Relation or Pattern between curves

I am having the following curves, and I am trying to find a relation between them, or a statistical factor to use it so i can predict curves just from one:

I would appreciate any help, Thanks!

[Star Strider]

The ‘Duplicate’ flag is likely not appropriate here. This is the original post. There are (at least) two other duplicate posts related to it that I saw.

[Hidd_1]

Sorry it was a mistake, I wanted to flag another post not this one!

[Image Analyst]

But if the question was posted multiple times, perhaps with slight modifications each time, the question becomes which is the one to answer. Presumably the latest, most recent one is the final/best one is the one to answer, rather than the first/oldest one.

Answer 1

[Star Strider]

The plot appears to be incomplete.

What do the curves look like between 0 (or whatever the minimum independent value is) and 150? What are the independent variable values?

The parts of the curves displayed appear to be inverted Logistic function curves, so one option might be to estimate the parameters of each of them and then compare those parameters.

That objective function could be —

lgfcn = @(b,x) b(1) - b(2)./(1 + exp(-b(3).*(x-b(4))));

.

[John D'Errico]

@Star Strider - good idea, but wrong model. The logistic you suggest will be symmetrical around the midpoint. So the left and right tails will look like each other in that model. But the curves shown are strongly skewed. So that cannot be a viable model for this process.

[Star Strider]

I decided to go with ga for this, since gradient-descent approaches weren’t working.

This gives reasonably decent fits to the data, considering that we have no information on the process that created it. (A mathematical description of that process would be the best model for it.) This uses a slightly changed version of the modified logistic function that I originally proposed.

The funciton fits are plotted in the same colours as the data. It might be best to plot the data and the fit each together in a subplot or tiledlayout array. I plotted them on the same axes here since that’s how they were originally presented.

Try this —

LD = load('Data.mat')

LD = struct with fields:
    Inter_cubic: [5×249 double]

Data = LD.Inter_cubic;
x = 1:size(Data,2);

lgfcn = @(b,x) b(1) .* -b(2)./(1 + exp(-b(3).*((x-max(x)).*b(4)-b(5))));

PopSz = 100;
Parms =   5;

for k = 1:size(Data,1)
    optsAns = optimoptions('ga', 'PopulationSize',PopSz, 'InitialPopulationMatrix',randi(1E+4,PopSz,Parms).*1E-3.*[Data(1,1) -1 -1 0.01 -1], 'MaxGenerations',5E3, 'FunctionTolerance',1E-10);                                                                  % Options Structure For 'Answers' Problems
    B(:,k) = ga(@(b) norm(Data(k,:)-lgfcn(b,x)), Parms, [],[],[],[],[],[],[],[],optsAns);
end

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

figure
hold on
for k = 1:size(Data,1)
    hp1 = plot(x, Data(k,:), 'd', 'MarkerSize',2, 'DisplayName',"Inter\_cubic Row "+k);
    hp2 = plot(x, lgfcn(B(:,k),x), 'DisplayName',"Function Fit "+k);
    hp2.Color = hp1.Color;
end
hold off
grid
legend('Location','best', 'NumColumns',size(Data,1))

Parameters = array2table(B, 'VariableNames',compose('Inter_cubic Row %d',1:size(Data,1)), 'RowNames',compose('b(%d)',1:size(Data,1)))

Parameters = 5×5 table
            Inter_cubic Row 1    Inter_cubic Row 2    Inter_cubic Row 3    Inter_cubic Row 4    Inter_cubic Row 5
            _________________    _________________    _________________    _________________    _________________

    b(1)          8.0385               9.4816               3.8418               63.688               6.8332     
    b(2)          -1.819               -2.116               -5.281               -0.321               -2.953     
    b(3)          -0.791               -0.806               -2.233              -2.8395              -4.7949     
    b(4)         0.09692              0.09419              0.06454              0.06038              0.04561     
    b(5)         -3.7631              -3.4574               -2.163              -1.9266               -1.095     

The fitted lines aren’t perfect, however they are quite close.

.

[Hidd_1]

Why did you multiply b1 and -b2, you could just use one variable?

[Star Strider]

I was experimenting with the function. The first parameter was originally added, and I forgot to simplify it when I changed the code for ‘lgfcn’.

Simplifying it —

LD = load('Data.mat')

LD = struct with fields:
    Inter_cubic: [5×249 double]

Data = LD.Inter_cubic;
x = 1:size(Data,2);

lgfcn = @(b,x)  b(1)./(1 + exp(-b(2).*((x-max(x)).*b(3)-b(4))));

PopSz = 100;
Parms =   4;

for k = 1:size(Data,1)
    optsAns = optimoptions('ga', 'PopulationSize',PopSz, 'InitialPopulationMatrix',randi(1E+4,PopSz,Parms).*1E-3.*[Data(1,1) -1 0.01 -1], 'MaxGenerations',5E3, 'FunctionTolerance',1E-10);                                                                  % Options Structure For 'Answers' Problems
    B(:,k) = ga(@(b) norm(Data(k,:)-lgfcn(b,x)), Parms, [],[],[],[],[],[],[],[],optsAns);
end

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

figure
hold on
for k = 1:size(Data,1)
    hp1 = plot(x, Data(k,:), 'd', 'MarkerSize',2, 'DisplayName',"Inter\_cubic Row "+k);
    hp2 = plot(x, lgfcn(B(:,k),x), 'DisplayName',"Function Fit "+k);
    hp2.Color = hp1.Color;
end
hold off
grid
legend('Location','northoutside', 'NumColumns',size(Data,1))

Parameters = array2table(B, 'VariableNames',compose('Inter_cubic Row %d',1:size(Data,1)), 'RowNames',compose('b(%d)',1:size(B,1)))

Parameters = 4×5 table
            Inter_cubic Row 1    Inter_cubic Row 2    Inter_cubic Row 3    Inter_cubic Row 4    Inter_cubic Row 5
            _________________    _________________    _________________    _________________    _________________

    b(1)          14.236               20.072                20.15               20.369               20.375     
    b(2)          -0.824              -1.6437              -1.5384              -1.9802               -2.289     
    b(3)         0.08379              0.04973             0.090338               0.0763               0.0942     
    b(4)         -2.7504              -1.8677               -2.939              -2.3413              -2.4704     

.

[Star Strider]

I devised a different model using tanh that provides a much better (and more robust) fit —

load('Data.mat')

Data = Inter_cubic;

x = 1:size(Data,2);
objfcn = @(b,x) b(1) + b(2).*tanh(b(3).*(x+b(4)));

for k = 1:size(Data,1)
    B0 = [Data(k,1); 10; 0.1; -200];
    mdl = fitnlm(x,Data(k,:), objfcn, B0)
    B(:,k) = mdl.Coefficients.Estimate;
    R2(k) = mdl.Rsquared.Adjusted;
end

mdl = 
Nonlinear regression model:
    y ~ b1 + b2*tanh(b3*(x + b4))

Estimated Coefficients:
          Estimate       SE         tStat       pValue   
          ________    _________    _______    ___________

    b1      2.0155       0.7002     2.8785      0.0043485
    b2     -12.956      0.71557    -18.106      4.399e-47
    b3    0.018988    0.0006045     31.411     6.9664e-88
    b4     -236.73       2.8622    -82.709    5.1885e-181


Number of observations: 249, Error degrees of freedom: 245
Root Mean Squared Error: 0.381
R-Squared: 0.992,  Adjusted R-Squared 0.992
F-statistic vs. constant model: 1e+04, p-value = 4.3e-256

mdl = 
Nonlinear regression model:
    y ~ b1 + b2*tanh(b3*(x + b4))

Estimated Coefficients:
          Estimate        SE         tStat       pValue   
          ________    __________    _______    ___________

    b1       7.892       0.19599     40.267    5.1454e-110
    b2     -12.314       0.20445    -60.232    7.2457e-149
    b3    0.029224    0.00060136     48.597    8.5211e-128
    b4     -219.01       0.69285     -316.1    6.1264e-322


Number of observations: 249, Error degrees of freedom: 245
Root Mean Squared Error: 0.385
R-Squared: 0.996,  Adjusted R-Squared 0.996
F-statistic vs. constant model: 1.92e+04, p-value = 2e-290

mdl = 
Nonlinear regression model:
    y ~ b1 + b2*tanh(b3*(x + b4))

Estimated Coefficients:
          Estimate        SE         tStat       pValue   
          ________    __________    _______    ___________

    b1      9.9895      0.038228     261.31    9.6776e-302
    b2     -10.287      0.039946    -257.53     3.388e-300
    b3    0.059716    0.00062939      94.88    3.5387e-195
    b4     -216.29       0.11342    -1906.9              0


Number of observations: 249, Error degrees of freedom: 245
Root Mean Squared Error: 0.199
R-Squared: 0.999,  Adjusted R-Squared 0.999
F-statistic vs. constant model: 7.36e+04, p-value = 0

mdl = 
Nonlinear regression model:
    y ~ b1 + b2*tanh(b3*(x + b4))

Estimated Coefficients:
          Estimate        SE         tStat       pValue   
          ________    __________    _______    ___________

    b1      10.258      0.023407     438.26              0
    b2     -10.028       0.02426    -413.36              0
    b3    0.079354    0.00063438     125.09    5.4588e-224
    b4     -218.66      0.061486    -3556.3              0


Number of observations: 249, Error degrees of freedom: 245
Root Mean Squared Error: 0.142
R-Squared: 0.999,  Adjusted R-Squared 0.999
F-statistic vs. constant model: 1.39e+05, p-value = 0

mdl = 
Nonlinear regression model:
    y ~ b1 + b2*tanh(b3*(x + b4))

Estimated Coefficients:
          Estimate        SE         tStat       pValue   
          ________    __________    _______    ___________

    b1      10.56       0.015929     662.98              0
    b2    -9.7661       0.016335    -597.86              0
    b3    0.11607     0.00079811     145.43    8.3072e-240
    b4    -222.21       0.035038      -6342              0


Number of observations: 249, Error degrees of freedom: 245
Root Mean Squared Error: 0.108
R-Squared: 1,  Adjusted R-Squared 1
F-statistic vs. constant model: 2.19e+05, p-value = 0

Parameters = array2table(B, 'VariableNames',compose('Data Row %d',1:size(Data,1)), 'RowNames',compose('b(%d)',1:size(B,1)))

Parameters = 4×5 table
            Data Row 1    Data Row 2    Data Row 3    Data Row 4    Data Row 5
            __________    __________    __________    __________    __________

    b(1)       2.0155         7.892        9.9895        10.258        10.56  
    b(2)      -12.956       -12.314       -10.287       -10.028      -9.7661  
    b(3)     0.018988      0.029224      0.059716      0.079354      0.11607  
    b(4)      -236.73       -219.01       -216.29       -218.66      -222.21  

figure
tiledlayout(size(Data,1),1)
for k = 1:size(Data,1)
    nexttile
    plot(x, Data(k,:), 'd', 'MarkerSize',1)
    hold on
    plot(x,objfcn(B(:,k),x), '-r')
    hold off
    grid
    ylim([0 30])
    ylabel("Row "+k)
    text(215,25,sprintf('R^2 = %.2f',R2(k)), 'FontSize',9)
end

The tanh function is a ratio of two exponential functions so it lends itself to this sort of function fit. The first parameter shifts it in the ‘y’ direction, the second parameter scales its amplitude, the third parameter scales its slope, and the fourth parameter scales its location in the ‘x’ direction.

.

[Hidd_1]

Thank a lot Star Strider! I really do appreciate your help.

Btw how did you found the initial values B0?

[Star Strider]

I initially just used randn. When I found values that privded the best fit, I used approximations of those for the initial values. I suspect ga would have found them quickly, and in this instance, it did. The only specific requirement was to set the scale for the fourth parameter to be negative, and larger than the other values, and that only because the magnitude of ‘x’ was much greater than the magnitudes of the other parameter values. Dividing ‘x’ by 100 and changing the sign of ‘b(4)’ in ‘objfcn’ could make even that not be necessary. It comes up with a reasonably accurate estimate for ‘B0’ in every instance, that fitnlm then refines.

Try this —

load('Data.mat')
Data = Inter_cubic;

x = 1:size(Data,2);
objfcn = @(b,x) b(1) + b(2).*tanh(b(3).*(x+b(4)));

PopSz = 100;
Parms =   4;

for k = 1:size(Data,1)
    optsAns = optimoptions('ga', 'PopulationSize',PopSz, 'InitialPopulationMatrix',randi(1E+4,PopSz,Parms).*[ones(1,3)*1E-4 -0.1], 'MaxGenerations',5E3, 'FunctionTolerance',1E-10);                                                                  % Options Structure For 'Answers' Problems
    B0 = ga(@(b) norm(Data(k,:)-objfcn(b,x)), Parms, [],[],[],[],[],[],[],[],optsAns)
    mdl = fitnlm(x,Data(k,:), objfcn, B0)
    B(:,k) = mdl.Coefficients.Estimate;
    R2(k) = mdl.Rsquared.Adjusted;
end

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

B0 = 1×4

8.5122 5.6045 -0.1005 -209.5831

mdl = 
Nonlinear regression model:
    y ~ b1 + b2*tanh(b3*(x + b4))

Estimated Coefficients:
          Estimate         SE         tStat       pValue   
          _________    __________    _______    ___________

    b1       2.0156       0.70014     2.8788      0.0043446
    b2       12.956       0.71551     18.107     4.3494e-47
    b3    -0.018988    0.00060451    -31.411     6.9762e-88
    b4      -236.73         2.862    -82.714    5.1157e-181


Number of observations: 249, Error degrees of freedom: 245
Root Mean Squared Error: 0.381
R-Squared: 0.992,  Adjusted R-Squared 0.992
F-statistic vs. constant model: 1e+04, p-value = 4.3e-256

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

B0 = 1×4

10.0702 9.8639 -0.0432 -212.1799

mdl = 
Nonlinear regression model:
    y ~ b1 + b2*tanh(b3*(x + b4))

Estimated Coefficients:
          Estimate         SE         tStat       pValue   
          _________    __________    _______    ___________

    b1        7.892         0.196     40.266    5.1596e-110
    b2       12.314       0.20445     60.231    7.2637e-149
    b3    -0.029224    0.00060135    -48.597    8.5125e-128
    b4      -219.01       0.69286     -316.1    6.1264e-322


Number of observations: 249, Error degrees of freedom: 245
Root Mean Squared Error: 0.385
R-Squared: 0.996,  Adjusted R-Squared 0.996
F-statistic vs. constant model: 1.92e+04, p-value = 2e-290

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

B0 = 1×4

10.4556 -9.8869 0.0612 -215.3375

mdl = 
Nonlinear regression model:
    y ~ b1 + b2*tanh(b3*(x + b4))

Estimated Coefficients:
          Estimate        SE         tStat       pValue  
          ________    __________    _______    __________

    b1      9.9895      0.038229     261.31    9.702e-302
    b2     -10.287      0.039946    -257.53    3.397e-300
    b3    0.059716    0.00062938     94.881    3.532e-195
    b4     -216.29       0.11343    -1906.9             0


Number of observations: 249, Error degrees of freedom: 245
Root Mean Squared Error: 0.199
R-Squared: 0.999,  Adjusted R-Squared 0.999
F-statistic vs. constant model: 7.36e+04, p-value = 0

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

B0 = 1×4

10.9978 9.2900 -0.1028 -217.2195

mdl = 
Nonlinear regression model:
    y ~ b1 + b2*tanh(b3*(x + b4))

Estimated Coefficients:
          Estimate         SE         tStat       pValue   
          _________    __________    _______    ___________

    b1       10.258      0.023407     438.26              0
    b2       10.028       0.02426     413.36              0
    b3    -0.079354    0.00063438    -125.09    5.4558e-224
    b4      -218.66      0.061486    -3556.3              0


Number of observations: 249, Error degrees of freedom: 245
Root Mean Squared Error: 0.142
R-Squared: 0.999,  Adjusted R-Squared 0.999
F-statistic vs. constant model: 1.39e+05, p-value = 0

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

B0 = 1×4

10.3432 10.0783 -0.0899 -222.7092

mdl = 
Nonlinear regression model:
    y ~ b1 + b2*tanh(b3*(x + b4))

Estimated Coefficients:
          Estimate        SE         tStat       pValue  
          ________    __________    _______    __________

    b1       10.56      0.015929     662.98             0
    b2      9.7661      0.016335     597.86             0
    b3    -0.11607    0.00079812    -145.43    8.331e-240
    b4     -222.21      0.035038      -6342             0


Number of observations: 249, Error degrees of freedom: 245
Root Mean Squared Error: 0.108
R-Squared: 1,  Adjusted R-Squared 1
F-statistic vs. constant model: 2.19e+05, p-value = 0

Parameters = array2table(B, 'VariableNames',compose('Data Row %d',1:size(Data,1)), 'RowNames',compose('b(%d)',1:size(B,1)))

Parameters = 4×5 table
            Data Row 1    Data Row 2    Data Row 3    Data Row 4    Data Row 5
            __________    __________    __________    __________    __________

    b(1)       2.0156         7.892        9.9895        10.258         10.56 
    b(2)       12.956        12.314       -10.287        10.028        9.7661 
    b(3)    -0.018988     -0.029224      0.059716     -0.079354      -0.11607 
    b(4)      -236.73       -219.01       -216.29       -218.66       -222.21 

figure
tiledlayout(size(Data,1),1)
for k = 1:size(Data,1)
    nexttile
    plot(x, Data(k,:), 'd', 'MarkerSize',1)
    hold on
    plot(x,objfcn(B(:,k),x), '-r')
    hold off
    grid
    ylim([0 30])
    ylabel("Row "+k)
    text(215,25,sprintf('R^2 = %.2f',R2(k)), 'FontSize',9)
end

.

[Hidd_1]

Can there be other functions to fit the same Data with the same or better accuracy?

[Star Strider]

There could be. I initially used the logistic function, then found that tanh provided an even better and more robust fit.

You are certainly free to experiment with them, however with an adjusted R² of close to 1, I doubt other functions can improve on tanh, at least with these data. Other functions may be more appropriate to different data. My code should be robust to other functions as coded in ‘objfcn’, although it may require changing some of the parameters, such as ‘Parms’.

[Hidd_1]

I have seen that the b(3) is approx 0.01 in all the cases, is it possible to ignore it?! or put it as fixed number so I can get less variables.

[Star Strider]

The ‘b(3)’ parameter (that controls the slope) varies between about -0.01 and -0.11, so I would be reluctant to eliminate it.

The ‘B0’ result is quite similar in every call, so perhaps only one ga call (before the loop, using ‘x’ and the first row of the matrix) would be necessary, using that ‘B0’ for all the fitnlm calls.

[Hidd_1]

From the results of the fitting, do you think that the system can be described mathematically?

[Star Strider]

From just the fit, no. The fit is empirical.

Describing it mathematically would require knowing the process that created it.

[Hidd_1]

In case I wanted to do a Curve fitting between the curves, is it easier to interpolate between the b's or should I interpolate between the data sets?

[Star Strider]

It wouild be best to interpolate between the data sets. They asll have the same numbers of points.

One possible way of interpolating them would be to initially plot them as a surface to understand how they relate to each other —

load(websave('Data','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1387204/Data.mat'))
Data = Inter_cubic;
x = 1:size(Data,2);
r = (1:size(Data,1));

xr = repmat(x,1,size(Data,1)).';
Datar = reshape(Data.',[],1);
rr = repmat(r,1,size(Data,2));
DataF = scatteredInterpolant(xr(:), rr(:), Datar);

rq = 3.5*ones(size(x));
NewData = DataF(x(:), rq(:));


figure
surfc(x, r, Data)
hold on
plot3(x, rq, NewData, '-r', 'LineWidth',2)
hold off
colormap(turbo)
view(40,30)
xlabel('X')
ylabel('Row')

figure
plot(x, NewData)
hold on
plot(x, mean(Data([3 4],:)))
hold off

There are then a number of ways to interpolate this, my favourite being scatteredInterpolant. I am at something of a loss to explain the reason it is oscillating like that, unless the data oscillates more than I thought it does. It normally produces a smooth interpolation. The mean of the neighbouring data do not oscillate.

.

[Hidd_1]

In case I have another variable, which is related to this process, would that help to kind gets what's behind the b's, right? I mean would I able to get a clearer picture about what the b's reflect physically

[Star Strider]

To fully understand the parameters (b’s), it would be necessary to know what process created the data in the first place. Then, they would likely have some physical meaning (and the objective function would likely reflect that process, not simply reflect the curve fitting).

In my ‘objfcn’, the parameters translate and scale the tanh function, so that ‘b(1)’ translates (shifts) it with respect to its dependent variable (‘y’) value, ‘b(2)’ scales its amplitude with respect to the dependent variable value, ‘b(3)’ scales the slope of the tanh function, and ‘b(4)’ translates (shifts) the tanh function with respect to the independent variable ‘x’ value. They have no physical significance beyond that, and they all relate only to the tanh function, not to the process that created the data.

To fully understand what the ‘b’ parameters mean physically, (with respect to the data) it would be necessary to understand how they relate to the process that created the data. Since I do not know what that is, I cannot comment on it.

[Hidd_1]

Will it be possible to do nl fitting over 3 or more Data set (variables)?

I mean if I defined a function objfcn = @(b,x,z) , will the fitnlm works?

[Star Strider]

It depends on what ‘z’ is, what your objective function is, and what it does with ‘z’.

If ‘z’ is an extra parameter, and is already defined in the workspace, and ‘objfcn’ was defined as:

objfcn = @(b,x,z) ... ;

it would be prresented to fitnlm as:

@(b,x)objfcn(b,x,z)

So in that instance, ‘z’ would not be a parameter to be optimised, or a specific independent variable.

[Hidd_1]

Do you know how to perform a validation test on this non-linear regression model?

[Star Strider]

I am not certain what sort of ‘validation test’ you want. The fitnlm function provides several relevant statistics on the goodness-of-fit, as illustrated in my earlier Comment.

[Hidd_1]

How can we validate a non-linear regression model when it only fits the measured data but may not accurately represent the underlying mathematical description of the physical process?

for instance: Should a cross-correlation analysis be performed to validate a non-linear regression model on additional measurements?

[Star Strider]

‘How can we validate a non-linear regression model when it only fits the measured data but may not accurately represent the underlying mathematical description of the physical process?’

Without knowing the process that created the data (and I do not know what it is here), and creating a mathematical model of it, there only way to model the data is empirically, that is, fitting the data to a function that just looks like it would work. There is no way to ‘validate’ it in the sense of determining how well the estimated parameters describe the underlying process, if the underlying process is not actually known.

‘Should a cross-correlation analysis be performed to validate a non-linear regression model on additional measurements?’

I would simply fit the new measurments to the existing empirical model and see how well the empirical model fits the data. I am not certain there is any other way to determine that.

If the process that created the data was known, then it might be possible to see how well the estimated parameters of the empirical model modeled the actual mathematical model of the process (that is, compared to its parameters). Without having the mathematical model of the process that created the data, we are restricted to using an empirical model. There is simply no other option.

In the absence of a mathematical model of the process that created the data, other empirical models can certainly be hypothesized and tested. For example, if a different input produced a different output, procedures described in the System Identification Toolbox could be used to derive different models, or perhaps one model that could fit outputs from other inputs, providing that both the inputs and corresponding outputs were available. I am not certain what benefit a cross-correlation analysis would have, when it is possible to fit the data, however I may not understand the sort of analysis you want to do.

Answer 2

[Image Analyst]

You forgot to attach your data! It should be possible to fit any particular curve to some sigmoid kind of curve, if you have the model for it, with fitnlm. I'm attaching some demos using various models so you can see how it's done.

[Hidd_1]

Here is the 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)

I don't have a mathematical model for the data, I've tried curve fitting but than found difficulties finding the relation between the curves.

I have tried the following code for a sigmoid function but I got an error:

modelfun = @(x)(1/(1+exp(-x)));
beta0 = Data(1,1);
mdl = fitnlm(Data(1,:),modelfun,beta0)

I would appreciate any feedback, & Thanks a lot!

[Alex Sha]

Try the fitting function below:

For the first row data:

Sum Squared Error (SSE): 0.10950384442613
Root of Mean Square Error (RMSE): 0.0209708005475935
Correlation Coef. (R): 0.999987584910591
R-Square: 0.999975169975316

Parameter	Best Estimate      
---------	-------------      
p1       	-0.0249084190435688
p2       	-0.0763243107482052
p3       	0.142316610539568  
p4       	6.04837338439208   
p5       	29.8257426089756   
p6       	15.1516810377477   
p7       	4.69340638351262   
p8       	18.1352678248419   
p9       	-34.4425824172933  
p10      	-20.5645162561211  

For the second row data:

Sum Squared Error (SSE): 0.490607218014365
Root of Mean Square Error (RMSE): 0.0443881753680807
Correlation Coef. (R): 0.999971420379912
R-Square: 0.999942841576619

Parameter	Best Estimate      
---------	-------------      
p1       	0.0929778051317028 
p2       	-0.0761628900340691
p3       	-0.178882792529087 
p4       	-8.72339392625596  
p5       	2.37142383251581   
p6       	8.93100805985586   
p7       	-18.9633995662387  
p8       	11.9590466793433   
p9       	39.6871533513552   
p10      	9.02043998188305 

For the third row data:

Sum Squared Error (SSE): 0.112293028574181
Root of Mean Square Error (RMSE): 0.0212361959486675
Correlation Coef. (R): 0.999993593414753
R-Square: 0.999987186870549

Parameter	Best Estimate     
---------	-------------     
p1       	-0.122327538545226
p2       	0.365168326229641 
p3       	-0.413118508834921
p4       	15.4162121444737  
p5       	-0.230218900430568
p6       	4.24989701962221  
p7       	25.9289257568436  
p8       	-61.9155528043671 
p9       	91.7605255786753  
p10      	0.644726102395323 

For the fourth row data:

Sum Squared Error (SSE): 0.130148263082422
Root of Mean Square Error (RMSE): 0.0228622787026929
Correlation Coef. (R): 0.999992227866027
R-Square: 0.999984455792459

Parameter	Best Estimate     
---------	-------------     
p1       	-0.620509642952949
p2       	-0.831361297046103
p3       	-0.158884676824082
p4       	2.7433055327425   
p5       	0.22148229931868  
p6       	16.8000287134801  
p7       	138.373410204698  
p8       	152.0209560771    
p9       	34.3737366450204  
p10      	0.545395269938482 

For the fifth row data:

Sum Squared Error (SSE): 0.0393241104660582
Root of Mean Square Error (RMSE): 0.0125669469037695
Correlation Coef. (R): 0.999997424563047
R-Square: 0.999994849132727

Parameter	Best Estimate     
---------	-------------     
p1       	-0.599320668258529
p2       	-0.44341678007072 
p3       	-0.279977917807812
p4       	3.70706464742905  
p5       	-7.18206037822055 
p6       	23.1054679279924  
p7       	127.864208391053  
p8       	100.743337230697  
p9       	62.8901303777835  
p10      	0.679481485801022 

[Hidd_1]

Why excatley this fitting function, is there a specific reason?

[Alex Sha]

It likes Neural Network fitting, by using sigmoid transfer function in hide layer, and linear transfer function in output layer, like the picture below:

the corresponded function will then be:

[Hidd_1]

Thnak a lot Alex!

I've tried to implement your algorithm using your objective function, but I got bad results:

clc
clear all
close all

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

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

I coudn't achieve your result, can you please check what went wrong, please!

[Alex Sha]

GA only has the so-called global optimization capability in theory, but in practice it will be far from the same, even Matlab's global optimization toolbox, the results are often unsatisfactory。