[datatable-help] Looking for a faster method

[Arunkumar Srinivasan]

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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