Using “…” and “replicate”

Question

In the documentation of sapply and replicate there is a warning regarding using ...

Now, I can accept it as such, but would li

Answer 1

An alternative way to do that:

g <- function(x, y) x + y

f <- function(a = 1, ...) {
    arg_list <- list(...)
    replicate(n = 3, expr = do.call(g, args = arg_list))
}

f(x = 1, y = 2)

Answer 2

?replicate, in the Examples section, tells us explicitly that what you are trying to do does not and will not work. In the Note section of ?replicate we have:

     If ‘expr’ is a function call, be aware of assumptions about where
     it is evaluated, and in particular what ‘...’ might refer to.  You
     can pass additional named arguments to a function call as
     additional named arguments to ‘replicate’: see ‘Examples’.

And if we look at Examples, we see:

 ## use of replicate() with parameters:
 foo <- function(x=1, y=2) c(x,y)
 # does not work: bar <- function(n, ...) replicate(n, foo(...))
 bar <- function(n, x) replicate(n, foo(x=x))
 bar(5, x=3)

My reading of the docs is that they do far more than warn you about using ... in replicate() calls; they explicitly document that it does not work. Much of the discussion in that help file relates to the ... argument of the other functions, not necessarily to replicate().

Answer 3

If you look at the code for replicate:

> replicate
function (n, expr, simplify = TRUE) 
sapply(integer(n), eval.parent(substitute(function(...) expr)), 
    simplify = simplify)
<environment: namespace:base>

You see that the function is evaluated in the parent frame, where the ... from your calling function no longer exists.

Answer 4

There actually is a way to do this:

# Simple function:
ff <- function(a,b) print(a+b)

# This will NOT work:
testf <- function(...) {
  replicate(expr = ff(...), n = 5)
}
testf(45,56) # argument "b" is missing, with no default

# This will:
testf <- function(...) {
  args <- as.list(substitute(list(...)))[-1L]
  replicate(expr = do.call(ff, args), n = 5)
}
testf(45,56) # 101