Wacek,
Peter gave you a full answer explaining it very well. If you really
want to be able to trace each instance yourself, you have to learn far
more about R internals than you apparently know (and Peter hinted at
that). Internally x=1 an x=c(1) are slightly different in that the
former has NAMED(x) = 2 whereas the latter has NAMED(x) = 0 which is
what causes the difference in behavior as Peter explained. The reason
is that c(1) creates a copy of the 1 (which is a constant [=unmutable]
thus requiring a copy) and the new copy has no other references and
thus can be modified and hence NAMED(x) = 0.
Cheers,
Simon
On Mar 10, 2009, at 18:16 , Wacek Kusnierczyk wrote:
i got an offline response saying that my original post may have not
been
clear as to what the problem was, essentially, and that i may need to
restate it in words, in addition to code.
the problem is: the performance of 'names<-' is incoherent, in that
in
some situations it acts in a functional manner, producing a copy of
its
argument with the names changed, while in others it changes the object
in-place (and returns it), without copying first. your explanation
below is of course valid, but does not seem to address the issue. in
the examples below, there is always (or so it seems) just one
reference
to the object.
why are the following functional:
x = 1; 'names<-'(x, 'foo'); names(x)
x = 'foo'; 'names<-'(x, 'foo'); names(x)
while these are destructive:
x = c(1); 'names<-'(x, 'foo'); names(x)
x = c('foo'); 'names<-'(x, 'foo'); names(x)
it is claimed that in r a singular value is a one-element vector, and
indeed,
identical(1, c(1))
# TRUE
all.equal(is(1), is(c(1)))
# TRUE
i also do not understand the difference here:
x = c(1); 'names<-'(x, 'foo'); names(x)
# "foo"
x = c(1); names(x); 'names<-'(x, 'foo'); names(x)
# "foo"
x = c(1); print(x); 'names<-'(x, 'foo'); names(x)
# NULL
x = c(1); print(c(x)); 'names<-'(x, 'foo'); names(x)
# "foo"
does print, but not names, increase the reference count for x when
applied to x, but not to c(x)?
if the issue is that there is, in those examples where x is left
unchanged, an additional reference to x that causes the value of x
to be
copied, could you please explain how and when this additional
reference
is created?
thanks,
vQ
Peter Dalgaard wrote:
is there something i misunderstand here?
Only the ideology/pragmatism... In principle, R has call-by-value
semantics and a function does not destructively modify its
arguments(*),
and foo(x)<-bar behaves like x <- "foo<-"(x, bar). HOWEVER, this has
obvious performance repercussions (think x <- rnorm(1e7); x[1] <-
0), so
we do allow destructive modification by replacement functions,
PROVIDED
that the x is not used by anything else. On the least suspicion that
something else is using the object, a copy of x is made before the
modification.
So
(A) you should not use code like y <- "foo<-"(x, bar)
because
(B) you cannot (easily) predict whether or not x will be modified
destructively
--------
(*) unless you mess with match.call() or substitute() and the like.
But
that's a different story.
--
-------------------------------------------------------------------------------
Wacek Kusnierczyk, MD PhD
Email: w...@idi.ntnu.no
Phone: +47 73591875, +47 72574609
Department of Computer and Information Science (IDI)
Faculty of Information Technology, Mathematics and Electrical
Engineering (IME)
Norwegian University of Science and Technology (NTNU)
Sem Saelands vei 7, 7491 Trondheim, Norway
Room itv303
Bioinformatics & Gene Regulation Group
Department of Cancer Research and Molecular Medicine (IKM)
Faculty of Medicine (DMF)
Norwegian University of Science and Technology (NTNU)
Laboratory Center, Erling Skjalgsons gt. 1, 7030 Trondheim, Norway
Room 231.05.060
______________________________________________
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
______________________________________________
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
