Simpler I think: ?all.vars
> all.vars(~A+B)
[1] "A" "B"
Note also:
> all.vars(~log(A))
[1] "A"
Cheers,
Bert
"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 Tue, Dec
> On Dec 6, 2016, at 7:33 AM, Rui Barradas wrote:
>
> 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
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)
myf
Ok, that's a way of seeing it.
Rui Barradas
Em 06-12-2016 14:28, John Sorkin escreveu:
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
Note that library has another argument, character.only=TRUE/FALSE,
to control whether the main argument should be regarded as a variable
or a literal. I think you need two arguments to handle this.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Dec 6, 2016 at 7:33 AM, Rui Barradas wrote:
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
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 re
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