[R] The opposite of "lag"

Dimitri Liakhovitski dimitri.liakhovitski at gmail.com
Wed Jul 21 16:14:49 CEST 2010


Hello!

I have a data frame A (below) with a grouping factor (group). I take
my DV and create the new, lagged DV by applying the function lag.it
(below). It works fine.

A <- data.frame(year=rep(c(1980:1984),3), group=
factor(sort(rep(1:3,5))), DV=c(rnorm(15)))
lag.it <- function(x) {
  DV <- ts(x$DV, start = x$year[1])
  idx <- seq(length = length(DV))
  DVs <- cbind(DV, lag(DV, -1))[idx,]
  out<-cbind(x, DVs[,2])  # wages[,2]
   names(out)[length(out)]<-"DV.lag"
   return(out)
}
A
A.lagged <- do.call("rbind", by(A, A$group, lag.it))
A.lagged


Now, I am trying to create the oppostive of lag for DV (should I call
it "lead"?)
I tried exactly the same as above, but with a different number under
lag function (below), but it's not working. I am clearly doing
something wrong. Any advice?
Thanks a lot!


lead.it <- function(x) {
  DV <- ts(x$DV, start = x$year[1])
  idx <- seq(length = length(DV))
  DVs <- cbind(DV, lag(DV, 2))[idx,]
 out<-cbind(x, DVs[,2])
   names(out)[length(out)]<-"DV.lead"
   return(out)
}
A
A.lead <- do.call("rbind", by(A, A$group, lead.it))
A.lead

-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com



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