Thanks, Gabor. Here are the problems I'm trying to solve.
*FIRST*, I run this to simulate a 20 years time series process. The data
from 1-15 years are used to estimate the model, and this model is used to
predict the year from 16-20. The following script works.

set.seed(123)
tt <- ts(cbind(Y = 1:20, x = rnorm(20), z = rnorm(20)))
L <- function(x, k = 1) lag(x, -k)
tt.zoo <- as.zoo(tt)
fit <- dyn$lm(Y ~ L(Y) + z + L(x, 0:1), tt.zoo[(1:15), ])
fit
pred <- predict(fit, tt.zoo[(16:20),])
pred

*SECONDLY*, I use similar script, but pretend that we do not know the Y data
from year 16-20. We know x and z for year 16-20, and use them predict Y
based on the model estimated from 1-15 years. So, in the "newdata" part, I
use tt.zoo[(16:20), (2:3)] to remove the Y out. here is the script.
Unfortunately, it does not work in that way.

pred1 <- predict(fit, tt.zoo[(16:20),(2:3)])
pred1

It will be greatly appreciated if you can give me some guide on this.
Thanks.

Harry






On Wed, Jul 22, 2009 at 10:04 PM, Gabor Grothendieck <
[email protected]> wrote:

> Use dyn.predict like this:
>
> > library(dyn)
> > x <- y <- zoo(1:5)
> > mod <- dyn$lm(y ~ lag(x, -1))
> > predict(mod, list(x = zoo(6:10, 6:10)))
>  7  8  9 10
>  7  8  9 10
>
>
> On Thu, Jul 23, 2009 at 12:54 AM, Hongwei Dong<[email protected]> wrote:
> > I have a dynamic time series model like this:
> > dyn$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]]
> >
> > ______________________________________________
> > [email protected] 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.
> >
>

        [[alternative HTML version deleted]]

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