[R] access to the current element of lapply

[Marc Schwartz]

On May 13, 2009, at 9:12 AM, Martial Sankar wrote:

>
> Dear All,
>
> I would like to use the 'split' function on the dataframe elements  
> contained in a list L.
>
> For example :
>
>> (df <- data.frame(cbind(c(rep('A',2), rep('B',2)), rep(1:4))))
>  X1 X2
> 1  A  1
> 2  A  2
> 3  B  3
> 4  B  4
>> (L<-split(df, df$X1))
> $A
>  X1 X2
> 1  A  1
> 2  A  2
>
> $B
>  X1 X2
> 3  B  3
> 4  B  4
>
> Now, I would like to split EACH data frame, ie, according to column  
> 2(X2).
>
>> lapply(L, split, df$X2)
>
> $A
> $A$`1`
>  X1 X2
> 1  A  1
>
> $A$`2`
>  X1 X2
> 2  A  2
>
> $A$`3`
> [1] X1 X2
> <0 rows> (or 0-length row.names)
>
> $A$`4`
> [1] X1 X2
> <0 rows> (or 0-length row.names)
>
>
> $B
> $B$`1`
>  X1 X2
> 3  B  3
>
> $B$`2`
>  X1 X2
> 4  B  4
>
> $B$`3`
> [1] X1 X2
> <0 rows> (or 0-length row.names)
>
> $B$`4`
> [1] X1 X2
> <0 rows> (or 0-length row.names)
>
>
> Warning messages:
> 1: In split.default(seq_len(nrow(x)), f, drop = drop, ...) :
>  data length is not a multiple of split variable
> 2: In split.default(seq_len(nrow(x)), f, drop = drop, ...) :
>  data length is not a multiple of split variable
>
>
>
> I works but it's dirty.
> How  could I do it properly, without warnings and 0 rows data frame  
> in output ?
> I thought accessing to the current element of 'lapply' to recuperate  
> the vector of the column 2 would work.
> i.e:
>
> lapply(L,split, L[[current]][,2])
>
>
> Is there a way to do something like that in R ?
>
>
> Thanks in advance !
>
> - Martial

# Split on BOTH columns and drop unused levels
L <- split(df, list(df$X1, df$X2), drop = TRUE)

 > L
$A.1
   X1 X2
1  A  1

$A.2
   X1 X2
2  A  2

$B.3
   X1 X2
3  B  3

$B.4
   X1 X2
4  B  4


Is that what you want?

HTH,

Marc Schwartz