[R] sapply behavior

[Prof Brian Ripley]

On Mon, 27 Sep 2004, Liaw, Andy wrote:

> The problem is that temp2 is a data frame, and the function you are
> sapply()ing to returns a row from a data frame.  A data frame is really a
> list, with each variable corresponding to a component.  If you extract a row
> of a data frame, you get another data frame, not a vector, even if all
> variables are the same type.  sapply() can really `simplify' the right way
> if it's given a vector (or matrix).  Consider:
> 
> > str(temp2)
> `data.frame':   6 obs. of  4 variables:
>  $ A1: int  5 2 4 6 1 3
>  $ A2: int  5 2 4 6 1 3
>  $ A3: int  5 2 4 6 1 3
>  $ A4: int  5 2 4 6 1 3
> > temp2 <- as.matrix(temp2)
> > str(temp2)
>  int [1:6, 1:4] 5 2 4 6 1 3 5 2 4 6 ...
>  - attr(*, "dimnames")=List of 2
>   ..$ : chr [1:6] "1" "2" "3" "4" ...
>   ..$ : chr [1:4] "A1" "A2" "A3" "A4"
> > str(sapply(1:6,function(x){xmat<-temp2[temp2[,1]==x,,drop=F]; xmat[1,]}))
>  int [1:4, 1:6] 1 1 1 1 2 2 2 2 3 3 ...
>  - attr(*, "dimnames")=List of 2
>   ..$ : chr [1:4] "A1" "A2" "A3" "A4"
>   ..$ : NULL
> 
> (The is.matrix() function probably just check whether the dim attribute is a
> vector of length 2, and not a data frame (as it says in ?is.matrix).  The
> newtemp2 object you get is a list with 24 components, each component is a
> vector of one integer, and has a dim attribute of c(4, 6).  Not what I would
> call a matrix.)

That *is* a matrix, though, and is useful for lists of length greater than
one.  A matrix in R is just a vector with a dim attribute (and length the
product of the dims's), so as well as any of the atomic vectors it can be
a generic vector aka list.

-- 
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)
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