Write a function that allows access to columns of a passeddataframe.

Over my almost 50 years programming, I have come to believe that if one
wants a program to be useful, one should write the program to do as much
work as possible and demand as little as possible from the user of the
program. In my opinion, one should not ask the person who uses my
function to remember to put the name of the data frame column in
quotation marks. The function should be written so that all that needs
to be passed is the name of the column; the function should take care of
the quotation marks.
Jihny

John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and
Geriatric Medicine
Baltimore VA Medical Center
10 North Greene Street <x-apple-data-detectors://12>
GRECC <x-apple-data-detectors://12> (BT/18/GR)
Baltimore, MD 21201-1524 <x-apple-data-detectors://13/0>
(Phone) 410-605-711 <tel:410-605-7119>9
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prior to faxing)

On Dec 6, 2016, at 3:17 AM, Rui Barradas <ruipbarradas at sapo.pt
<mailto:ruipbarradas at sapo.pt>> wrote:

Hello,

Just to say that I wouldn't write the function as John did. I would get
rid of all the deparse/substitute stuff and instinctively use a quoted
argument as a column name. Something like the following.

myfun <- function(frame, var){
   [...]
   col <- frame[, var]  # or frame[[var]]
   [...]
}

myfun(mydf, "age")  # much better, simpler, no promises.

Rui Barradas

Em 05-12-2016 21:49, Bert Gunter escreveu:

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
<mailto: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
<mailto: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
<mailto: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 <mailto: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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