On Sun, 24 Aug 2008, Berwin A Turlach wrote:
G'day Murray,
On Sun, 24 Aug 2008 18:34:39 +1200
Murray Jorgensen <[EMAIL PROTECTED]> wrote:
I want to extra the part of the formula not including the response
variable from an lm object. For example if the lm object ABx.lm was
created by the call
ABx.lm <- lm( y ~ A + B + x, ...)
Then ACx.lm is saved as part of a workspace.
I wish to extract "~ A + B + x". Later in my code I will fit
another linear model of the form z ~ A + B + x for some other
response variable z. I would be grateful for any suggestions of a
nice way to do this.
AFAIK, a formula is essentially a list of two or three components. The
first component is "~". The second is the LHS of the formula if there
are three components; otherwise the RHS of the formula. The third
component, if it exists, is the RHS of the formula.
So storing "~ A + B + x" and manipulating this part for different
responses could turn out to be painful; you would have to insert the
new LHS as the second component of the list. I would suggest that it
is easier to store the complete formula and just manipulate the LHS;
see:
An ulternative is to use update.formula on the formula extracted, or even
just use update() on the lm fit. For the information in this posting
(there could be more going on, like where to find 'z'),
update(ABX.lm, z ~ .)
should be all that is needed.
R> library(MASS)
R> fm <- lm(time~dist+climb, hills)
R> formula(fm)
time ~ dist + climb
R> formula(fm)[[1]]
`~`
R> formula(fm)[[2]]
time
R> formula(fm)[[3]]
dist + climb
R> tt <- formula(fm)
R> tt[[2]] <- NULL
R> tt
~dist + climb
R> tt <- formula(fm)
R> class(tt[[2]])
[1] "name"
R> typeof(tt[[2]])
[1] "symbol"
R> tt[[2]] <- as.name("y")
R> tt
y ~ dist + climb
R> tt <- formula(fm)
R> tt[[2]] <- as.symbol("z")
R> tt
z ~ dist + climb
HTH.
Cheers,
Berwin
=========================== Full address =============================
Berwin A Turlach Tel.: +65 6515 4416 (secr)
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
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