[R] Subsetting a list

[Prof Brian Ripley]

On Fri, 21 Oct 2005, Martin Maechler wrote:

>>>>>> "PaCo" == Patrick Connolly <p.connolly at hortresearch.co.nz>
>>>>>>     on Fri, 21 Oct 2005 13:26:08 +1300 writes:
>
>    PaCo> On Wed, 19-Oct-2005 at 05:09PM +0200, Martin Maechler wrote:
>    PaCo> |> Lists can have 'dim' attributes and hence be treated as arrays;
>    PaCo> |> Note that this is pretty rarely used and not too well supported
>    PaCo> |> by some tools, one could say even 'print()' :
>    PaCo> |>
>    PaCo> |> > set.seed(0); L0 <- L <- lapply(rpois(12, lambda=3), seq); dim(L) <- 3:4; L
>    PaCo> |>      [,1]      [,2]      [,3]      [,4]
>    PaCo> |> [1,] Integer,5 Integer,3 Integer,5 Integer,3
>    PaCo> |> [2,] Integer,2 Integer,5 Integer,6 1
>    PaCo> |> [3,] Integer,2 Integer,2 Integer,4 Integer,2
>
>    PaCo> for an occasion such as this, it can be clearer to do:
>
>    >> transform(L)
>    PaCo> X1            X2               X3      X4
>    PaCo> 1 1, 2, 3, 4, 5       1, 2, 3    1, 2, 3, 4, 5 1, 2, 3
>    PaCo> 2          1, 2 1, 2, 3, 4, 5 1, 2, 3, 4, 5, 6       1
>    PaCo> 3          1, 2          1, 2       1, 2, 3, 4    1, 2
>
> wow!
> Thank you Patrick!
>
> Actually, that's identical to
>
>   data.frame(L)
>
> which I honestly wouldn't have had expected to work for L.

or even as.data.frame(L).

Now replace lambda=3 by lambda=30 and see which one is legible (let alone 
clearer).  I prefer the internal matrix printing in almost all cases.

It works for data frames because print.data.frame uses format.data.frame 
which uses format, and that's how format works on a 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)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595