[Rd] Quiz: How to get a "named column" from a data frame
Joshua Ulrich
josh.m.ulrich at gmail.com
Wed Aug 22 16:15:00 CEST 2012
On Tue, Aug 21, 2012 at 2:34 PM, Duncan Murdoch
<murdoch.duncan at gmail.com> wrote:
> On 12-08-18 12:33 PM, Martin Maechler wrote:
>>>>>>>
>>>>>>> Joshua Ulrich <josh.m.ulrich at gmail.com>
>>>>>>> on Sat, 18 Aug 2012 10:16:09 -0500 writes:
>>
>>
>> > I don't know if this is better, but it's the most obvious/shortest
>> I
>> > could come up with. Transpose the data.frame column to a 'row'
>> vector
>> > and drop the dimensions.
>>
>> R> identical(nv, drop(t(df)))
>> > [1] TRUE
>>
>> Yes, that's definitely shorter,
>> congratulations!
>>
>> One gotta is that I'd want a solution that also works when the
>> df has more columns than just one...
>>
>> Your idea to use t(.) is nice and "perfect" insofar as it
>> coerces the data frame to a matrix, and that's really the clue:
>>
>> Where as df[,1] is losing the names,
>> the matrix indexing is not.
>> So your solution can be changed into
>>
>> t(df)[1,]
>>
>> which is even shorter...
>> and slightly less efficient, at least conceptually, than mine, which has
>> been
>>
>> as.matrix(df)[,1]
>>
>> Now, the remaining question is: Shouldn't there be something
>> more natural to achieve that?
>> (There is not, currently, AFAIK).
>
>
> I've been offline, so I'm a bit late to this game, but the examples above
> fail when df contains a character column as well as the desired one, because
> everything gets coerced to a character matrix. You need to select the
> column first, then convert to a matrix, e.g.
>
> drop(t(df[,1,drop=FALSE]))
>
That's true, but I was assuming a one-column data.frame, which can be
achieved via:
df <- data.frame(VAR=nv,CHAR=letters[1:3],stringsAsFactors=FALSE)
drop(t(df[1]))
That said, I prefer the setNames() solution for its efficiency.
Best,
Josh
> Duncan Murdoch
>
>
>>
>> Martin
>>
>>
>> > Best,
>> > --
>> > Joshua Ulrich | about.me/joshuaulrich
>> > FOSS Trading | www.fosstrading.com
>>
>>
>> > On Sat, Aug 18, 2012 at 10:03 AM, Martin Maechler
>> > <maechler at stat.math.ethz.ch> wrote:
>> >> Today, I was looking for an elegant (and efficient) way to get a
>> named
>> >> (atomic) vector by selecting one column of a data frame. Of
>> course,
>> >> the vector names must be the rownames of the data frame.
>> >>
>> >> Ok, here is the quiz, I know one quite "cute"/"slick" answer, but
>> was
>> >> wondering if there are obvious better ones, and also if this
>> should
>> >> not become more idiomatic (hence "R-devel"):
>> >>
>> >> Consider this toy example, where the dataframe already has only
>> one
>> >> column :
>> >>
>> >>> nv <- c(a=1, d=17, e=101); nv
>> >> a d e
>> >> 1 17 101
>> >>
>> >>> df <- as.data.frame(cbind(VAR = nv)); df
>> >> VAR
>> >> a 1
>> >> d 17
>> >> e 101
>> >>
>> >> Now how, can I get 'nv' back from 'df' ? I.e., how to get
>> >>
>> >>> identical(nv, .......)
>> >> [1] TRUE
>> >>
>> >> where ...... only uses 'df' (and no non-standard R packages)?
>> >>
>> >> As said, I know a simple solution (*), but I'm sure it is not
>> >> obvious to most R users and probably not even to the majority of
>> >> R-devel readers... OTOH, people like Bill Dunlap will not take
>> >> long to provide it or a better one.
>> >>
>> >> (*) In my solution, the above '.......' consists of 17 letters.
>> >> I'll post it later today (CEST time) ... or confirm
>> >> that someone else has done so.
>> >>
>> >> Martin
>> >>
>> >> ______________________________________________
>> >> R-devel at r-project.org mailing list
>> >> https://stat.ethz.ch/mailman/listinfo/r-devel
>>
>> ______________________________________________
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>
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