[Rd] Attributes of a vector element?

[Iñaki Ucar]

On Tue, 22 Feb 2022 at 19:29, J C Nash <profjcnash using gmail.com> wrote:
>
> I'm sending this to r-devel rather than r-help as I suspect I'm pushing the boundaries
> of the R language. In that I get no errors but things don't "work", I may have revealed
> a gap in the attributes structure.
>
> In displaying results of some optimization calculations (in the optimx package I maintain),
> I'd like to add indicators F(ree), L(ower) and U(pper) to indicate if parameters returned are
> on bounds. I've got my optimr() routine to put the character indicators in the list() structure of
> the answer to each individual solver result. However, when I collect these into a data frame
> for the results of several solvers (run by the opm() function), it seems that attributes
> can't be carried through.
>
> Here is a small example.
>
> pp<-c(1,2)
> attr(pp[1], "trial")<-"first"
> attributes(pp)
> attributes(pp[1])
> attr(pp[1],"trial")
> pp[1]
> pp[2]
> attr(pp, "name")<-"rubbish"
> attributes(pp)
> str(pp)
>
> The output is:
>
>  > pp<-c(1,2)
>
>  > attr(pp[1], "trial")<-"first"  ## APPEARS TO ACCEPT ATTRIBUTE
>
>  > attributes(pp)
> NULL
>
>  > attributes(pp[1])
> NULL
>
>  > attr(pp[1],"trial")            ## FAILS TO FIND ATTRIBUTE
> NULL
>
>  > pp[1]
> [1] 1
>
>  > pp[2]
> [1] 2
>
>  > attr(pp, "name")<-"rubbish"
>
>  > attributes(pp)
> $name
> [1] "rubbish"
>
>
>  > str(pp)
>   num [1:2] 1 2
>   - attr(*, "name")= chr "rubbish"
>
> Similarly,
>
> pl<-list(one=1, two=2)
> attr(pl[1],"trial")<- "lfirst"
> attr(pl[1], "trial")
> attributes(pl)
> attributes(pl[1])
> attributes(pl$one)
> str(pl)
>
> This also fails.
>
> Am I doing something silly?
>
> Thanks in advance for any pointers.

In short, this is not how it works. In both cases, you are assigning
an attribute to a temporary subset that is immediately destroyed,
because it's not assigned to anything. So you are doing this:

pp<-c(1,2)
tmp <- pp[1]
attr(tmp, "trial")<-"first"
rm(tmp)

The assignment of course succeeds, but that's not what you want. What
you can do is to add a data.frame of size length(pp) as an attribute
to pp, with as many columns as you need (lower, upper...).

-- 
Iñaki Úcar