Hello,
In the mean time, I've "discovered" an R package that might do the job,
chemCal. It only implements first order linear regression (weighted),
and like I'd said in my first reply, and was repeated several times, the
standard errors are not reversed, the prediction bands follow Massart et
al. (1988).
Take a look at it, it's a very simple package.
Rui Barradas
Em 02-09-2012 18:07, John Thaden escreveu:
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 <pda...@gmail.com> wrote:
From: peter dalgaard <pda...@gmail.com>
Subject: Re: [R] predict.lm(...,type="terms") question
To: "David Winsemius" <dwinsem...@comcast.net>
Cc: "Rui Barradas" <ruipbarra...@sapo.pt>, r-help@r-project.org, "jjthaden"
<jjtha...@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....@cbs.dk
Priv: pda...@gmail.com
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