How does predict deal with models that include the AsIs function?

Question

I have the model lm(y~x+I(log(x)) and I would like to use predict to get predictions of a new data frame containing new values of x, based on my model. How does predict deal with the AsIs function I in the model? Does the I(log(x)) need to be extra specified in the newdata argument of predict or does predict understand that it should construct and use I(log(x)) from x?

UPDATE

@DWin: The way the variables enter in the model affect the coefficients especially for interactions. My example is simplistic but try this out

x<-rep(seq(0,100,by=1),10)
y<-15+2*rnorm(1010,10,4)*x+2*rnorm(1010,10,4)*x^(1/2)+rnorm(1010,20,100)
z<-x^2

plot(x,y)
lm1<-lm(y~x*I(x^2))
lm2<-lm(y~x*x^2)
lm3<-lm(y~x*z)


summary(lm1)
summary(lm2)
summary(lm3)

You see that lm1=lm3, but lm2 is something different (only 1 coefficient). Assuming you don't want to create the dummy variable z (computationally inefficient for large datasets), the only way to build an interaction model like lm3 is with I. Again this is a very simplistic example (that may make no statistical sense) however it makes sense in complicated models.

@Ben Bolker: I would like to avoid guessing and try to ask for an authoritative answer (I can't direct check this with my models since they are much more complicated than the example). My guess is that predict correctly assumes and constructs the I(log(x))

Answer 1

You do not need to make your variable names look like the term I(x). Just use "x" in the newdata argument.

The reason lm(y~x*I(x^2)) and lm(y~x*x^2) are different is that "^" and "*" are reserved symbols for formula in R. That's not the case with the log function. It is also incorrect that interactions can only be constructed with I(). If you wanted a second degree polynomial in R you should use poly(x, 2). If you build with I(log(x)) or with just log(x) you should get the same model. Both of them will get transformed to the predictor value properly with predict if you use:

newdata=dataframe( x=seq( min(x), max(x), length=10) )

Using poly will protect you from incorrect inferences that are so commonly caused by the use of I(x^2).