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 <[EMAIL PROTECTED]> 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@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. > ______________________________________________ R-help@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.