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
(Fax) 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> 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> 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.