Using the pipe, |>, syntax with "names<-"
Iris's reply is what I was looking for. Many thanks -- I can now sleep tonight! Both Rui's and Duncan's responses merely hid what I wanted to avoid. I hope that I did not occupy much of your times on my useless question and rather pathetic attempts at an answer. Cheers, Bert
On Sat, Jul 20, 2024 at 3:02?PM Iris Simmons <ikwsimmo at gmail.com> wrote:
It should be written more like this:
```R
z <- data.frame(a = 1:3, b = letters[1:3])
z |> names() |> _[2] <- "foo"
z
```
Regards,
Iris
On Sat, Jul 20, 2024 at 4:47?PM Bert Gunter <bgunter.4567 at gmail.com> wrote:
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
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