Please provide your code in a reproducible form (as requested previously). Here we fit with first 9 points and then add a point for prediction. (Of course your model can only predict the current value of Y so you may have to rethink your model even aside from the implementation if you really want to predict future values of Y.)
> library(dyn) > set.seed(123) > tt <- ts(cbind(Y = 1:10, x = rnorm(10), z = rnorm(10))) > L <- function(x, k = 1) lag(x, -k) > tt.zoo <- as.zoo(tt) > fit <- dyn$lm(Y ~ L(Y) + L(x, 0:2) + z, tt.zoo[-10, ]) > predict(fit, tt.zoo) 1 2 3 4 5 6 7 8 9 10 11 12 NA NA 3 4 5 6 7 8 9 10 NA NA On Thu, Jul 23, 2009 at 1:01 PM, Hongwei Dong<[email protected]> wrote: > Hi, Gabor and Other R users, > I'm re-posting my script and the results I got. > > here is the dynamic model I used to estimate in-sample model (1996-2006) > and it works: > > fit<-dyn$lm(Y~lag(Y,-1)+z+x+lag(x,-1)+lag(x,-2)+lag(x,-3)+lag(x,-4)) > Then I used this model to do out sample forecast with the following > scripts, which do not work: > > z<-ts(Z[41:52],start=2006,frequency=4) > x<-ts(X[41:52],start=2006,frequency=4) > newdata<-data.frame(cbind(z,x)) > newdata<-ts(newdata,start=2006,frequency=4) > pred<-predict(fit,newdata) > Here is the results I got from R: > Qtr1 Qtr2 Qtr3 Qtr4 > 2006 NA NA NA NA > 2007 3083.362 NA NA NA > 2008 NA NA NA NA > 2009 NA NA NA NA > > I got only one prediction for the first quarter in 2007. Intuitively, there > might be two problems: the definition of the newdata and how to define Y in > newdata. But I just can't figure this out. It will greatly appreciated if > someone can give me some help. Thanks. > Harry > > > On Thu, Jul 23, 2009 at 5:15 AM, Gabor Grothendieck > <[email protected]> wrote: >> >> You have to use consistent classes. You can't start out using >> ts class and then suddenly switch to zoo class as if you had been >> using zoo all along. Either use ts everywhere or zoo everywhere. >> >> Also in the future please post reproducible examples as requested >> at the bottom of every message to r-help. That means include >> a minimal amount of data so we can get exactly what you >> are getting. >> >> On Thu, Jul 23, 2009 at 4:48 AM, Hongwei Dong<[email protected]> wrote: >> > Thanks, Gabor. This is really helpful. >> > When the regressive part, lag(Y,-1), is not included, my sytax works >> > well. >> > However, when I include the lag(Y) in the model, the prediction part >> > does >> > not work. Here is my sytax for in-sample estimation and it works well: >> > fit<-dyn$lm(Y~lag(Y,-1)+x+lag(x,-1)+lag(x,-2)+lag(x,-3)+lag(x,-4)+z). >> > Then I use this model to do out of sample prediction: >> > x<-ts(X[41:52],start=2006,frequency=4) >> > z<-ts(Z[41:52],start=2006,frequency=4) >> > newdata<-data.frame(cbind(x,z)) >> > newdata<-zooreg(newdata) >> > pred<-predict(fit,newdata) >> > With these, I got weird result, a prediction for each year from year 1 >> > to >> > the first quarter of year 2007 (all "NA"). What I expect is a prediction >> > for >> > the 8 quarters from 2007 to 2008. Intuitively, I know there must be >> > something wrong with my newdata definition. But I can't figure it out. >> > I'll >> > appreciate it very much if you can give some suggestions to modify my >> > syntax. Thanks. >> > >> > Harry >> > >> > >> > >> > >> > >> > >> > >> > >> > >> > >> > >> > On Wed, Jul 22, 2009 at 10:53 PM, Gabor Grothendieck >> > <[email protected]> wrote: >> >> >> >> Here is an example closer to yours. >> >> >> >> > library(dyn) >> >> > set.seed(123) >> >> > x <- zooreg(rnorm(10)) >> >> > y <- zooreg(rnorm(10)) >> >> > L <- function(x, k = 1) lag(x, k = -k) >> >> > mod <- dyn$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<[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. >> >> >> >> >> > >> > >> > > > ______________________________________________ [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.
