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 ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
