[R] creating individual records from a frequency distribution

[Sven E. Templer]

With melt and rep you are close. If you combine them it works:

library(reshape)
# your data:
df1 <- data.frame(area=c(1,2),group1=c(2,3),group2=c(1,5),group3=c(4,0))
df2<-data.frame(person_id=seq(1:15),area=c(rep(1,7),rep(2,8)),group_num=c(1,1,2,3,3,3,3,1,1,1,2,2,2,2,2))
# first melt
d <- melt(df1,"area",2:4)
# then repeat each row by 'counts'
d <- d[rep(seq(nrow(d)), times=d$value),]
# then order (if order of id's is not arbitrary), and add ids
d <- d[order(d$area,d$variable),]
d$value <- seq(nrow(d))
# compare
cbind(df2,"---",d)

Best, Sven.

On 22 October 2014 17:48, Gavin Rudge <G.Rudge at bham.ac.uk> wrote:
> I've got data frame containing a simple frequency distribution of numbers of people in three age groups by area.
>
> df1<-data.frame(area=c(1,2),group1=c(2,3),group2=c(1,5),group3=c(4,0))
> df1
>
> I want to get a data frame with one record per person (in my case 15 of them) which would look like this, with variables indicating the area and age group to which each belongs
>
> df2<-data.frame(person_id=seq(1:15),area=c(rep(1,7),rep(2,8)),group_num=c(1,1,2,3,3,3,3,1,1,1,2,2,2,2,2))
> df2
>
> This is not the same as melting wide to long data as in reshape2, as I'm melting from aggregated data. I can get vectors of columns values using the rep command, but sewing them together and allowing for zeros looks a bit cumbersome. I'm assuming there is a simple command that does this sort of thing.
>
> Any help gratefully received,
>
> GavinR
>
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