How to use use an anonymous function to fit data based on the fittype and fit functions.
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Hi there,
I have some sort of data point and I want to use an anonymous function to fit this data based on the fittype and fit functions.
Also, the function is look like this:
F(x)=a*x^2+b*x+c+d*diff(x^2)+e*diff(x)
Any suggestions?
Thanks
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採用された回答
William Rose
2022 年 6 月 24 日
I think the manual page examples for this are pretty good.
a=1; b=2; c=-1; d=-2; e=3;
xdata=-3:.03:3;
%ydata=a*xdata(2:end)+b*xdata(2:end).^2+c+d*diff(xdata)+e*diff(xdata).^2+randn(1,length(xdata)-1);
fun=@(p,x) p(1)*x(2:end)+p(2)*x(2:end).^2+p(3)+p(4)*diff(x)+p(5)*diff(x).^2;
yclean=fun([a,b,c,d,e],xdata);
ydata=yclean+randn(1,length(xdata)-1);
p0=[1,1,1,1,1];
p = lsqcurvefit(fun,p0,xdata,ydata)
yfit=fun(p,xdata);
plot(xdata(2:end),yclean,'-r',xdata(2:end),ydata,'rx',xdata(2:end),yfit,'-b')
legend('Clean Data','Noisy Data','Best Fit')
The fit to the data is excellent, and the two curves are so close that they overlap, but the fitted parameters do not match the original parameters. This is because the input diff(x) is a constant, and therefore is not independent of the "c" term, and the input diff(x^2) is a straight line, and therefore not indepndent of the "a" term.
Try this. Good luck.
7 件のコメント
William Rose
2022 年 6 月 27 日
編集済み: William Rose
2022 年 6 月 27 日
[edit: correct typographical errors]
@payam samadi, I would also ask my student to tell me the p-values and 95% confidence intervals for each fitted parameter. To get this info easily, and to do the whole regression easily (with the potential to add stepwise regression), use the Statistics & Machine Learning Toolbox. See how easy it is:
load('xdata'); load('ydata');
%Next: Create array with the four predictor variables as columns.
X=[xdata(2:end),xdata(2:end).^2,diff(xdata),diff(xdata).^2];
y=ydata(2:end);
mdl1=fitlm(X,y) %fit the linear model
coefCI(mdl1)
The values intercept,x1,x2,x3,x4 correspond to c,b,a,d,e in your original model. If you look back, you will see that the parameter estimates here match the values obtained above, when we used lsqcurvefit(). Because we used the Statistics toolbox, we get the p values and confidence intervals (with coefCI()) for each fitted parameter. The p-values show that all five fitted parameters are highly significant.
We can do a stepwise linear regression to see if all four predictors are useful in addition to the intercept. Given the highly significant p values above, I expect that all four will turn out to be significant.
%If you don't include 'Upper','linear' below, stepwiselm() will also try to fit
%interaction terms in the model. YOu did not request those.
mdl2=stepwiselm(X,y,'Upper','linear')
The result above shows that stepwiselm() keeps all four predictor variables in the model - so you are justified in including all four, plus the intercept.
Try it.
その他の回答 (2 件)
Walter Roberson
2022 年 6 月 24 日
There is no chance of fitting that function.
fit() and fittype() pass in numeric data, and diff() of numeric data is always smaller than the original data. Therefore your a*x^2+b*x+c is an incompatible size to be added to d*diff(x^2)+e*diff(x)
It might make sense to rewrite in terms of gradient()
Does diff(x) happen to be constant (points are equally spaced)? If so then I suspect you can rewrite the formulas for more efficient fitting
2 件のコメント
Walter Roberson
2022 年 6 月 25 日
fit() needs to work mathematically when the order of the input vectors is changed. fitting x, y needs to give you the same sum of squared error as fitting x(P) y(P) gives, where P is a permutation of the indices.
If you were to reorder the inputs, e*diff(x) might balance out, but d*diff(x^2) cannot balance out.
Also could you confirm for us that you mean diff(x.^2) and not diff(x).^2 ?
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