[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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