Here is an example closer to yours.
> library(dyn)
> set.seed(123)
> x <- zooreg(rnorm(10))
> y <- zooreg(rnorm(10))
> L <- function(x, k = 1) lag(x, k = -k)
> mod <- dyn$lm(y ~ L(y) + L(x, 0:2))
> mod
Call:
lm(formula = dyn(y ~ L(y) + L(x, 0:2)))
Coefficients:
(Intercept) L(y) L(x, 0:2)1 L(x, 0:2)2 L(x, 0:2)3
0.06355 -0.74540 0.63649 0.44957 -0.41360
> newdata <- cbind(x = c(coredata(x), rnorm(1)), y = c(coredata(y), rnorm(1)))
> newdata <- zooreg(newdata)
> predict(mod, newdata)
1 2 3 4 5 6 7
NA NA 0.9157808 0.6056333 -0.5496422 1.5984615 -0.2574875
8 9 10 11 12 13
-1.6148859 0.3329285 -0.5284646 -0.1799693 NA NA
On Thu, Jul 23, 2009 at 1:04 AM, Gabor
Grothendieck<ggrothendi...@gmail.com> wrote:
> Use dyn.predict like this:
>
>> library(dyn)
>> x <- y <- zoo(1:5)
>> mod <- dyn$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<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.
>>
>
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