Thank a lot for these answers.
Some of you wondered why I needed
to do that!
In fact, I have not so big data.frame whith many
columns (98) and many of them are similar (many binary answers, some
factor data and a few quantitative datas).
As I am a lazy guy, I
wanted to do a sort of function that would apply some descriptive
statistics (and maybe some graphics) to each on them regarding to its
class.
And I wanted the results to be outputed as :
"Descriptive
statistics for : Bar"
results....
then, this would allow me to do a :
For (i in names(mydata)) desc(mydata[[i]])
In
fact, I was testing this function with a : desc(mydata$Bar)
where
desc was my lazy guy function defined as:
desc <- function(x) {
print(paste("descriptive
statistics for ",x))
if (is.factor(x)) {
print("Raw count:")
print(freq(x, total=T))
print("Net counts:")
print(freq(x, total=T, exclude=NA))
}
else print(summary(x))
}
PS: freq is another function defined above
Regards
________________________________
De : Prof Brian Ripley <rip...@stats.ox.ac.uk>
Cc : R-help@r-project.org
Envoyé le : Lun 3 mai 2010, 13 h 40 min 15 s
Objet : Re: [R] Extract a 'data.frame$column' column name
> as.character(quote(foo$Bar)[[3]])
[1] "Bar"
Hint: this is nothing to do with data frames ($ applies to lists). $ is an
operator, so foo$Bar is a call. quote() stops it being evaluated, [[3]]
selects the third of the elements (which are $, foo, Bar) and as.character
turns the name into a character string.
On Mon, 3 May 2010, adrien Penac wrote:
> Hello,
> I can't find how to get de column name from a data.frame dollar reference.
>
> To make it simple, I'd like to obtain "Bar" from a "foo$Bar" notation.
> I've tried col.names(foo$Bar), names(foo$Bar) and so on without sucess.
>
> Regards
>
> Blaise
-- Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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