[R] Use predict.lm

[Gabor Grothendieck]

Sorry, I don't think that my earlier reply was what you wanted.
Try this instead:

# fit using first 100 points
iris.lm <- lm(Sepal.Length ~., iris[1:100,c(1,2,4)])

# predict using coefficients from above and variables from next 50 points
predict(iris.lm, iris[101:150, c(1,2,4)])


On 5/2/06, Gabor Grothendieck <ggrothendieck at gmail.com> wrote:
> Try this:
>
> # regression of Sepal.Length on cols 2 and 4 using first 100 rows
> iris.lm <- lm(Sepal.Length ~ ., iris[,c(1,2,4)], subset = 1:100)
>
> # now do it with next 50 rows
> predict(update(iris.lm, subset = 101:150))
>
> # double check - this gives same result as last line
> predict(lm(Sepal.Length ~ ., iris[,c(1,2,4)], subset = 101:150))
>
>
> On 5/2/06, Jiang, Jincai (Institutional Securities Management)
> <Jincai.Jiang at morganstanley.com> wrote:
> > Hi All,
> >
> > I created a two variable lm() model
> >
> > slm<-lm(y[1:3000,8]~y[1:3000,12]+y[1:3000,15])
> >
> > I made two predictions
> >
> > predict(slm,newdata=y[201:3200,])
> > predict(slm,newdata=y[601:3600,])
> >
> > there is no error message for either of these.
> > the results are identical, and identical to slm$fitted as well.
> >
> > if this is not the right way to apply the model coefficients to a new
> > set of inputs, what is the right way?
> >
> > Thank you
> >
> > Regards,
> >
> > Jincai Jiang
> > (Office) 212-761-3984
> >
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