Apply functions along "layers" of a data matrix

Hello

How can I apply functions along "layers" of a data matrix?

Example:

daf <- data.frame(
?'id' = rep(1:5, 3),
?matrix(1:60, nrow=15, dimnames=list( NULL, paste('v', 1:4, sep='') )),
?rep = rep(1:3, each=5)
)

The data frame "daf" contains 3 repetitions/layers (rep) of 4 variables of 5
persons (id). For some reason, I want to calculate various statistics (e.g.,
mean, median) *along* the repetitions. The "mean" calculation, for example,
would produce the means of daf[1, 'v1'] *along* the 3 repetition:

(daf[1, 'v1'] + daf[6, 'v1'] + daf[11, 'v1']) / 3

That is to say, each of the calculations would result in a data frame with 4
variables (and the id) of the 5 persons:

?id v1 v2 v3 v4
1 ?1 ?6 21 36 51
2 ?2 ?7 22 37 52
3 ?3 ?8 23 38 53
4 ?4 ?9 24 39 54
5 ?5 10 25 40 55

Currently, I do this in a loop, but I was wondering about a quick and
ressource-friendly way to achieve this?

Thanks
*S*

--
Sascha Vieweg, saschaview at gmail.com

daf <- data.frame(
  'id' = rep(1:5, 3),
  matrix(1:60, nrow=15, dimnames=list( NULL, paste('v', 1:4, sep='') )),
  rep = rep(1:3, each=5)
)

Hello

How can I apply functions along "layers" of a data matrix?

Example:

daf <- data.frame(
 'id' = rep(1:5, 3),
 matrix(1:60, nrow=15, dimnames=list( NULL, paste('v', 1:4,  
sep='') )),
 rep = rep(1:3, each=5)
)

The data frame "daf" contains 3 repetitions/layers (rep) of 4  
variables of 5 persons (id). For some reason, I want to calculate  
various statistics (e.g., mean, median) *along* the repetitions. The  
"mean" calculation, for example, would produce the means of daf[1,  
'v1'] *along* the 3 repetition:

(daf[1, 'v1'] + daf[6, 'v1'] + daf[11, 'v1']) / 3

That is to say, each of the calculations would result in a data  
frame with 4 variables (and the id) of the 5 persons:

 id v1 v2 v3 v4
1  1  6 21 36 51
2  2  7 22 37 52
3  3  8 23 38 53
4  4  9 24 39 54
5  5 10 25 40 55

Currently, I do this in a loop, but I was wondering about a quick  
and ressource-friendly way to achieve this?

Thanks
*S*

-- 
Sascha Vieweg, saschaview at gmail.com