Ted,
Brilliant explanation (as usual)
I'm back in school, just starting on a post-graduate degree in stats so
the help is really appreciated.
Now, I have a slightly trickier question about the same model.
I've seen more than one way to get "values" out of the glm model.
i.e. If we're looking at the 10th item in the dataset:
note: "m" is the model
fitted(m)[10]
predict(m,dataset[10,])
Give me different results. From my data, I get the following real results:
> predict(m,data[100,])
100
7.727999
> fitted(m)[100]
179
3956.637
From my understanding, the exp of the prediction should be equal to the
fitted value. Here it is not. I don't understand why. Any insight?
-N
On 3/2/10 12:47 AM, (Ted Harding) wrote:
On 02-Mar-10 08:02:27, Noah Silverman wrote:
Hi,
I'm just learning about poison links for the glm function.
One of the data sets I'm playing with has several of the
variables as factors (i.e. month, group, etc.)
When I call the glm function with a formula that has a factor
variable, R automatically converts the variable to a series of
variables with unique names and binary values.
For example, with this pseudo data:
y v1 month
2 1 january
3 1.4 februrary
1.5 6.3 february
1.2 4.5 january
5.5 4.0 march
I use this call:
m<- glm(y ~ v1 + month, family="poisson")
R gives me back a model with variables of
Intercept
v1
monthJanuary
monthFebruary
monthMarch
I'm concerned that this might be doing some strange things
to my model.
Can anyone offer some enlightenment?
Thanks!
The creation of auxiliary variables is the way to incorporate
a factor variable into a model. These are usually called
"dummy variables", and are essentially indicator variables.
Your data above would correspond to variables I (for Intercept),
J (for January), F (for February) and M (for March) in addition
to the other variables y and v1 as below:
y v1 I J F M # month
2 1 1 1 0 0 # january
3 1.4 1 0 1 0 # februrary
1.5 6.3 1 0 1 0 # february
1.2 4.5 1 1 0 0 # january
5.5 4.0 1 0 0 1 # march
The linear predictor L in the model for y would then be
L = a*I + b*v1 + c1*J + c2*F + c3*J
evaluated arithmetically; e.g. for row 2 of the data it is
a + b*1.4 + c2
However, as given, J + F + M = I, so there is redundancy in
the variables, since there are only three independent values
there (not so if you exclude the Intercept using a model
formula y ~ v1 + month - 1), so R will provide estimates
which are computed in terms of some pattern of differences
between these four variables called contrasts. Different
patterns of difference present different representations
of the three independent aspects.
There are many different kinds of contrasts available.
One of these will be chosen as default by R (depending in
particular on whether the factor variable is being used
as an ordered factor or an unordered factor). See ?contrasts
for an outline of what is there, ?contrast for more detail,
and look at the help for particular contrasts such as
?contr.helmert, ?contr.poly, ?contr.sum, ?contr.treatment.
After all that: No, R is not doing strange things to your model!
ted.
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E-Mail: (Ted Harding)<ted.hard...@manchester.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 02-Mar-10 Time: 08:47:11
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