Using fitglm for the generalized linear model in matlab

4 ビュー (過去 30 日間)
MINA
MINA 2017 年 1 月 11 日
回答済み: Philip 2018 年 5 月 3 日
I have two question regarding fitglm and glmfit.
1- After doing some research, I realized that t is better to use fitglm instead of glmfit. The reason is that the output of the first function is more complete. I wanted to know if that is correct!
2- Instead of giving the model I have the design matirx which I would like to give as an input to fitglm. Here is my code
x=rand(3000,1);
X=[ones(size(x)) ,x, x.^2, x.^3];
a=5;
b=3.4;
c=-3;
d=20;
func=[d a b c]*X';
fitglm(X,fun);
This is what I get as the coefficient
Estimate SE tStat pvalue
(Intercept) 20 2.21336782492520e-15 9.03600376529188e+15 0
x1 0 0 NaN NaN
x2 4.99999999999998 1.46537110155627e-14 341210495736530 0
x3 3.40000000000005 2.14250387190849e-14 158692828730872 0
x4 -3.00000000000002 1.10439613833154e-14 -271641659715711 0
I don't understand why it has five Coefficients which one of them is zero. And if I remove ones from the design matrix I don't get any Intercept.
Additional, it also gives the NumPredictors =4, namely x1 x2 x3 x4. I am assuming that it uses this number for computing the AIC but isn't the NumPredictors=3?
Thanks

回答 (1 件)

Philip
Philip 2018 年 5 月 3 日
The fitglm function assumes an intercept, so your design matrix is actually: [1s, 1s, x, x.^2, x.^3].
You need to turn the intercept parameter to false in order to manually manipulate the constants evaluation. The fitglm function is newer (r2013b I think?) and does have more stuff baked into it, so I would recommend using it over glmfit.
The older glmfit function has the same parameter except I think it was called 'constant'. I've only recently started converting much of my code from glmfit to fitglm and am finding the new baked in calculations to be hugely helpful.

カテゴリ

Help Center および File ExchangeDimensionality Reduction and Feature Extraction についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by