxy(k1) = length(x(x<=(B(2,k1-1))));
end
figure(1)
semilogy(1:25, B(1,:),  1:25, B(2,:))
legend('c_0', 'a_0')
xlabel('Iteration #')
ylabel('Parameter Value')

The ‘B’ matrix contains the parameter values for each iteration, and the ‘Lxy’ vector (length of the data vectors at each step) demonstrates that the data are indeed being truncated. The preceding parameter values are used as the initial estimates for the subsequent estimation. The results are yours to interpret.

To plot the estimated y-values for a particular index of parameter values, you can use this code:

Lidx = 9;                           % Choose Parameter Set By Index
xv = linspace(min(x),x(Lidx));      % Create x-Vector
yfit = gam2(B(:,Lidx), xv);         % Calculate Estimated Fit
figure(2)
plot(x(1:Lidx), y(1:Lidx), '+')
hold on
plot(xv, yfit, '-r')
hold off
grid
xlabel('\ith\rm')
ylabel('\gamma_{2}(\ith\rm)')

Have fun with it!

Mani Ahmadian 2014-10-15

Great

Thanks for your nice job. I'll check it as soon as possible.

Have a nice time

Mani

Star Strider 2014-10-15

My pleasure!

You too!

请先登录，再进行评论。

set parameters of nlinfit function

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

10 个评论
显示 8更早的评论隐藏 8更早的评论

更多回答（1 个）

12 个评论
显示 10更早的评论隐藏 10更早的评论

另请参阅

类别

标签

Community Treasure Hunt

set parameters of nlinfit function

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

10 个评论 显示 8更早的评论隐藏 8更早的评论

更多回答（1 个）

12 个评论 显示 10更早的评论隐藏 10更早的评论

另请参阅

类别

标签

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

10 个评论
显示 8更早的评论隐藏 8更早的评论

12 个评论
显示 10更早的评论隐藏 10更早的评论