[R] do.call environment misunderstanding

[Duncan Murdoch]

On 25/06/2013 9:32 AM, Dan Murphy wrote:
> I am having difficulty understanding the envir argument of do.call.
> The help page says
>
> envir  an environment within which to evaluate the call.
>
> so I thought that in the following toy example x would be found in the
> environment e and f would return 4 via do.call:
>
> > e <- new.env()
> > e$x <- 2
> > f <- function() x^2
> > do.call(f, list(), envir = e)
> Error in (function ()  : object 'x' not found
>
> Thanks in advance for clarifying my misunderstanding.

do.call will construct the expression f(), then evaluate it in e. It 
will try to look up f there, and not finding it, will go to the parent 
environment and find it.

When evaluating the function, the environment in which it was evaluated 
is used for looking up arguments, but f() has none, so e is not used at 
all.  R will use the environment attached to f, which is the global 
environment, since you created f by evaluating its definition there.

To get what you want, you could use the sequence

e <- new.env()
e$x <- 2
f <- function() x^2
environment(f) <- e
f()

An alternative way to do the 3rd and 4th lines is

f <- with(e, function() x^2)

because that would evaluate the creation of f within e.

A third approach (which might be the nicest one, depending on what else 
you are doing) is never to name e:

f <- local({
   x <- 2
   function() x^2
})

Duncan Murdoch