optimization data fitting error
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I want find a model fit for a set of data. I'm just adapting the example code for my model, and there is just one line (my model equation) which is causing an error. Please could somebody tell me what about this line needs adjusting?
The error message:
"Error using lsqcurvefit (line 251)
Function value and YDATA sizes are not equal.
Error in curvefittingpractice2 (line 39)
[x,resnorm,~,exitflag,output] = lsqcurvefit(F,x0,t,y)"
Code:
Data = ...
[0.0000 5.8955
0.1000 3.5639
0.2000 2.5173
0.3000 1.9790
0.4000 1.8990
0.5000 1.3938
0.6000 1.1359
0.7000 1.0096
0.8000 1.0343
0.9000 0.8435
1.0000 0.6856
1.1000 0.6100
1.2000 0.5392
1.3000 0.3946
1.4000 0.3903
1.5000 0.5474
1.6000 0.3459
1.7000 0.1370
1.8000 0.2211
1.9000 0.1704
2.0000 0.2636];
plot(t,y,'ro')
title('Data points')
%F = @(x,xdata)x(1)*exp(-x(2)*xdata); --for this I get no error, for the
%following line I get an error message
F = @(x,xdata)x(1)*exp(x(2)/xdata).^x(3);
x0 = [1 1 1 0];
[x,resnorm,~,exitflag,output] = lsqcurvefit(F,x0,t,y)
hold on
plot(t,F(x,t))
hold off
3 个评论
Alex Sha
2020-5-3
Hi, Em, in your function: F = @(x,xdata)x(1)*exp(x(2)/xdata).^x(3);
note the part: x(2)/xdata, the first data value for xdata is 0, so this will cause a math logical error (divided by zero)
Star Strider
2020-5-3
... as I wrote two hours earlier.
Em
2020-5-3
采纳的回答
Star Strider
2020-5-3
For the second ‘F’, as written, returns a row vector rather than a column vector. The solution is to use element-wise division:
F = @(x,xdata)x(1)*exp(x(2)./xdata).^x(3);
(The first ‘F’ returns a column vector as written, so no problem.)
However the other problem is that the second ‘F’ produces an infinite result with the first value of ‘t’, since it is 0. One possible way to address that is to eliminate the first row of ‘Data’, however the fit is inferior to the first ‘F’ with the full data set.
I would go with the first ‘F’ and be done with it.
12 个评论
Em
2020-5-3
Star Strider
2020-5-3
If you use the first ‘F’, there should be no problems. The problems occur with the second ‘F’.
Em
2020-5-3
Star Strider
2020-5-3
The problem is the division-by-zero. There is no way to get around that.
The fit is much better for the first ‘F’ with a y-offset parameter:
F = @(x,xdata)x(1)*exp(-x(2)*xdata) + x(3);
A better fit is likely not possible without a mathematical model of the process that created those data. If available, it would be best to use it for the objective function.
Em
2020-5-3
Star Strider
2020-5-3
My pleasure!
I came up with a ‘sort of’ solution that avoids the divide-by-zero problem by using a logic test to substitute a suitably small number for 0 in the independent variable within the ‘F’ function. This is definitely not a perfect solution, however it has the advantage of producing a reasonable fit even in the presence of 0 values in the independent variable, and does not require prior editing of the data set.
Try this:
F = @(x,xdata) (x(1)*exp(x(2)./(xdata.*(xdata ~= 0)+(0.1).*(xdata==0))).^x(3));
This logic test for ‘xdata’ does the substitution only when the independent variable is 0:
(xdata.*(xdata ~= 0)+(0.1).*(xdata==0))
Experiment with it to get different results.
Em
2020-5-3
Star Strider
2020-5-3
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.
Em
2020-5-5
Star Strider
2020-5-5
My code fits the original data because that was all it was designed to do.
Your model does not even begin to describe the data in that file. It is simply the wrong model, and nothing can be done to fix it. I even tried to estimate the parameters with a genetic algorithm and failed, although most of the time, the genetic algorithm can fit an appropriate model to appropriate data.
I will delete my Answer in a few hours.
Good luck with whatever it is that you're doing!
Em
2020-5-5
Star Strider
2020-5-5
As always, my pleasure!
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