How to get the namespace of a function?

library(zoo)
f <- function(x) eval(substitute(index(x), list(x = x)),

environment(f) <- NULL
x <- ts(1:4)
f(x)

library(zoo)
f <- function(x) eval.parent(substitute(index(x)))
environment(f) <- NULL
x <- ts(1:4)
f(x)

Thanks, Gabor. I tried this without success:

z <- ts(1:5)
f <- function(x) {f1 <- function() get("index", "package:zoo"); f1()(x)}
environment(f) <- NULL
f(z)

Error in f1()(x) : no applicable method for "index"

I also tried this, which seemed simpler, with the same outcome:

z <- ts(1:5)
f <- function(x) {f1 <- get("index", "package:zoo"); f1(x)}
environment(f) <- NULL
f(z)

Error in f1(x) : no applicable method for "index"

FS



On 2/2/06, Gabor Grothendieck <ggrothendieck at gmail.com> wrote:

Try:

f <- function() get("index", "package:zoo")


On 2/2/06, Fernando Saldanha <fsaldan1 at gmail.com> wrote:

I declared the environment of the function myfun to be NULL as follows:

environment(myfun) <- NULL

Later on I called that myfun and got an error message because the
function index() in the zoo package was called inside myfun and was
not visible:

Error in myfun(args) : couldn't find function "index"

I tried to use zoo::index() instead of index(), but that did not work.
In fact, zoo::index does not work even in the command line:

z<-ts(1:5)
z

Time Series:
Start = 1
End = 5
Frequency = 1
[1] 1 2 3 4 5

index(z)

[1] 1 2 3 4 5

zoo::index(z)

Error in loadNamespace(name) : package 'zoo' does not have a name space

How can I qualify index() so that it is visible inside the body of myfun?

Thanks for any suggestions,

FS