Fit nonlinear regression model

I am trying the fit the eclosed data,

for the X-Axis:

x = 0:1:248;

I enclosed the data for the Y-axis. (Each row is a curve, there are 5 row = 5 curves)

with the following objective function:

here is my code:

DF = load('Inter_cubic.mat');
Data = DF.Inter_cubic(:,1:230);
Data1 = array2table(Data);

x = 1:size(Data,2);
Sig = @(p,x)  p(4)./(1 + exp(-p(1).*x + p(7))) + p(5)./(1 + exp(-p(2).*x + p(8))) + p(6)./(1 + exp(-p(3).*x + p(9))) + p(10);
beta0 = [Data(1,1) Data(2,1) Data(3,1) Data(4,1) Data(5,1)];


for k = 1:size(Data1,1)
    mdl(1,:) = fitnlm(Data1(k,:),Sig,beta0(1,k));
end

I am getting error message regarding using the function "fitnlm", and regarding the initial value beta0 are they the intial values of the data?

Answer 1

[the cyclist]

Answering your main question: beta0 is the initial guess at the coefficients of the fit. In your case, MATLAB is expecting a vector of length 10, because you have 10 parameters to fit.

There are lots of problems with both your MATLAB syntax, and how you are trying to fit your curves. In the code below, I have fixed the syntax problems, by

• fixing the beta0 problem you asked about
• creating separate tables for each fit
• storing the resulting models in a cell array

But, I did not try to fix the equation you are trying to fit to.

load Data

x = (1:size(Data,2))';

% Define the fitting function
Sig = @(p,x)  p(4)./(1 + exp(-p(1).*x + p(7))) + p(5)./(1 + exp(-p(2).*x + p(8))) + p(6)./(1 + exp(-p(3).*x + p(9))) + p(10);

% Initial guess of coefficients
beta0 = ones(1,10);

for k = 1:size(Data,1)
    
    % Put the data for this curve into a table
    y = Data(k,:)';
    tbl = table(x,y);

    % Fit the model
    mdl{k} = fitnlm(tbl,Sig,beta0);
    
    % Plot the fit against the data
    figure
    hold on
    plot(x,Data(k,:),'o')
    plot(x,predict(mdl{k},x))

end

Warning: Rank deficient, rank = 4, tol = 8.082340e-13.

Warning: Rank deficient, rank = 4, tol = 8.081976e-13.

Warning: Some columns of the Jacobian are effectively zero at the solution, indicating that the model is insensitive to some of its parameters. That may be because those parameters are not present in the model, or otherwise do not affect the predicted values. It may also be due to numerical underflow in the model function, which can sometimes be avoided by choosing better initial parameter values, or by rescaling or recentering. Parameter estimates may be unreliable.

Warning: Rank deficient, rank = 4, tol = 8.082340e-13.

Warning: Rank deficient, rank = 4, tol = 8.081976e-13.

Warning: Some columns of the Jacobian are effectively zero at the solution, indicating that the model is insensitive to some of its parameters. That may be because those parameters are not present in the model, or otherwise do not affect the predicted values. It may also be due to numerical underflow in the model function, which can sometimes be avoided by choosing better initial parameter values, or by rescaling or recentering. Parameter estimates may be unreliable.

Warning: Rank deficient, rank = 4, tol = 8.082340e-13.

Warning: Rank deficient, rank = 4, tol = 8.081976e-13.

Warning: Some columns of the Jacobian are effectively zero at the solution, indicating that the model is insensitive to some of its parameters. That may be because those parameters are not present in the model, or otherwise do not affect the predicted values. It may also be due to numerical underflow in the model function, which can sometimes be avoided by choosing better initial parameter values, or by rescaling or recentering. Parameter estimates may be unreliable.

Warning: Rank deficient, rank = 4, tol = 8.082340e-13.

Warning: Rank deficient, rank = 4, tol = 8.081976e-13.

Warning: Some columns of the Jacobian are effectively zero at the solution, indicating that the model is insensitive to some of its parameters. That may be because those parameters are not present in the model, or otherwise do not affect the predicted values. It may also be due to numerical underflow in the model function, which can sometimes be avoided by choosing better initial parameter values, or by rescaling or recentering. Parameter estimates may be unreliable.

Warning: Rank deficient, rank = 4, tol = 2.680475e-12.

Warning: Rank deficient, rank = 4, tol = 1.142958e-12.

Warning: Rank deficient, rank = 4, tol = 8.476406e-13.

Warning: Rank deficient, rank = 4, tol = 8.122245e-13.

Warning: Some columns of the Jacobian are effectively zero at the solution, indicating that the model is insensitive to some of its parameters. That may be because those parameters are not present in the model, or otherwise do not affect the predicted values. It may also be due to numerical underflow in the model function, which can sometimes be avoided by choosing better initial parameter values, or by rescaling or recentering. Parameter estimates may be unreliable.

[Hidd_1]

The data that you have plotted does't fit the curve stored in Data? it's just a straight line!

[the cyclist]

As I mentioned, I only fixed your syntax, not your fitting problem. Given your equation, and a very naive initial guess, MATLAB cannot do better.

Note the warning that MATLAB is giving: The model is insensitive to some of the parameters. If you look at the results of one of the fits, you'll see

Nonlinear regression model:
    y ~ F(p,x)

Estimated Coefficients:
           Estimate        SE          tStat         pValue   
           ________    __________    __________    ___________
    b1        9407              0           Inf              0
    b2        9407     4.1233e-17    2.2814e+20              0
    b3        9406              0           Inf              0
    b4      3.3047       0.046424        71.183    3.4629e-158
    b5      3.3047       0.046424        71.183    3.4629e-158
    b6      3.3047       0.046424        71.183    3.4629e-158
    b7      8681.2              0           Inf              0
    b8      8681.2              0           Inf              0
    b9      8681.4              0           Inf              0
    b10     3.3403       0.046424        71.951    3.2946e-159

with redundant info where you have the exact same functional term three times. Unless you give much closer initial guesses for each of those terms, I don't think MATLAB is going to be able to figure it out (even if this is a good function to try to fit the curve with).

[Hidd_1]

Should I change the model in your opinion?

[the cyclist]

I think a simplified model can still do a pretty good fit, if you choose better initial guesses. And you may need different initial guesses for each curve. For example, looking only at the 5th set of data ...

load Data

x = (1:size(Data,2))';

% Define the fitting function  
% Sig = @(p,x)  p(4)./(1 + exp(-p(1).*x + p(7))) + p(5)./(1 + exp(-p(2).*x + p(8))) + p(6)./(1 + exp(-p(3).*x + p(9))) + p(10);
Sig = @(p,x)  p(2)./(1 + exp(-p(1).*x + p(3))) + p(4); % <----- I used only one non-linear function here, not all three

% Initial guess of coefficients
beta0 = [0.01 1 0 19] % <------- I changed these

beta0 = 1×4

0.0100 1.0000 0 19.0000

for k = 5 % <------- I am ONLY looking at 5th curve
    
    % Put the data for this curve into a table
    y = Data(k,:)';
    tbl = table(x,y);

    % Fit the model
    mdl{k} = fitnlm(tbl,Sig,beta0);
    
    % Plot the fit against the data
    figure
    hold on
    plot(x,Data(k,:),'o')
    plot(x,predict(mdl{k},x))

end

[Hidd_1]

Thanks a lot!