[Rd] Confused about NAMED

[peter dalgaard]

On Nov 24, 2011, at 11:13 , Matthew Dowle wrote:

> Hi,
> 
> I expected NAMED to be 1 in all these three cases. It is for one of them,
> but not the other two?
> 
>> R --vanilla
> R version 2.14.0 (2011-10-31)
> Platform: i386-pc-mingw32/i386 (32-bit)
> 
>> x = 1L
>> .Internal(inspect(x))   # why NAM(2)? expected NAM(1)
> @2514aa0 13 INTSXP g0c1 [NAM(2)] (len=1, tl=0) 1
> 
>> y = 1:10
>> .Internal(inspect(y))   # NAM(1) as expected but why different to x?
> @272f788 13 INTSXP g0c4 [NAM(1)] (len=10, tl=0) 1,2,3,4,5,...
> 
>> z = data.frame()
>> .Internal(inspect(z))   # why NAM(2)? expected NAM(1)
> @24fc28c 19 VECSXP g0c0 [OBJ,NAM(2),ATT] (len=0, tl=0)
> ATTRIB:
>  @24fc270 02 LISTSXP g0c0 []
>    TAG: @3f2120 01 SYMSXP g0c0 [MARK,gp=0x4000] "names"
>    @24fc334 16 STRSXP g0c0 [] (len=0, tl=0)
>    TAG: @3f2040 01 SYMSXP g0c0 [MARK,gp=0x4000] "row.names"
>    @24fc318 13 INTSXP g0c0 [] (len=0, tl=0)
>    TAG: @3f2388 01 SYMSXP g0c0 [MARK,gp=0x4000] "class"
>    @25be500 16 STRSXP g0c1 [] (len=1, tl=0)
>      @1d38af0 09 CHARSXP g0c2 [MARK,gp=0x21,ATT] "data.frame"
> 
> It's a little difficult to search for the word "named" but I tried and
> found this in R-ints :
> 
>    "Note that optimizing NAMED = 1 is only effective within a primitive
> (as the closure wrapper of a .Internal will set NAMED = 2 when the
> promise to the argument is evaluated)"
> 
> So might it be that just looking at NAMED using .Internal(inspect()) is
> setting NAMED=2?  But if so, why does y have NAMED==1?

This is tricky business... I'm not quite sure I'll get it right, but let's try

When you are assigning a constant, the value you assign is already part of the assignment expression, so if you want to modify it, you must duplicate. So NAMED==2 on z <- 1 is basically to prevent you from accidentally "changing the value of 1". If it weren't, then you could get bitten by code like for(i in 1:2) {z <- 1; if(i==1) z[1] <- 2}.

If you're assigning the result of a computation, then the object only exists once, so 
z <- 0+1  gets NAMED==1.

However, if the computation is done by returning a named value from within a function, as in

> f <- function(){v <- 1+0; v}
> z <- f()

then again NAMED==2. This is because the side effects of the function _might_ result in something having a hold on the function environment, e.g. if we had 

e <- NULL
f <- function(){e <<-environment(); v <- 1+0; v}
z <- f()

then z[1] <- 5 would change e$v too. As it happens, there aren't any side effects in the forme case, but R loses track and assumes the worst.


> 
> Thanks!
> Matthew
> 
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-- 
Peter Dalgaard, Professor
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com