[R] do.call environment misunderstanding

[David Winsemius]

On Jun 25, 2013, at 6:05 PM, Dan Murphy wrote:

> My problem is to evaluate a function/model whose definition and
> parameters (I'll put x into the arguments) and other data are saved by
> someone else in an Rdata file, but I don't know the function name,
> definition or data. Nevertheless, I need to save whatever functional
> values/model solutions are so determined and to compare and contrast
> solutions of same and similar problems over time. Environments seemed
> to hold promise, but understanding the relationship between f and e
> was my first hurdle. Thanks again.
> 

The load() function has an optional 'envir' parameter. Then you can just use ls().

> xx <- pi; f <- function() print(xx) 
> save(xx, f,  file= "test.RData")
> rm(list=c('xx','f') 
+ )
> load("test.RData", envir=env)
> ls(env)
[1] "f"  "xx"
> find('f')
character(0)
> grep("f", ls(envir=env, all.names=TRUE))
[1] 1
> grep("f", ls(envir=env))
[1] 1
> apropos("f", where=env)
Error in if (where) structure(li, names = rep.int(i, length(li))) else li : 
  argument is not interpretable as logical
In addition: Warning message:
In if (where) structure(li, names = rep.int(i, length(li))) else li :
  the condition has length > 1 and only the first element will be used
> grep("xx", ls(envir=env))
[1] 2

-- 
David.
> On Tue, Jun 25, 2013 at 9:54 AM, Duncan Murdoch
> <murdoch.duncan at gmail.com> wrote:
>> On 25/06/2013 11:56 AM, Dan Murphy wrote:
>>> 
>>> So the trick is to put the function f into e and define its environment to
>>> be e:
>> 
>> 
>> Putting f into e, and defining the environment of f to be e solve different
>> problems.  Your toy example has both problems so it's a reasonable solution
>> there, but most real examples don't, so I wouldn't think of those two
>> solutions as being connected.
>> 
>> Duncan Murdoch
>> 
>> 
>>>> e <- new.env()
>>>> e$f <- function() x^2
>>>> environment(e$f) <- e
>>>> e$x <- 2
>>>> do.call("f", list(), envir = e)
>>> [1] 4
>>> 
>>> Thanks, Duncan.
>>> 
>>> On Tue, Jun 25, 2013 at 6:49 AM, Duncan Murdoch
>>> <murdoch.duncan at gmail.com> wrote:
>>>> On 25/06/2013 9:32 AM, Dan Murphy wrote:
>>>>> 
>>>>> I am having difficulty understanding the envir argument of do.call.
>>>>> The help page says
>>>>> 
>>>>> envir  an environment within which to evaluate the call.
>>>>> 
>>>>> so I thought that in the following toy example x would be found in the
>>>>> environment e and f would return 4 via do.call:
>>>>> 
>>>>>> e <- new.env()
>>>>>> e$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
>> 
>> 
> 
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David Winsemius
Alameda, CA, USA