On 09/12/2020 9:55 a.m., Jan van der Laan wrote:
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.
I agree that would be sufficient, but I don't see how it makes reasoning
easier. The transformation is trivial, so I'll assume that doesn't
consume any mental energy compared to understanding what the final
expression actually does. Using your currying example, the choice is
between
x |> mean(na.rm = TRUE)
which transforms to mean(x, na.rm = TRUE), or your proposed
x |> curry(mean, na.rm = TRUE)
which transforms to
curry(mean, na.rm = TRUE)(x)
To me curry(mean, na.rm = TRUE)(x) looks a lot more complicated than
mean(x, na.rm = TRUE), especially since it has the additional risk that
users can define their own function called "curry".
Duncan Murdoch
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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