Using which() to subset t$id should do the trick:
sapply(levels(t$A), function(x) t$id[which(t$A==x)])
Ivan
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On 07/06/2018 14:21, Massimo Bressan wrote:
sorry, but by further looking at the example I just realised that the posted
solution it's not completely what I need because in fact I do not need to get
back the 'indices' but instead the corrisponding values of column A
#please consider this new example
t<-data.frame(id=c(18,91,20,68,54,27,26,15,4,97),A=c(123,345,123,678,345,123,789,345,123,789))
t
# I need to get this result
r<-data.frame(unique_A=c(123, 345, 678,
789),list_id=c('18,20,27,4','91,54,15','68','26,97'))
r
# any help for this, please?
Da: "Massimo Bressan" <massimo.bres...@arpa.veneto.it>
A: "r-help" <R-help@r-project.org>
Inviato: Giovedì, 7 giugno 2018 10:09:55
Oggetto: Re: aggregate and list elements of variables in data.frame
thanks for the help
I'm posting here the complete solution
t<-data.frame(id=1:10,A=c(123,345,123,678,345,123,789,345,123,789))
t$A <- factor(t$A)
l<-sapply(levels(t$A), function(x) which(t$A==x))
r<-data.frame(list_id=unlist(lapply(l, paste, collapse = ", ")))
r<-cbind(unique_A=row.names(r),r)
row.names(r)<-NULL
r
best
Da: "Massimo Bressan" <massimo.bres...@arpa.veneto.it>
A: "r-help" <R-help@r-project.org>
Inviato: Mercoledì, 6 giugno 2018 10:13:10
Oggetto: aggregate and list elements of variables in data.frame
#given the following reproducible and simplified example
t<-data.frame(id=1:10,A=c(123,345,123,678,345,123,789,345,123,789))
t
#I need to get the following result
r<-data.frame(unique_A=c(123, 345, 678,
789),list_id=c('1,3,6,9','2,5,8','4','7,10'))
r
# i.e. aggregate over the variable "A" and list all elements of the variable "id"
satisfying the criteria of having the same corrisponding value of "A"
#any help for that?
#so far I've just managed to "aggregate" and "count", like:
library(sqldf)
sqldf('select count(*) as count_id, A as unique_A from t group by A')
library(dplyr)
t%>%group_by(unique_A=A) %>% summarise(count_id = n())
# thank you
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