David L Carlson dcarlson at tamu.edu
Thu Sep 25 23:05:17 CEST 2014

```Another approach

fun <- function(i, dat=x) {
grp <- rep(1:(nrow(dat)/i), each=i)
aggregate(dat[1:length(grp),]~grp, FUN=sum)
}

lapply(2:6, fun, dat=TT)

-------------------------------------
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Rui Barradas
Sent: Thursday, September 25, 2014 3:34 PM
To: eliza botto; r-help at r-project.org

Hello,

Try the following.

fun <- function(x, r){
if(r > 0){
m <- length(x) %/% r
y <- numeric(m)
for(i in seq_len(m)){
y[i] <- sum(x[((i - 1)*r + 1):(i*r)])
}
y
}else{
NULL
}
}

apply(TT, 2, fun, r = 2)
apply(TT, 2, fun, r = 3)
etc

Hope this helps,

Em 25-09-2014 20:50, eliza botto escreveu:
> Dear useRs,
> Here is my data with two columns and 20 rows.
>> dput(TT)
> structure(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288, 312, 336, 360, 384, 408, 432, 456, 480), .Dim = c(20L, 2L), .Dimnames = list(NULL, c("", "SS")))
> I first of all want to sum up continuously  two rows (1 & 2, 3 & 4, 5 & 6 and so on) of each column.
> Then I want to sum up 3 rows as (1-2-3,4-5-6,..... 16-17-18) and since 19th and 20th rows do not up 3 rows, so they should be ignored.
> Similarly with 4 sets of rows and 5 sets of rows and even 6.
> I hope I was clear.
> Thankyou so very much in advance,
> Eliza
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help