Re: [Rd] full copy on assignment?
Thanks very much. By the way, I tried setting a GDB breakpoint at duplicate1(), with the following: x - 1:1000 x[3] - 8 x[33] - 88 I found that duplicate1() was called on both of the latter two lines. I was a bit surprised, since change-on-write would seem to imply that copying would be done in that second line but NOT on the third. Moreover, system.time() gave 0.284 user time for the second and 0 on the third. YET duplicate1() WAS called on the third, and in stepping through the code, there didn't seem to be an immediate exit. Thanks to both John and Duncan for their comment on the fact that using [- directly is a very different situation. That's not what I asked, but the comment is useful to me for other reasons. Norm Message: 4 Date: Sat, 03 Apr 2010 17:54:58 -0700 From: John Chambers j...@r-project.org To: r-devel@r-project.org Subject: Re: [Rd] full copy on assignment? ... ... How often does y get duplicated? Hopefully not a million times. One can look at this in gdb, by trapping calls to duplicate1. The answer is: just once, to ensure that the object is local. Then the duplicated version has only one reference and the primitive replacement doesn't copy it. ... __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] full copy on assignment?
On 04/04/2010 05:27 PM, Norm Matloff wrote: Thanks very much. By the way, I tried setting a GDB breakpoint at duplicate1(), with the following: x - 1:1000 x[3] - 8 x[33] - 88 Here's how I investigated this, with the last line somewhat surprising R -d gdb gdb r ... cntrl-C gdb break duplicate1 gdb commands call Rf_PrintValue(s) c end gdb c then x=5:1 x[1L] = 1L # no copy x[1L] = 10 # type coercion, new alloc but no copy x[1] = 20 # copy of index (!) Martin I found that duplicate1() was called on both of the latter two lines. I was a bit surprised, since change-on-write would seem to imply that copying would be done in that second line but NOT on the third. Moreover, system.time() gave 0.284 user time for the second and 0 on the third. YET duplicate1() WAS called on the third, and in stepping through the code, there didn't seem to be an immediate exit. Thanks to both John and Duncan for their comment on the fact that using [- directly is a very different situation. That's not what I asked, but the comment is useful to me for other reasons. Norm Message: 4 Date: Sat, 03 Apr 2010 17:54:58 -0700 From: John Chambers j...@r-project.org To: r-devel@r-project.org Subject: Re: [Rd] full copy on assignment? ... ... How often does y get duplicated? Hopefully not a million times. One can look at this in gdb, by trapping calls to duplicate1. The answer is: just once, to ensure that the object is local. Then the duplicated version has only one reference and the primitive replacement doesn't copy it. ... __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel -- Martin Morgan Computational Biology / Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: Arnold Building M1 B861 Phone: (206) 667-2793 __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
[Rd] full copy on assignment?
Here's a basic question that doesn't seem to be completely answered in the docs, and which unfortunately I've not had time to figure out by wading through the R source code: In a vector (or array) element assignment such as z[3] - 8 is there in actuality a full rewriting of the entire vector pointed to by z, as implied by z - [-(z,3,value=8) Assume that an element of z has already being changed previously, so that copy-on-change issues don't apply, with z being reassigned back to the same memory address. I seem to recall reading somewhere that recent R versions make some attempt to avoid rewriting the entire vector, and my timing experiments seem to suggest that it's true. So, is a full rewrite avoided? And where in the source code is this done? Thanks. Norm Matloff __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] full copy on assignment?
On 03/04/2010 6:34 PM, Norm Matloff wrote: Here's a basic question that doesn't seem to be completely answered in the docs, and which unfortunately I've not had time to figure out by wading through the R source code: In a vector (or array) element assignment such as z[3] - 8 is there in actuality a full rewriting of the entire vector pointed to by z, as implied by z - [-(z,3,value=8) Assume that an element of z has already being changed previously, so that copy-on-change issues don't apply, with z being reassigned back to the same memory address. I seem to recall reading somewhere that recent R versions make some attempt to avoid rewriting the entire vector, and my timing experiments seem to suggest that it's true. So, is a full rewrite avoided? And where in the source code is this done? It depends. User-written assignment functions can't avoid the copy. They act like the expansion z - [-(z,3,value=8) and in that, R can't tell that the newly created result of [-(z,3,value=8) will later overwrite z. However, if z is a regular vector without a class and you're using the built-in version of z[3] - 8, it can take some shortcuts. This happens in multiple places; one is around line 488 of subassign.c another is around line 1336. In each of these places copies are made in some circumstances, but not in general. Duncan Murdoch __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] full copy on assignment?
In particular, Duncan's comment applies in situations where the
replacement is in a loop, obviously the case one worries about.
What happens in the stupid little function:
foo - function(x) { for(i in seq_along(x)) x[i] - x[i] +1; x}
for the case
y - 1:1e6
y1 - foo(y)
How often does y get duplicated? Hopefully not a million times. One can
look at this in gdb, by trapping calls to duplicate1. The answer is:
just once, to ensure that the object is local. Then the duplicated
version has only one reference and the primitive replacement doesn't
copy it.
Unfortunately, as Duncan said, changing the definition to a user-written
replacement function:
sub- - function(x,i, value){x[i]- value; x}
foo - function(x) { for(i in seq_along(x)) sub(x,i) - x[i]+1; x}
does duplicate a million times, since every call to `sub-` gets an
argument with two references.
John
On 4/3/10 4:42 PM, Duncan Murdoch wrote:
On 03/04/2010 6:34 PM, Norm Matloff wrote:
Here's a basic question that doesn't seem to be completely answered in
the docs, and which unfortunately I've not had time to figure out by
wading through the R source code:
In a vector (or array) element assignment such as
z[3] - 8
is there in actuality a full rewriting of the entire vector pointed to
by z, as implied by
z - [-(z,3,value=8)
Assume that an element of z has already being changed previously, so
that copy-on-change issues don't apply, with z being reassigned back to
the same memory address.
I seem to recall reading somewhere that recent R versions make some
attempt to avoid rewriting the entire vector, and my timing experiments
seem to suggest that it's true.
So, is a full rewrite avoided? And where in the source code is this
done?
It depends. User-written assignment functions can't avoid the copy.
They act like the expansion
z - [-(z,3,value=8)
and in that, R can't tell that the newly created result of
[-(z,3,value=8) will later overwrite z.
However, if z is a regular vector without a class and you're using the
built-in version of z[3] - 8, it can take some shortcuts. This
happens in multiple places; one is around line 488 of subassign.c
another is around line 1336. In each of these places copies are made
in some circumstances, but not in general.
Duncan Murdoch
__
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
Re: [Rd] full copy on assignment?
Thanks, Martin and Duncan, for the quick, cleary replies. Norm __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
