Matlab fit to three dimensions function

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standy
standy il 28 Ago 2019
Commentato: the cyclist il 28 Ago 2019
Hi
I got three input vectors: x1, x2, x3 and output vectory y. How do i fit to my custom function?
y=a0+a1*x1+a2*x2+a3*x3+a11*x1^2+a22*x2^2+a33*x3^2+a12*x1*x2+a13*x1*x3+a23*x2*x3
Obviously I want to get parameters a1, a2, ...
For two variables (x1, x2) i can do it like:
f = fit([x1, x2], y, 'poly33');
But I'm struggling to do that for the function above.
Any help appreciated.
  2 Commenti
the cyclist
the cyclist il 28 Ago 2019
Do you also have the Statistics and Machine Learning Toolbox available, or only the Curve Fitting Toolbox?
standy
standy il 28 Ago 2019
Yes, i got this toolbox.

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Risposte (2)

Bruno Luong
Bruno Luong il 28 Ago 2019
Modificato: Bruno Luong il 28 Ago 2019
n = length(y);
A = [ones(n,1) x1(:) x2(:) x3(:) x1(:).^2 x2(:).^2 x3(:).^2 x1(:).*x2(:) x1(:).*x3(:) x2(:).*x3(:)] \ y(:)

the cyclist
the cyclist il 28 Ago 2019
Modificato: the cyclist il 28 Ago 2019
I would do it like this:
% Set the random number generator seed, for reproducibility
rng default
% Create some random data
N = 1000;
x1 = randn(N,1);
x2 = randn(N,1);
x3 = randn(N,1);
% Create a response variable with known coefficients, and some noise
y = 2 + 3*x1 + 5*x2 + 7*x3 ...
+ 11*x1.^2 + 13*x2.^2 + 17*x3.^2 ...
+ 19*x1.*x2 + 23*x1.*x3 + 29*x2.*x3 ...
+ 31*randn(N,1);
% Fit a quadratic model
mdl = fitlm([x1 x2 x3],y,'quadratic')
% % The above is equivalent to the following model, written out in full Wilkinson notation
% mdl = fitlm([x1,x2,x3],y,'y ~ x1 + x2 + x3 + x1^2 + x2^2 + x3^2 + x1:x2 + x1:x3 + x2:x3');
Almost all of this code is me creating the data, to illustrate everything. Since you have the data already, you should only need
mdl = fitlm([x1 x2 x3],y,'quadratic')
The resulting model object, mdl, has methods for lots of information about the model fit.
  4 Commenti
standy
standy il 28 Ago 2019
Ahhh
Makes sense now. I got the output like:
Linear regression model:
y ~ 1 + x1*x2 + x1*x3 + x2*x3 + x1^2 + x2^2 + x3^2
Estimated Coefficients:
Estimate SE tStat pValue
__________ __________ ________ _______
(Intercept) 1.4267 1.1233 1.2701 0.23281
x1 -0.015338 0.024769 -0.61922 0.54962
x2 -0.90665 3.3137 -0.27361 0.78995
x3 0 0 NaN NaN
x1:x2 0.0031125 0.012077 0.25772 0.80185
x1:x3 0 0 NaN NaN
x2:x3 0.765 3.0553 0.25038 0.80736
x1^2 6.8008e-05 0.00011935 0.56982 0.58137
x2^2 0 0 NaN NaN
x3^2 0 0 NaN NaN
Does it mean that a0=1, a1=-0.015338 etc?

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