Hi,
if you are willing to use dplyr, you can do all in one line of code:
library(dplyr)
df<-data.frame(id=1:10,A=c(123,345,123,678,345,123,789,345,123,789))
df%>%group_by(unique_A=A)%>%summarise(list_id=paste(id,collapse=", "))->r
cheers
Am 06.06.2018 um 10:13 schrieb Massimo Bressan:
> #given
which() is unnecessary. Use logical subscripting:
... t$id[t$A ==x]
Further simplification can be gotten by using the with() function:
l <- with(t, sapply(unique(A), function(x) id[A ==x]))
Check this though -- there might be scoping issues.
Cheers,
Bert
On Thu, Jun 7, 2018, 6:49 AM Massimo
#ok, finally this is my final "best and more compact" solution of the problem
by merging different contributions (thanks to all indeed)
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))
l<-sapply(unique(t$A), function(x) t$id[which(t$A==x)])
r<-dat
vals<- lapply(idx, function(index) x$id[index])
data.frame(unique_A = uA, list_vals=unlist(lapply(vals, paste, collapse = ",
")))
best
Da: "Ben Tupper"
A: "Massimo Bressan"
Cc: "r-help"
Inviato: Giovedì, 7 giugno 2018 14:47:55
Oggetto: Re: [
Hi,
Does this do what you want? I had to change the id values to something more
obvious. It uses tibbles which allow each variable to be a list.
library(tibble)
library(dplyr)
x <- tibble(id=LETTERS[1:10],
A=c(123,345,123,678,345,123,789,345,123,789))
uA <- unique(x$
Using which() to subset t$id should do the trick:
sapply(levels(t$A), function(x) t$id[which(t$A==x)])
Ivan
--
Dr. Ivan Calandra
TraCEr, laboratory for Traceology and Controlled Experiments
MONREPOS Archaeological Research Centre and
Museum for Human Behavioural Evolution
Schloss Monrepos
56567
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
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)
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 Resea
#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
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