Hi Massimo,
Something along those lines could help you I guess:
t$A <- factor(t$A)
sapply(levels(t$A), function(x) which(t$A==x))
You can then play with the output using paste()
Ivan
--
Dr. Ivan Calandra
TraCEr, laboratory for Traceology and Controlled Experiments
MONREPOS Archaeological Research Centre and
Museum for Human Behavioural Evolution
Schloss Monrepos
56567 Neuwied, Germany
+49 (0) 2631 9772-243
https://www.researchgate.net/profile/Ivan_Calandra
On 06/06/2018 10:13, Massimo Bressan wrote:
#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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