[R] mob(party) formula question
Birgitle
birgit.lemcke at systbot.uzh.ch
Wed Aug 13 15:17:46 CEST 2008
Many thanks for your answer and the code that you offered me.
I get this error message after calling mob (look at my given example).
I guess it has something to do with the missings?
The iris example works also fine for me.
Sorry that I am not enough into statistics to really understand the
following:
Achim Zeileis wrote:
>
>
> .....
> For the variables for which a linear specification makes sense (at least
> in each component) then you should include them for modeling. And those
> variables for which it is not clear a priori what a useful parametric
> specification would be should be used as partitioning variables.
> ...
>
>
What do you mean with "linear specification"? I would be very happy if you
could explain.
Thanks again
B.
Achim Zeileis wrote:
>
> On Wed, 13 Aug 2008, Birgitle wrote:
>
>> I try tu use mob() with my data.frame ('data.frame': 288 obs. of 81
>> variables; factors, numerics and ordered factors)
>> My response is a binary variable and I should use for modelling a
>> logistic
>> regression (family=binomial).
>>
>> I read in the "MOB" Vignette that I could use a formula like this if I
>> would
>> like to have only partitioning variables apart from the response.
>>
>> Test.mob<-mob(Resp~1|Var1+Var2+...., data=dataframe, model=glinearModel,
>> family=binomial())
>
> This works for me. Considering an example that is easily reproducible:
> classifying just two (out of three) species in the iris data.
>
> iris2 <- iris[-(1:50),]
> iris2$Species <- factor(iris2$Species)
> mb <- mob(Species ~ 1 | Petal.Length + Petal.Width + Sepal.Length +
> Sepal.Width, data = iris2, model = glinearModel, family = binomial())
>
>
>
>
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