Or a slight refinement of that:
system.time(out2 <- data.table(id=seq(vec2), val=vec2,
key="val")[J(vec1), list(list(id))]$V1)
Matthew
On 25/08/13 21:58, Arunkumar Srinivasan wrote:
How about this?
system.time(out <- data.table(id=seq(vec2), val=vec2,
key="val")[J(vec1)][, list(list(id)), by=val]$V1)
user system elapsed
0.098 0.004 0.103
Arun
On Sunday, August 25, 2013 at 10:27 PM, arun wrote:
Hi,
I tried a ?data.table() method to solve the problem in the link below.
http://r.789695.n4.nabble.com/how-to-combine-apply-and-which-or-alternative-ways-to-do-so-td4674424.html#a4674434
But, it was not that fast.
set.seed(24)
vec1<- sample(1e5,1e3,replace=FALSE)
set.seed(48)
vec2<- sample(1e3,1e6,replace=TRUE)
system.time({res1<- tapply(vec1,1:1e3,FUN=function(i)
{which(vec2==i)})})?# user system elapsed
# 3.912 0.000 3.880
system.time(res2<- sapply(vec1,function(x) which(vec2%in%x)))
# user system elapsed
# 24.368 0.000 23.247
?vecR1<-unlist(res1)
names(vecR1)<-NULL
vecR2<- unlist(res2)
identical(vecR1,vecR2)
#[1]TRUE
library(data.table)
dt1<- data.table(vec1,Group=1:1e3,key='Group')
system.time({res3<-
dt1[,list(list(which(vec1==vec2))),by=Group]})##Not that fast?
# user system elapsed
# 3.756 0.120 3.886 ######
identical(vecR1,unlist(res3$V1))
#[1] TRUE
Is there a faster way?
Thanks.
A.K.
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