Re: [R] RandomForest question

From: Weiwei Shi <helprhelp_at_gmail.com>
Date: Fri 22 Jul 2005 - 01:16:57 EST

Hi,
I found the following lines from Leo's randomForest, and I am not sure if it can be applied here but just tried to help:

mtry0 = the number of variables to split on at each node. Default is the square root of mdim. ATTENTION! DO NOT USE THE DEFAULT VALUES OF MTRY0 IF YOU WANT TO OPTIMIZE THE PERFORMANCE OF RANDOM FORESTS. TRY DIFFERENT VALUES-GROW 20-30 TREES, AND SELECT THE VALUE OF MTRY THAT GIVES THE SMALLEST OOB ERROR RATE. mdim is the number of predicators.

HTH, weiwei

On 7/21/05, Liaw, Andy <andy_liaw@merck.com> wrote:
> > From: Arne.Muller@sanofi-aventis.com
> >
> > Hello,
> >
> > I'm trying to find out the optimal number of splits (mtry
> > parameter) for a randomForest classification. The
> > classification is binary and there are 32 explanatory
> > variables (mostly factors with each up to 4 levels but also
> > some numeric variables) and 575 cases.
> >
> > I've seen that although there are only 32 explanatory
> > variables the best classification performance is reached when
> > choosing mtry=80. How is it possible that more variables can
> > used than there are in columns the data frame?
>
> It's not. The code for randomForest.default() has:
>
> ## Make sure mtry is in reasonable range.
> mtry <- max(1, min(p, round(mtry)))
>
> so it silently sets mtry to number of predictors if it's too large.
> As an example:
>
> > library(randomForest)
> randomForest 4.5-12
> Type rfNews() to see new features/changes/bug fixes.
> > iris.rf = randomForest(Species ~ ., iris, mtry=10)
> > iris.rf$mtry
> [1] 4
>
> I should probably add a warning in such cases...
>
> Andy
>
>
> > thanks for your help
> > + kind regards,
> >
> > Arne
> >
> >
> >
> >
> > [[alternative HTML version deleted]]
> >
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> >
> >
> >
>
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-- 
Weiwei Shi, Ph.D

"Did you always know?"
"No, I did not. But I believed..."
---Matrix III

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Received on Fri Jul 22 01:50:28 2005

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