Matlab fit to three dimensions function
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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 Comments
the cyclist
on 28 Aug 2019
Do you also have the Statistics and Machine Learning Toolbox available, or only the Curve Fitting Toolbox?
standy
on 28 Aug 2019
Answers (2)
Bruno Luong
on 28 Aug 2019
Edited: Bruno Luong
on 28 Aug 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
on 28 Aug 2019
Edited: the cyclist
on 28 Aug 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 Comments
standy
on 28 Aug 2019
the cyclist
on 28 Aug 2019
Edited: the cyclist
on 28 Aug 2019
Unless I misunderstood, you want all possible linear terms and squared terms. That's a quadratic polynomial. Notice that right underneath, in some commented-out code, I show the full notation if I had written out all terms.
standy
on 28 Aug 2019
the cyclist
on 28 Aug 2019
Yes
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