[R] Use predict.lm

[Gabor Grothendieck]

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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