Thanks a ton Rolf,
I was using the coefficients as given from summary(fit), which have been
rounded. When I used coef(fit) as you've done, the poly function is the same
now in Octave and returned the R predicted values as expected. Your time is
appreciated,
C
> CC: r-help@r-project.org
> From: r.tur...@auckland.ac.nz
> Subject: Re: [R] Polynomial Fitting
> Date: Wed, 30 Sep 2009 08:30:15 +1300
> To: w_chris_carle...@hotmail.com
>
>
> On 30/09/2009, at 5:34 AM, chris carleton wrote:
>
> >
> > Thanks for the response. I'm sorry I didn't provide the code or
> > data example earlier. I was using the polynomial fitting technique
> > of this form;
> >
> > test <- lm(x[,34] ~ I(x[,1]) + I(x[,1]^2) + I(x[,1]^3))
> >
> > for the original fitting operation. I also tried to use;
> >
> > lm(y ~ poly(x,3,raw=TRUE))
> >
> > with the same results for the polynomial coefficients in both
> > cases. If my understanding is correct, both of the methods above
> > produce the coefficients of a polynomial based on the data in 'y'
> > as that data varies over 'x'. Therefore, I would assume that the
> > function of the polynomial should always produce the same results
> > as the predict() function in R produces. However, here are the raw
> > data for anyone that has the time to help me out.
> >
> > y:
> > [1] 9097 9074 9043 8978 8912 8847 8814 8786 8752 8722 8686 8657
> > 8610 8604 8554
> > [16] 8546 8496 8482 8479 8462 8460 8438 8428 8418 8384
> >
> > x:
> > [1] 17.50 NA 20.59 21.43 17.78 21.89 NA 22.86 NA 6.10
> > NA 5.37
> > [13] 3.80 NA 6.80 NA NA NA 5.80 NA NA NA
> > NA NA
> > [25] NA
> >
> > I think that R lm() just ignores the NA values, but I've also tried
> > this by first eliminating NAs and the corresponding x values from
> > the data before fitting the poly and the result was the same
> > coefficients. Thanks very much to anyone who is willing to provide
> > information.
>
> What's the problem?
>
> fit <- lm(y~x+I(x^2)+I(x^3))
> ccc <- coef(fit)
> X <- cbind(1,x,x^2,x^3)
> print(fitted(fit))
> chk <- X%*%ccc
> print(chk[!is.na(chk)])
> print(range(fitted(fit)-chk[!is.na(chk)]))
> [1] 0.000000e+00 5.456968e-12
>
> The answers are the same.
>
> cheers,
>
> Rolf Turner
>
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