[Rd] New pipe operator

[Deepayan Sarkar]

On Mon, Dec 7, 2020 at 9:23 PM Gabor Grothendieck
<ggrothendieck using gmail.com> wrote:
>
> One could examine how magrittr works as a reference implementation if
> there is a question on how something should function.  It's in
> widespread use and seems to work well.

Yes, but it has many inconsistencies (including for the example I
gave). Do you want a magrittr clone, or do you want consistency? It's
OK to want either, but I don't think you can get both.

What we actually end up with is another matter, depending on many
other factors. I was just trying to understand your consistency
argument.

-Deepayan

> On Mon, Dec 7, 2020 at 10:20 AM Deepayan Sarkar
> <deepayan.sarkar using gmail.com> wrote:
> >
> > On Mon, Dec 7, 2020 at 6:53 PM Gabor Grothendieck
> > <ggrothendieck using gmail.com> wrote:
> > >
> > > On Mon, Dec 7, 2020 at 5:41 AM Duncan Murdoch <murdoch.duncan using gmail.com> wrote:
> > > > 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
> > > > other calls, but I think it's a justified one.
> > >
> > > This admitted inconsistency is justified by what?  No argument has been
> > > presented.  The justification seems to be implicitly driven by implementation
> > > concerns at the expense of usability and language consistency.
> >
> > Sorry if I have missed something, but is your consistency argument
> > basically that if
> >
> > foo <- function(x) x + 1
> >
> > then
> >
> > x |> foo
> > x |> function(x) x + 1
> >
> > should both work the same? Suppose it did. Would you then be OK if
> >
> > x |> foo()
> >
> > no longer worked as it does now, and produced foo()(x) instead of foo(x)?
> >
> > If you are not OK with that and want to retain the current behaviour,
> > what would you want to happen with the following?
> >
> > bar <- function(x) function(n) rnorm(n, mean = x)
> >
> > 10 |> bar(runif(1))() # works 'as expected' ~ bar(runif(1))(10)
> > 10 |> bar(runif(1)) # currently bar(10, runif(1))
> >
> > both of which you probably want. But then
> >
> > baz <-  bar(runif(1))
> > 10 |> baz
> >
> > (not currently allowed) will not be the same as what you would want from
> >
> > 10 |> bar(runif(1))
> >
> > which leads to a different kind of inconsistency, doesn't it?
> >
> > -Deepayan
>
>
>
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