[R] More on for() Loop...

[ripley@stats.ox.ac.uk]

On Sat, 8 Jun 2002, Frank E Harrell Jr wrote:

> On Sat, 8 Jun 2002 22:44:07 +1200 (NZST)
> Ko-Kang Kevin Wang <kwan022 at stat.auckland.ac.nz> wrote:
>
> > Hi,
> >
> > Say I want to do something like fitting 10 different sized trees with
> > rpart() function.  The only modification I need to do is to set 10
> > different cp's, which I have in a vector called foo.
> >
> > Can I do something like:
> >
> > for(i in 1:10) {
> >   rpart(y ~ ., cp = foo[i], data = mydata)
> > }
> >
> > My problem is, I wish to save the 10 rpart objects into 10 different
> > names, my.rpart1 ~ my.rpart10, for example.  But I'm not sure how to do
> > this...
> >
>
> fits <- vector('list',10)
> names(fits) <- format(foo)  # optional - will label list components
> for(i in 1:10) fits[[i]] <- rpart(...)
>
> Fetch the ith fit by using fits[[i]] (or fits[['formatted cp value']]).

Two refinements:

1) Use substitute() to get the call information in the fit as a value
and not fun[i].  The Programmer's Niche in the next R-news is all about
that, or see `S Programming'.

2) You only need to call prune.rpart to change the amount of pruning
in an rpart fit.  Even if you didn't, I would advise setting the random
seed so you got the same cross-validation partition both times.

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272860 (secr)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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