[R] Using the pipe, |>, syntax with "names<-"

Bert Gunter
Sun, 21 Jul 2024 15:10:45 -0000

hmmm...
But note that you still used the nested assignment, names()[2] <-
"foo", to circumvent R's pipe limitations, which is exactly what
Iris's solution avoids. So I think I was overawed by your cleverness
;-)

Best,
Bert


On Sun, Jul 21, 2024 at 8:01?AM Bert Gunter <bgunter.4567 at gmail.com>
wrote:
>
> Wow!
> Yes, this is very clever -- way too clever for me -- and meets my
> criteria for a solution.
>
> I think it's also another piece of evidence of why piping in base R is
> not suited for complex/nested assignments, as discussed in Deepayan's
> response.
>
> Maybe someone could offer a better Tidydata piping solution just for
> completeness?
>
> Best,
> Bert
>
> On Sun, Jul 21, 2024 at 7:48?AM Gabor Grothendieck
> <ggrothendieck at gmail.com> wrote:
> >
> > This
> > - is non-destructive (does not change z)
> > - passes the renamed z onto further pipe legs
> > - does not use \(x)...
> >
> > It works by boxing z, operating on the boxed version and then unboxing
it.
> >
> >   z <- data.frame(a = 1:3, b = letters[1:3])
> >   z |> list(x = _) |> within(names(x)[2] <- "foo")
|> _$x
> >   ##   a foo
> >   ## 1 1   a
> >   ## 2 2   b
> >   ## 3 3   c
> >
> > On Sat, Jul 20, 2024 at 4:07?PM Bert Gunter <bgunter.4567 at
gmail.com> wrote:
> > >
> > > This post is likely pretty useless;  it is motivated by a recent
post
> > > from "Val" that was elegantly answered using Tidyverse
constructs, but
> > > I wondered how to do it using base R only. Along the way, I ran
into
> > > the following question to which I think my answer (below) is
pretty
> > > awful. I would be interested in more elegant base R approaches.
So...
> > >
> > > z <- data.frame(a = 1:3, b = letters[1:3])
> > > > z
> > >   a h
> > > 1 1 a
> > > 2 2 b
> > > 3 3 c
> > >
> > > Suppose I want to change the name of the second column of z from
'b'
> > > to 'foo' . This is very easy using nested function syntax
by:
> > >
> > > names(z)[2] <- "foo"
> > > > z
> > >   a foo
> > > 1 1   a
> > > 2 2   b
> > > 3 3   c
> > >
> > > Now suppose I wanted to do this using |> syntax, along the
lines of:
> > >
> > > z |> names()[2] <- "foo"  ## throws an error
> > >
> > > Slightly fancier is:
> > >
> > > z |> (\(x)names(x)[2] <- "b")()
> > > ## does nothing, but does not throw an error.
> > >
> > > However, the following, which resulted from a more careful read
of
> > > ?names works (after changing the name of the second column back
to "b"
> > > of course):
> > >
> > > z |>(\(x) "names<-"(x,value =
"[<-"(names(x),2,'foo')))()
> > > >z
> > >   a foo
> > > 1 1   a
> > > 2 2   b
> > > 3 3   c
> > >
> > > This qualifies to me as "pretty awful." I'm sure
there are better ways
> > > to do this using pipe syntax, so I would appreciate any better
> > > approaches.
> > >
> > > Best,
> > > Bert
> > >
> > > ______________________________________________
> > > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more,
see
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible
code.
> >
> >
> >
> > --
> > Statistics & Software Consulting
> > GKX Group, GKX Associates Inc.
> > tel: 1-877-GKX-GROUP
> > email: ggrothendieck at gmail.com