[R] The opposite of "lag"

[Dimitri Liakhovitski]

Thank you, Gabor, but sorry - what is the exact order of those rows again?
Thank you!
Dimitri

On Wed, Jul 21, 2010 at 11:59 AM, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:
> 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))
>



-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com