[R] Write a function that allows access to columns of a passed dataframe.

[Bert Gunter]

Typo: "lazy evaluation" not "lay evaluation."

-- Bert



Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Mon, Dec 5, 2016 at 1:46 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote:
> Sorry, hit "Send" by mistake.
>
> Inline.
>
>
>
> On Mon, Dec 5, 2016 at 1:34 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote:
>> Inline.
>>
>> -- Bert
>>
>>
>> Bert Gunter
>>
>> "The trouble with having an open mind is that people keep coming along
>> and sticking things into it."
>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>
>>
>> On Mon, Dec 5, 2016 at 9:53 AM, Rui Barradas <ruipbarradas at sapo.pt> wrote:
>>> Hello,
>>>
>>> Inline.
>>>
>>> Em 05-12-2016 17:09, David Winsemius escreveu:
>>>>
>>>>
>>>>> On Dec 5, 2016, at 7:29 AM, John Sorkin <jsorkin at grecc.umaryland.edu>
>>>>> wrote:
>>>>>
>>>>> Rui,
>>>>> I appreciate your suggestion, but eliminating the deparse statement does
>>>>> not solve my problem. Do you have any other suggestions? See code below.
>>>>> Thank you,
>>>>> John
>>>>>
>>>>>
>>>>> mydf <-
>>>>> data.frame(id=c(1,2,3,4,5),sex=c("M","M","M","F","F"),age=c(20,34,43,32,21))
>>>>> mydf
>>>>> class(mydf)
>>>>>
>>>>>
>>>>> myfun <- function(frame,var){
>>>>>   call <- match.call()
>>>>>   print(call)
>>>>>
>>>>>
>>>>>   indx <- match(c("frame","var"),names(call),nomatch=0)
>>>>>   print(indx)
>>>>>   if(indx[1]==0) stop("Function called without sufficient arguments!")
>>>>>
>>>>>
>>>>>   cat("I can get the name of the dataframe as a text string!\n")
>>>>>   #xx <- deparse(substitute(frame))
>>>>>   print(xx)
>>>>>
>>>>>
>>>>>   cat("I can get the name of the column as a text string!\n")
>>>>>   #yy <- deparse(substitute(var))
>>>>>   print(yy)
>>>>>
>>>>>
>>>>>   # This does not work.
>>>>>   print(frame[,var])
>>>>>
>>>>>
>>>>>   # This does not work.
>>>>>   print(frame[,"var"])
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>   # This does not work.
>>>>>   col <- xx[,"yy"]
>>>>>
>>>>>
>>>>>   # Nor does this work.
>>>>>   col <- xx[,yy]
>>>>>   print(col)
>>>>> }
>>>>>
>>>>>
>>>>> myfun(mydf,age)
>>>>
>>>>
>>>>
>>>> When you use that calling syntax, the system will supply the values of
>>>> whatever the `age` variable contains. (And if there is no `age`-named
>>>> object, you get an error at the time of the call to `myfun`.
>>>
>>>
>>> Actually, no, which was very surprising to me but John's code worked (not
>>> the function, the call). And with the change I've proposed, it worked
>>> flawlessly. No errors. Why I don't know.
>
> See ?substitute and in particular the example highlighted there.
>
> The technical details are explained in the R Language Definition
> manual. The key here is the use of promises for lay evaluations. In
> fact, the expression in the call *is* available within the functions,
> as is (a pointer to) the environment in which to evaluate the
> expression. That is how substitute() works. Specifically, quoting from
> the manual,
>
> *****
> It is possible to access the actual (not default) expressions used as
> arguments inside the function. The mechanism is implemented via
> promises. When a function is being evaluated the actual expression
> used as an argument is stored in the promise together with a pointer
> to the environment the function was called from. When (if) the
> argument is evaluated the stored expression is evaluated in the
> environment that the function was called from. Since only a pointer to
> the environment is used any changes made to that environment will be
> in effect during this evaluation. The resulting value is then also
> stored in a separate spot in the promise. Subsequent evaluations
> retrieve this stored value (a second evaluation is not carried out).
> Access to the unevaluated expression is also available using
> substitute.
> ********
>
> -- Bert
>
>
>
>
>>>
>>> Rui Barradas
>>>
>>>  You need either to call it as:
>>>>
>>>>
>>>> myfun( mydf , "age")
>>>>
>>>>
>>>> # Or:
>>>>
>>>> age <- "age"
>>>> myfun( mydf, age)
>>>>
>>>> Unless your value of the `age`-named variable was "age" in the calling
>>>> environment (and you did not give us that value in either of your postings),
>>>> you would fail.
>>>>
>>>
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