Re: [R] Extract lags from a formula

From: Gabor Grothendieck <ggrothendieck_at_gmail.com>
Date: Fri, 02 May 2008 21:43:44 -0400

If x is a zoo object note that zoo (and therefore dyn) allows the more compact form lag(x, -(1:2)) so if we write:

mod.eq <- x ~ lag(x, -(1:2))

then mod.eq[[3]][[3]] is the vector -(1:2) or if you like you can define Lag <- function(x, k) lag(x, -k) in which case you can write it:

mod.eq <- x ~ Lag(x, 1:2)

and the same expression extracts 1:2.

On Fri, May 2, 2008 at 2:35 PM, Kerpel, John <John.Kerpel_at_infores.com> wrote:
> Hi folks!
>
>
>
> How do I extract lags from a formula? An example:
>
>
>
> mod.eq<-formula(x~lag(x,-1)+lag(x,-2))
>
> > mod.eq
>
> x ~ lag(x, -1) + lag(x, -2)
>
> > mod.eq[1]
>
> "~"()
>
> > mod.eq[2]
>
> x()
>
> > mod.eq[3]
>
> lag(x, -1) + lag(x, -2)()
>
>
>
> I'm trying to extract the lags into a vector that would be simply [1,2].
> How do I do this? I'm using the dyn package to do dynamic regression.
>
>
>
> John
>
>
> [[alternative HTML version deleted]]
>
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>



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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Sat 03 May 2008 - 01:46:27 GMT

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