From: Gabor Grothendieck <ggrothendieck_at_gmail.com>

Date: Fri, 02 May 2008 21:43:44 -0400

R-help_at_r-project.org mailing list

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

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]]
**>
**> ______________________________________________
**> R-help_at_r-project.org mailing list
**> 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.
**>
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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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