[R] How to create unique factor from two factors? + Boostrap Q
kjetil@entelnet.bo
kjetil at entelnet.bo
Sun Nov 9 15:51:57 CET 2003
On 9 Nov 2003 at 13:29, Prof Brian Ripley wrote:
> Factor3 <- factor(unclass(Factor1) + nlevels(Factor1)*(unclass(Factor2)-1))
>
Cannot this be done even easier by calculating the interaction?
> a <- factor(rep(1:3,rep(3,3)))
> b <- factor(rep(1:3,3))
> ab <- a:b
> ab
[1] 1:1 1:2 1:3 2:1 2:2 2:3 3:1 3:2 3:3
Levels: 1:1 1:2 1:3 2:1 2:2 2:3 3:1 3:2 3:3
Kjetil Halvorsen
> will give you the unique combinations, not labelled as you do but then I
> don't think you need that.
>
> On Sun, 9 Nov 2003, Scott Norton wrote:
>
> > This might be easy but I'm very new to R and this question doesn't seem to
> > have any nice keywords that bring up relevant search results when I search
> > the CRAN search engine. Therefore, I'll plead (as I have in the recent
> > past) Newbie status.
> >
> >
> >
> > I have a data frame with two factors (Factor 1 and 2) which together specify
> > another unique level. I want to create a third factor in the data frame
> > that captures this uniqueness.
> >
> > For example, say I had dataframe, Df, with Factors, 1 and 2. I want to
> > create Factor 3 and add it to my Df dataframe.
> >
> > i.e.
> >
> > Df dataframe: WANT TO
> >
> > Row# Factor1 Factor2 CREATE THIS: Factor 3 Data
> >
> > 1 1 1 1 23
> >
> > 2 1 2 2 43
> >
> > 3 1 2 2 19
> >
> > 4 1 2 2 11
> >
> > 5 1 4 3 3
> >
> > 6 1 4 3 13
> >
> > 7 3 1 4 52
> >
> > 8 3 1 4 12
> >
> > 9 3 1 4 9
> >
> > 10 3 3 5 21
> >
> > 11 3 3 5 43
> >
> >
> > 12 4 1 6 32
> >
> > 13 4 1 6 18
> >
> > 14 4 2 7 52
> >
> > 15 4 2 7 21
> >
> >
> >
> >
> > and of course, I'm trying to create Factor 3 without loops..
> >
> >
> >
> > My end goal here (which I add because maybe I don't need to create Factor 3
> > (although I'm still curious)), is to bootstrap "sample" Factor 3. I want to
> > repeatedly grab, say, 3 levels of Factor 3, and take the mean of those
> > levels (e.g. say in my first bootstrap sample, I grab levels 2,4, and 7 from
> > Factor 3, then I want to take the mean of rows, 2,3,4,7,8,9,14,15). Of
> > course, each sample from Factor 3 for my bootstrap will most likely have a
> > differing number of rows since my experiment is not balanced. I'm not sure
> > if this is an issue yet when I try to implement the "boot" function in R (I
> > haven't gotten to that point yet).
>
> The boot package will easily do this for you.
>
> --
> 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 272866 (PA)
> Oxford OX1 3TG, UK Fax: +44 1865 272595
>
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