Matlab fit to three dimensions function

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standy
standy 2019 年 8 月 28 日
コメント済み: the cyclist 2019 年 8 月 28 日
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 件のコメント
the cyclist
the cyclist 2019 年 8 月 28 日
Do you also have the Statistics and Machine Learning Toolbox available, or only the Curve Fitting Toolbox?
standy
standy 2019 年 8 月 28 日
Yes, i got this toolbox.

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回答 (2 件)

Bruno Luong
Bruno Luong 2019 年 8 月 28 日
編集済み: Bruno Luong 2019 年 8 月 28 日
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 2019 年 8 月 28 日
編集済み: the cyclist 2019 年 8 月 28 日
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 件のコメント
standy
standy 2019 年 8 月 28 日
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?
the cyclist
the cyclist 2019 年 8 月 28 日
Yes

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