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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