Problem with function handle for Basis function in fitrgp
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Stefan Hövel
el 27 de En. de 2020
Comentada: Adam Danz
el 21 de Nov. de 2020
I like to use a custom Basis function for fitrgp.
beta0 = [1.5,2.0];
hfcn = @(X) (beta0(1).*(X).^beta0(2));
This should be a powerMean-function.
If I call:
gprMdl1 = fitrgp(x,y,'BasisFunction',hfcn,'Beta',beta0)
with dummy-data from:
load(fullfile(matlabroot,'examples','stats','gprdata2.mat'))
I get
gprMdl1.Beta = 0.0019
Why is gprMdl1.Beta not 1x2 like beta0? Is it posible to optimize custom parameters in Basis function (like beta0)?
2 comentarios
Adam Danz
el 28 de En. de 2020
"Why is gprMdl1.Beta not 1x2 like beta0?"
That basis function will return as many values as provided in the input X. X is multiplied by 1 value (beta0(1)) and added to one value (beta0(2)). So if X is only 1 value, the output should only be one value. If x is 1x5, the output will be 1x5; if x is 3x4, so is the output.
Adam Danz
el 21 de Nov. de 2020
>gprMdl1.Beta is not 3x1, there is still only one beta. Could you help me check this?
Check this out.....
% Your function
beta0 = [1;2;3];
hfcn = @(X) ([ones(size(X,1),1),X,X.^2]*beta0);
% Test some inputs-outputs
hfcn(1)
hfcn([1;2;3])
Respuesta aceptada
Ridwan Alam
el 28 de En. de 2020
Editada: Ridwan Alam
el 28 de En. de 2020
load(fullfile(matlabroot,'examples','stats','gprdata2.mat'))
In this case, your x is n x 1;
hfcn returns H, which has size n x 1; i.e. p = 1;
So if you want to provide 'Beta' to fitrgp(), it should be of size p x 1, i.e 1 x 1.
fitrgp() is only using your beta0(1) as Adam mentioned in the comment.
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