how to use all the data for a nonlinear regression?

Hi everyone,

I'm modeling some of my data. As you can see from the attached file, there are 4 columns: time, biomass, hydrogen, acetate. In reality I have more data than hydrogen, but I don't know how to do the integration, since biomass and acetate have less data.

Attached you will find the model and complete data. The model works if you remove the additional data from the third column.

Thanks everyone for the advice!

Answer 1

[Star Strider]

You can only do the regression on complete data rows, so I used the rmmissing function to accomplish that.

Try something like this —

type('A.m')

clear all
load data1.txt

ld=size(data1,1);
y=[data1;data1];

[t,ix]=sort(y(:,1));

y=y(ix,:);

t=y(:,1);
x=y(:,2);
s=y(:,3);
p=y(:,4);
c = [x s p];


options = optimset('MaxFunEvals',100000,'MaxIter',100000);
theta0=[0.03 0.1 0.3 0.1]'; % il simbolo ' significa trasposto
options = optimoptions('lsqcurvefit','Algorithm','levenberg-marquardt');
[theta,Rsdnrm,Rsd,ExFlg,OptmInfo,Lmda,Jmat]=lsqcurvefit(@kinetics,theta0,t,c,zeros(1,numel(theta0)));
ci = nlparci(theta,Rsd,'Jacobian',Jmat);

format shortE
ParametersCI = table(ci(:,1), theta, ci(:,2), 'VariableNames',{'Lower 95% CI','Parmaeter Estimate','Upper 95% CI'})
C = kinetics(theta,t);
hold on

yyaxis right
set(gca, 'ycolor', 'k');  % Imposta il colore dell'asse sinistro a nero
plot(t, c(:,1), 'ok', 'MarkerFaceColor', 'k')  % Pallini neri ripieni

xlabel('Time (days)')
ylabel('Concentration (mmol/L)')


yyaxis left
set(gca, 'ycolor', 'k');  % Imposta il colore dell'asse destro a nero
plot(t,c(:,2),'ok')  % Pallini neri

yyaxis right
set(gca, 'ycolor', 'k');  % Imposta il colore dell'asse destro a nero
plot(t,c(:,3),'^k', 'MarkerFaceColor', [0.5 0.5 0.5], 'MarkerEdgeColor', 'k')

yyaxis right
set(gca, 'ycolor', 'k');  % Imposta il colore dell'asse destro a nero
plot(t, C(:,1), '--k', 'LineWidth', 2)  % Linea nera tratteggiata più spessa

yyaxis left
set(gca, 'ycolor', 'k');  % Imposta il colore dell'asse sinistro a nero
plot(t, C(:,2), '-.k', 'LineWidth', 1)  % Linea nera a puntini più spessa

yyaxis right
set(gca, 'ycolor', 'k');  % Imposta il colore dell'asse destro a nero
plot(t, C(:,3), '-k', 'LineWidth', 1)   % Linea grigia con contorno nero

hold off




legend('biomassa','idrogeno','acetato')
hold off
function C=kinetics(theta,t)

c0 = [0.77;25.1;0.0];
u=theta(1);
YH=theta(2);
YAC=theta(3);
Ks = theta(4)

lt=length(t); TF=zeros(lt,1); YF=zeros(lt,3);

YF(1,:)=c0;

for i=1:lt-1

if t(i+1)>t(i)

[T,Cv]=ode45(@DifEq,t(i:i+1),c0); 

TF(i+1)=t(i+1); YF(i+1,:)=Cv(end,:); c0=YF(i+1,:);

else

TF(i+1)=t(i+1); YF(i+1,:)=YF(i,:); c0=YF(i+1,:);

end

end


    function dC=DifEq(t,c)
        dcdt=zeros(3,1);
        Kh=0.000072; %mmol/L atm
        Vg= 0.1;
        Vl = 0.15;
        Te=298.15;
       R=0.082; %[(l ∙ atm) / (mmol ∙ K)] 
        K = Kh*R*Te;
        a=(Vg+Vl*K)/Vl;
        ua=u*(c(2)/(c(2)+(Ks/Kh)));
        dcdt(1)=ua*c(1);
        dcdt(2)= -1/YH*ua*c(1)/a;
        dcdt(3)= 1/YAC*ua*c(1);
        dC=dcdt;
    end
C=YF;
end

T1 = readtable('data1.txt')

T1 = 15×4 table
    Var1    Var2    Var3     Var4
    ____    ____    _____    ____

      0     0.77    20.03       0
      3      NaN    19.89     NaN
      4     0.89     24.2    0.55
      6      NaN     21.6     NaN
      7        1     20.8    0.61
     10     0.95     18.9    0.84
     11      NaN    17.51     NaN
     12     1.32    15.87    1.22
     13      NaN       15     NaN
     14     1.27    11.72    1.39
     17      NaN    10.76     NaN
     18     1.35     7.77    1.42
     19      NaN     7.64     NaN
     20      NaN     6.69     NaN
     21      1.4     5.47    1.48

T1rm = rmmissing(T1)

T1rm = 8×4 table
    Var1    Var2    Var3     Var4
    ____    ____    _____    ____

      0     0.77    20.03       0
      4     0.89     24.2    0.55
      7        1     20.8    0.61
     10     0.95     18.9    0.84
     12     1.32    15.87    1.22
     14     1.27    11.72    1.39
     18     1.35     7.77    1.42
     21      1.4     5.47    1.48

writematrix(table2array(T1rm),'data1.txt')

run('A.m')

Ks = 0.1000

Ks = 0.1000

Ks = 0.1000

Ks = 0.1000

Ks = 0.1000

Ks = 0.1000

Ks = 0.1000

Ks = 0.1000

Ks = 0.1000

Ks = 0.1000

Ks = 0.0491

Ks = 0.0491

Ks = 0.0491

Ks = 0.0491

Ks = 0.0491

Ks = 0.0922

Ks = 0.0922

Ks = 0.0922

Ks = 0.0922

Ks = 0.0922

Ks = 0.0724

Ks = 0.0724

Ks = 0.0724

Ks = 0.0724

Ks = 0.0724

Ks = 0.0882

Ks = 0.0882

Ks = 0.0882

Ks = 0.0882

Ks = 0.0882

Ks = 0.0781

Ks = 0.0781

Ks = 0.0781

Ks = 0.0781

Ks = 0.0781

Ks = 0.0624

Ks = 0.0624

Ks = 0.0624

Ks = 0.0624

Ks = 0.0624

Ks = 0.0744

Ks = 0.0744

Ks = 0.0744

Ks = 0.0744

Ks = 0.0744

Ks = 0.0674

Ks = 0.0674

Ks = 0.0674

Ks = 0.0674

Ks = 0.0674

Ks = 0.0586

Ks = 0.0586

Ks = 0.0586

Ks = 0.0586

Ks = 0.0586

Ks = 0.0573

Ks = 0.0573

Ks = 0.0573

Ks = 0.0573

Ks = 0.0573

Ks = 0.0520

Ks = 0.0520

Ks = 0.0520

Ks = 0.0520

Ks = 0.0520

Ks = 0.0344

Ks = 0.0344

Ks = 0.0344

Ks = 0.0344

Ks = 0.0344

Ks = 0.0058

Ks = 0.0058

Ks = 0.0058

Ks = 0.0058

Ks = 0.0058

Ks = 0.0054

Ks = 0.0054

Ks = 0.0054

Ks = 0.0054

Ks = 0.0054

Ks = 2.7015e-04

Ks = 2.7015e-04

Ks = 2.7015e-04

Ks = 2.7015e-04

Ks = 2.7017e-04

Ks = 2.7792e-04

Ks = 2.7792e-04

Ks = 2.7792e-04

Ks = 2.7792e-04

Ks = 2.7794e-04

Ks = 2.7794e-04

Ks = 2.7794e-04

Ks = 2.7794e-04

Ks = 2.7794e-04

Ks = 2.7795e-04

Ks = 2.1860e-04

Ks = 2.1860e-04

Ks = 2.1860e-04

Ks = 2.1860e-04

Ks = 2.1862e-04

Ks = 1.8626e-04

Ks = 1.8626e-04

Ks = 1.8626e-04

Ks = 1.8626e-04

Ks = 1.8627e-04

Ks = 1.5710e-04

Ks = 1.5710e-04

Ks = 1.5710e-04

Ks = 1.5710e-04

Ks = 1.5712e-04

Ks = 1.6609e-04

Ks = 1.6609e-04

Ks = 1.6609e-04

Ks = 1.6609e-04

Ks = 1.6611e-04

Ks = 1.6240e-04

Ks = 1.6240e-04

Ks = 1.6240e-04

Ks = 1.6240e-04

Ks = 1.6241e-04

Ks = 1.6397e-04

Ks = 1.6397e-04

Ks = 1.6397e-04

Ks = 1.6397e-04

Ks = 
   1.6399e-04

Local minimum possible.

lsqcurvefit stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

ParametersCI = 4×3 table
    Lower 95% CI    Parmaeter Estimate    Upper 95% CI
    ____________    __________________    ____________

     1.6521e-03         5.5335e-02         1.0902e-01 
     1.6049e-02         9.7913e-02         1.7978e-01 
    -3.1424e-02         7.0951e-01         1.4504e+00 
    -4.8476e-04         1.6397e-04         8.1271e-04 

Ks = 
   1.6397e-04

for sorting the matrix (that actually does not need to be sorted), the sortrows function may be more efficient here.

Thank you for quotimg my code!

.

[Torsten]

@Star Strider

All data that are not NaN data can be used in the regression. Of course this will make it more difficult to construct the C to be returned to "lsqcurvefit". Why do you think that only complete rows could be used ?

Of course, if the OP uses the measurement data as initial values for each time step, all 3 measurement data have to be known. But this is cheating at its best.

[Star Strider]

‘All data that are not NaN data can be used in the regression. Of course this will make it more difficult to construct the C to be returned to "lsqcurvefit". Why do you think that only complete rows could be used ?’

Exactly. Only complete rows have no NaN elements, and all columns are necessary here. If the regression used only column 1 as the independent variable and column 3 as the dependent variable, the entire data set could be used. The alternative would be to use fillmissing rather than rmmissing, however that creates data where no data previusly existed.

In my revision of the code quoted here (that I remember posting as a much later Comment to my original Answer), I add the initial conditions as parameters to be estimated, along with other updates.

[Torsten]

Why not just integrating from t(1) to t(end) in one step with output times taken as the times of the measurements and defining the residuals for lsqcurvefit for all indices (i,j) where the (nx3) matrix c has no NaN values ? This way, also rows with NaN values were used in the fitting process.

[Star Strider]

I am not following that. Only the third column (second data column) has no NaN values, however all the columns are necessary to calculate the parameters. The only option to use all the rows is to use fillmissing.

[Torsten]

From ode45, you get a nx3 matrix of values Cv of simulated values. From the measurements, you also have a nx3 matrix c of values, but some of them are NaN. Now form c-Cv for all values c(i,j) for which c(i,j) is not NaN, make these values a vector and return it to lsqnonlin. This way, you consider all values in the c-matrix that are not NaN, not only those where all three values in a row are not NaN.

[Star Strider]

To me, that means using the entire time vector to simulate the original matrix, and plotting that. That is what I usually do to create the fitted data for the plot. I believe that was done here as well, since the plotted fit lines appear smooth.

Answer 2

[Mr. Pavl M.]

Hi,

I carefully attain further requirements/demands:

It's okay, normal right, you do there curve-fitting by l.-m. algorithm, 3 values in c matrix as a functions of time,I checked the data file timestamps to values consistency ( number of quantities in each column must be equal number of timestamps (sampling points in time - length of the first column)), then you solve custom differential equation.

My answer: remove NaN and Inf values from input by:

c(isnan(c))=0; % (or value to impute) and input can be your c matrix or y(:,i) vectors,

%c(isinf(c)) = max_num; % if there are Inf in inputs

and run, what d'you get?