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

With further fooling around, I realized that explicitly assigning my
last "solution" 'works'; i.e.

names(z)[2] <- "foo"

can be piped as:

  z <- z |>(\(x) "names<-"(x,value = "[<-"(names(x),2,'foo')))()

z

   a foo
1 1   a
2 2   b
3 3   c

This is even awfuller than before. So my query still stands.

-- Bert

On Sat, Jul 20, 2024 at 1:14?PM Bert Gunter <bgunter.4567 at gmail.com> wrote:

Nope, I still got it wrong: None of my approaches work.  :(

So my query remains: how to do it via piping with |> ?

Bert


On Sat, Jul 20, 2024 at 1:06?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