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?
Thanks for your time!
Francesca
----------------------------------
Francesca Pancotto, PhD
Università di Modena e Reggio Emilia
Viale A. Allegri, 9
40121 Reggio Emilia
Office: +39 0522 523264
Web: https://sites.google.com/site/francescapancotto/
----------------------------------
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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[[5]] <- 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,
Rui Barradas
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?
>
> Thanks for your time!
>
> Francesca
>
> ----------------------------------
> Francesca Pancotto, PhD
> Universit? di Modena e Reggio Emilia
> Viale A. Allegri, 9
> 40121 Reggio Emilia
> Office: +39 0522 523264
> Web: https://sites.google.com/site/francescapancotto/
> ----------------------------------
>
> [[alternative HTML version deleted]]
>
>
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
> 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)))) > > } >I think I'd do it this way (correcting for the presumed error that Rui Barradas noted):> dim( p ) = c(25,4,3)> p2 = apply(p, c(2,3), sum) > p3 = t(apply(p2, 1, function(fa) 100-(100*abs(fa/sum(fa)-(1/3))) ) )p3 now contains all your results except the one including all the data, which is trivial to compute. -- Richard D. Morey Assistant Professor Psychometrics and Statistics Rijksuniversiteit Groningen / University of Groningen http://drsmorey.org/research/rdmorey