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')))()
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])
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"
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')))()
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