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
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Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com