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.
?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()
.
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.
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
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