Às 21:46 de 20/07/2024, Bert Gunter escreveu:
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.4...@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.4...@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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Hello,
This is not exactly the same but in one of your attempts all you have to
do is to return x.
The following works and does something.
z |> (\(x){names(x)[2] <- "foo";x})()
# a foo
# 1 1 a
# 2 2 b
# 3 3 c
Hope this helps,
Rui Barradas
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