[Rd] brief update on the pipe operator in R-devel

[Gabor Grothendieck]

These are documented but still seem like serious deficiencies:

> f <- function(x, y) x + 10*y
> 3 |> x => f(x, x)
Error in f(x, x) : pipe placeholder may only appear once

> 3 |> x => f(1+x, 1)
Error in f(1 + x, 1) :
  pipe placeholder must only appear as a top-level argument in the RHS call

Also note:

 ?"=>"
No documentation for ‘=>’ in specified packages and libraries:
you could try ‘??=>’

On Tue, Dec 22, 2020 at 5:28 PM <luke-tierney using uiowa.edu> wrote:
>
> It turns out that allowing a bare function expression on the
> right-hand side (RHS) of a pipe creates opportunities for confusion
> and mistakes that are too risky. So we will be dropping support for
> this from the pipe operator.
>
> The case of a RHS call that wants to receive the LHS result in an
> argument other than the first can be handled with just implicit first
> argument passing along the lines of
>
>      mtcars |> subset(cyl == 4) |> (\(d) lm(mpg ~ disp, data = d))()
>
> It was hoped that allowing a bare function expression would make this
> more convenient, but it has issues as outlined below. We are exploring
> some alternatives, and will hopefully settle on one soon after the
> holidays.
>
> The basic problem, pointed out in a comment on Twitter, is that in
> expressions of the form
>
>      1 |> \(x) x + 1 -> y
>      1 |> \(x) x + 1 |> \(y) x + y
>
> everything after the \(x) is parsed as part of the body of the
> function.  So these are parsed along the lines of
>
>      1 |> \(x) { x + 1 -> y }
>      1 |> \(x) { x + 1 |> \(y) x + y }
>
> In the first case the result is assigned to a (useless) local
> variable.  Someone writing this is more likely to have intended to
> assign the result to a global variable, as this would:
>
>      (1 |> \(x) x + 1) -> y
>
> In the second case the 'x' in 'x + y' refers to the local variable 'x'
> in the first RHS function. Someone writing this is more likely to have
> meant
>
>      (1 |> \(x) x + 1) |> \(y) x + y
>
> with 'x' in 'x + y' now referring to a global variable:
>
>      > x <- 2
>      > 1 |> \(x) x + 1 |> \(y) x + y
>      [1] 3
>      > (1 |> \(x) x + 1) |> \(y) x + y
>      [1] 4
>
> These issues arise with any approach in R that allows a bare function
> expression on the RHS of a pipe operation. It also arises in other
> languages with pipe operators. For example, here is the last example
> in Julia:
>
>      julia> x = 2
>      2
>      julia> 1 |> x -> x + 1 |> y -> x + y
>      3
>      julia> ( 1 |> x -> x + 1 ) |> y -> x + y
>      4
>
> Even though proper use of parentheses can work around these issues,
> the likelihood of making mistakes that are hard to track down is too
> high. So we will disallow the use of bare function expressions on the
> right hand side of a pipe.
>
> Best,
>
> luke
>
> --
> Luke Tierney
> Ralph E. Wareham Professor of Mathematical Sciences
> University of Iowa                  Phone:             319-335-3386
> Department of Statistics and        Fax:               319-335-3017
>     Actuarial Science
> 241 Schaeffer Hall                  email:   luke-tierney using uiowa.edu
> Iowa City, IA 52242                 WWW:  http://www.stat.uiowa.edu
>
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