predict.lm(...,type="terms") question
Thank you all. My muddle about predict.lm(..., type = "terms") was evident even in my first sentence of my original posting
How can I actually use the output of predict.lm(..., type="terms") to predict new term values from new response values?
the answer being that I cannot; that new response values, if included in newdata, will simply be ignored by predict.lm, as well they should.
As for the calibration issue, I am reviewing literature now as suggested.
Though predict.lm performed to spec (no bug), may I suggest a minor
change to ?predict.lm text?
Existing:
newdata An optional data frame in which to
look for variables with
which to predict. If omitted,
the fitted values are used.
Proposed:
newdata An optional data frame in which to
look for new values of terms with
which to predict. If omitted, the
fitted values are used.
-John Thaden, Ph.D.
College Station, TX
--- On Sun, 9/2/12, peter dalgaard <pdalgd at gmail.com> wrote:
From: peter dalgaard <pdalgd at gmail.com> Subject: Re: [R] predict.lm(...,type="terms") question To: "David Winsemius" <dwinsemius at comcast.net> Cc: "Rui Barradas" <ruipbarradas at sapo.pt>, r-help at r-project.org, "jjthaden" <jjthaden at flash.net> Date: Sunday, September 2, 2012, 1:35 AM On Sep 2, 2012, at 03:38 , David Winsemius wrote:
Why should predict not complain when it is offered a
newdata argument that does no contain a vector of values for "x"? The whole point of the terms method of prediction is to offer estimates for specific values of items on the RHS of the formula. The OP seems to have trouble understanding that point. Putting in a vector with the name of the LHS item makes no sense to me. I certainly cannot see that any particular behavior for this pathological input is described for predict.lm in its help page, but throwing an error seems perfectly reasonable to me. Yes. Lots of confusion going on here. First, data= is _always_ used as the _first_ place to look for variables, if things are not in it, search continues into the formula's environment. To be slightly perverse, notice that even this works:
y <- rnorm(10) x <- rnorm(10) d <- data.frame(z=rnorm(9)) lm(y ~ x, d)
Call: lm(formula = y ~ x, data = d) Coefficients: (Intercept)? ? ? ? ? ? x? ? ? -0.2760? ? ???0.2328? Secondly, what is predict(..., type="terms") supposed to have to do with inverting a regression equation? That's just not what it does, it only splits the prediction formula into its constituent terms. Thirdly; no, you do not invert a regression equation by regressing y on x. That only works if you can be sure that your new (x, y) are sampled from the same population as the data, which is not going to be the case if you are fitting to data with, say, selected equispaced x values. There's a whole literature on how to do this properly, Google e.g. "inverse calibration" for enlightenment.? -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd.mes at cbs.dk? Priv: PDalgd at gmail.com