Relation or Pattern between curves

Question

Hidd_1 2023-5-17

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评论： Star Strider 2023-6-11

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!

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Hidd_1 2023-5-20

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

Image Analyst 2023-5-20

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.

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Answer 1

Star Strider 2023-5-17

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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))));

.

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Star Strider 2023-5-18

在 MATLAB Online 中打开

Data.mat

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.

.

Star Strider 2023-5-18

在 MATLAB Online 中打开

Data.mat

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 2023-5-19

在 MATLAB Online 中打开

Data.mat

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.

.

Star Strider 2023-5-19

在 MATLAB Online 中打开

Data.mat

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