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

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

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