[R] How to create unique factor from two factors? + Boostrap Q

[kjetil@entelnet.bo]

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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