Yes, odd that ?Constants doesn't work, given that there is a page labeled
Constants {base}.
On a happier note, these all work now on my Mac R version 2.11.1
> ?base
> ?stats
> ?utils
though Constants is not an entry in the index.
R help gurus - is it possible to have entries such as "Constan
I just tried ?Constants at the console and was disappointed that the
so-named base help page would not come up.
> ?Constants
No documentation for 'Constants' in specified packages and libraries:
you could try '??Constants'
Seems like there should have been a match. I was looking for the mont
In R 2.12.0 I get
> d <- data.frame(x=1:10, y=log(1:10), f3=LETTERS[rep(1:3,c(3,3,4))])
> m <- model.frame(y~.^2, data=d)
> formula(m)
y ~ x + f3
In S+ formula(m) gives formula given to model.frame(),
but in R you have to do the following get that formula:
> formula(attr(m, "terms"))
y
on line 1714 of R-lang.texi (current SVN of R-devel, 53919)
@code{x$aa} will match @code{x$aabb} if @code{x} does not a component
should probably have "contain" or "have" inserted after "not"
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R-devel@r-project.org mailing list
https://stat.eth
Kurt Hornik writes
> >> if (!all(sapply(args, is.ordered)) ||
> >> !all(sapply(level.list, identical, y = level.set))) {
>
> I think it would be better to use something like
>
> ll <- lapply(args, levels)
>
> !all(sapply(ll, identical, ll[[1L]]))
>
> [using union() is not quite right]
Yes
On Jan 6, 2011, at 13:11 , Kenn Konstabel wrote:
> the following seems an easy solution:
>
> f1 <- function(mod){
> subs <- 1:10
> toeval <- quote(update(mod, subset=subs))
> toeval$subset<-subs
> eval(toeval)
> }
>
> f1(mod.2)
Tere, Kenn!
Yes, enforcing pass-by-value by p
the following seems an easy solution:
f1 <- function(mod){
subs <- 1:10
toeval <- quote(update(mod, subset=subs))
toeval$subset<-subs
eval(toeval)
}
f1(mod.2)
When experimenting with this I once had (by mistake):
mod.2 <- lm(update(mod.1, . ~ . - Year + Year)) # instead
> Martin Maechler writes:
I have 3 comments:
> Thaler, Thorn, LAUSANNE, Applied Mathematics
>
> on Wed, 5 Jan 2011 11:20:47 +0100 writes:
>> Hi everybody, Is there a particular reason, why this code
>> does not work as intended:
>> z <- factor(LETTERS[1:3], ordered = TRUE)
