Using “…” and “replicate”

问题

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

Now, I can accept it as such, but would like to understand what is behind it. So I've created this little contrived example:

innerfunction<-function(x, extrapar1=0, extrapar2=extrapar1)
{
    cat("x:", x, ", xp1:", extrapar1, ", xp2:", extrapar2, "\n")
}

middlefunction<-function(x,...)
{
    innerfunction(x,...)
}

outerfunction<-function(x, ...)
{
    cat("Run middle function:\n")
    replicate(2, middlefunction(x,...))
    cat("Run inner function:\n")
    replicate(2, innerfunction(x,...))
}

outerfunction(1,2,3)
outerfunction(1,extrapar1=2,3)
outerfunction(1,extrapar1=2,extrapar2=3)

Perhaps I've done something obvious horribly wrong, but I find the result of this rather upsetting. So can anyone explain to me why, in all of the above calls to outerfunction, I get this output:

Run middle function:
x: 1 , xp1: 0 , xp2: 0 
x: 1 , xp1: 0 , xp2: 0 
Run inner function:
x: 1 , xp1: 0 , xp2: 0 
x: 1 , xp1: 0 , xp2: 0

Like I said: the docs seem to warn for this, but I do not see why this is so.

回答1:

?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().

回答2:

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.

回答3:

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

回答4:

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)

来源：https://stackoverflow.com/questions/6704536/using-and-replicate

标签

ellipsis

replicate

sapply