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

Perhaps the best way is the one used by library(), where both
library(package) and library("package") work. It uses
as.charecter/substitute, not deparse/substitute, as follows.

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)
str(mydf)

myfun <- function(frame,var){
        yy <- as.character(substitute(var))
        frame[, yy]
}

myfun(mydf, age)
myfun(mydf, "age")

Rui Barradas

Em 06-12-2016 15:03, William Dunlap escreveu:

I basically agree with Rui - using substitute will cause trouble.  E.g.,
how
would the user iterate over the columns, calling your function for each?
      for(column in dataFrame) func(column)
would fail because dataFrame$column does not exist.  You need to provide
an extra argument to handle this case. something like the following:
      func <- function(df,
          columnAsName,,
          columnAsString = deparse(substitute(columnAsName))[1])
          ...
      }
The default value of columnAsString should also deal with the case that
the user supplied something like log(Conc.) instead of Conc.

I think that using a formula for the lazily evaluated argument
(columnAsName)
works well.  The user then knows exactly how it gets evaluated.

Bill Dunlap
TIBCO Software
wdunlap tibco.com <http://tibco.com>

On Tue, Dec 6, 2016 at 6:28 AM, John Sorkin <jsorkin at grecc.umaryland.edu
<mailto:jsorkin at grecc.umaryland.edu>> wrote:

    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
    > GRECC (BT/18/GR)
    > Baltimore, MD 21201-1524
    > (Phone)410-605-7119 <tel:410-605-7119>
    > (Fax)410-605-7913 <tel:410-605-7913> (Please call phone number

above
    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.
     >>>>>>

     >>>>>
     >>>>> ______________________________________________
     >>>>> R-help at r-project.org <mailto:R-help at r-project.org> mailing

    list -- To UNSUBSCRIBE and more, see

     >>>>> https://stat.ethz.ch/mailman/listinfo/r-help

    <https://stat.ethz.ch/mailman/listinfo/r-help>

     >>>>> PLEASE do read the posting guide