The opposite of "lag"

On Wed, Jul 21, 2010 at 10:50 AM, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:

On Wed, Jul 21, 2010 at 10:14 AM, Dimitri Liakhovitski
<dimitri.liakhovitski at gmail.com> wrote:

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

Try this:

library(zoo)
z <- read.zoo(A, index = 1, split = "group", frequency = 1)
lag(z, c(-1, 0, 1))

The ### line was missing:

library(zoo)
z <- read.zoo(A, index = 1, split = "group", frequency = 1)
z <- as.zooreg(z) ###
lag(z, c(-1, 0, 1))