Use dyn.predict like this: > library(dyn) > x <- y <- zoo(1:5) > mod <- dyn$lm(y ~ lag(x, -1)) > predict(mod, list(x = zoo(6:10, 6:10))) 7 8 9 10 7 8 9 10
On Thu, Jul 23, 2009 at 12:54 AM, Hongwei Dong<pdxd...@gmail.com> wrote: > I have a dynamic time series model like this: > dyn$lm( y ~ lag(y,-1) + x + lag(x,-1)+lag(x,-2) ) > > I need to do an out of sample forecast with this model. Is there any way I > can do this with R? > It would be greatly appreciated if some one can give me an example. Thanks. > > > Harry > > [[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.
