# [R] Loops

Sun Jan 27 12:17:36 CET 2013

```Hello,

I think there is an error in the expression

100-(100*abs(fa1[i]/sum(fa1[i])-(1/3)))

Note that fa1[i]/sum(fa1[i]) is always 1. If it's fa1[i]/sum(fa1), try
the following, using lists to hold the results.

# Make up some data
set.seed(6628)
p <- matrix(runif(300), nrow = 100)

idx <- seq(1, 100, by = 25)
fa <- lapply(idx, function(i) colSums(p[i:(i + 24), ]))
fa[] <- colSums(p)

fab <- lapply(fa, function(x) 100 - 100*abs(x/sum(x) - 1/3))
fab

You can give names to the lists elements, if you want to.

names(fa) <- paste0("fa", 1:5)
names(fab) <- paste0("fa", 1:5, "b")

Hope this helps,

Em 27-01-2013 08:02, Francesca escreveu:
> Dear Contributors,
> I am asking help on the way how to solve a problem related to loops for
> that I always get confused with.
> I would like to perform the following procedure in a compact way.
>
> Consider that p is a matrix composed of 100 rows and three columns. I need
> to calculate the sum over some rows of each
> column separately, as follows:
>
> fa1<-(colSums(p[1:25,]))
>
> fa2<-(colSums(p[26:50,]))
>
> fa3<-(colSums(p[51:75,]))
>
> fa4<-(colSums(p[76:100,]))
>
> fa5<-(colSums(p[1:100,]))
>
>
>
> and then I need to  apply to each of them the following:
>
>
> fa1b<-c()
>
> for (i in 1:3){
>
> fa1b[i]<-(100-(100*abs(fa1[i]/sum(fa1[i])-(1/3))))
>
> }
>
>
> fa2b<-c()
>
> for (i in 1:3){
>
> fa2b[i]<-(100-(100*abs(fa2[i]/sum(fa2[i])-(1/3))))
>
> }
>
>
> and so on.
>
> Is there a more efficient way to do this?
>
>
> Francesca
>
> ----------------------------------
> Francesca Pancotto, PhD
> Università di Modena e Reggio Emilia
> Viale A. Allegri, 9
> 40121 Reggio Emilia
> Office: +39 0522 523264
> ----------------------------------
>
> 	[[alternative HTML version deleted]]
>
>
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help