On 08-12-2020 12:46, Gabor Grothendieck wrote:
Duncan Murdoch:
I agree it's all about call expressions, but they aren't all being
treated equally:

x |> f(...)

expands to f(x, ...), while

x |> `function`(...)

expands to `function`(...)(x).  This is an exception to the rule for

Yes, this is the problem.  It is trying to handle two different sorts of right
hand sides, calls and functions, using only syntax level operations and
it really needs to either make use of deeper information or have some
method that is available at the syntax level for identifying whether the
right hand side is a call or function.  In the latter case having two
operators would be one way to do it.

   f <- \(x) x + 1
   x |> f()  # call
   x |:> f  # function
   x |:> \(x) x + 1  # function

In the other case where deeper information is used there would only be one
operator and it would handle all cases but would use more than just syntax
level knowledge.

R solved these sorts of problems long ago using S3 and other object oriented
systems which dispatch different methods based on what the right hand side is.
The attempt to avoid using the existing or equivalent mechanisms seems to have
led to this problem.




I think only allowing functions on the right hand side (e.g. only the |> operator and not the |:>) would be enough to handle most cases and seems easier to reason about. The limitations of that can easily be worked around using existing functionality in the language.

The problem with only allowing

x |> mean

and not

x |> mean()

is with additional arguments. However, this can be solved with a currying function, for example:

x |> curry(mean, na.rm = TRUE)

The cost is a few additional characters.

In the same way it is possible to write a function that accepts an expression and returns a function containing that expression. This can be used to have expressions on the right-hand side and reduces the need for anonymous functions.

x |> fexpr(. + 10)
dta |> fexpr(lm(y ~ x, data = .))

You could call this function .:

x |> .(. + 10)
dta |> .(lm(y ~ x, data = .))


Dummy example code (thanks to  a colleague of mine)


fexpr <- function(expr){
  expr <- substitute(expr)
  f <- function(.) {}
  body(f) <- expr
  f
}
. <- fexpr

curry <- function(fun,...){
  L <- list(...)
  function(...){
    do.call(fun, c(list(...),L))
  }
}

`%|>%` <- function(e1, e2) {
  e2(e1)
}


1:10 %>% mean
c(1,3,NA) %|>% curry(mean, na.rm = TRUE)
c(1,3,NA) %|>% .( mean(., na.rm = TRUE) ) %>% identity
c(1,3,NA) %|>% .( . + 4)
c(1,3,NA) %|>% fexpr( . + 4)
c(1,3,NA) %|>% function(x) mean(x, na.rm = TRUE) %>% fexpr(. + 1)

--
Jan

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