From murdoch.duncan at gmail.com  Tue Aug 21 21:34:13 2012
From: murdoch.duncan at gmail.com (Duncan Murdoch)
Date: Tue, 21 Aug 2012 15:34:13 -0400
Subject: [Rd] Quiz: How to get a "named column" from a data frame
In-Reply-To: <20527.50144.797572.795413@stat.math.ethz.ch>
References: <20527.44736.486888.854733@stat.math.ethz.ch>
	<CAPPM_gRmc+VE3ZGJ3TzTn6m4qb4ToUq30X0USATmmEjxq3u4dA@mail.gmail.com>
	<20527.50144.797572.795413@stat.math.ethz.ch>
Message-ID: <5033E2B5.8000308@gmail.com>

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]))

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
>
> ______________________________________________
> R-devel at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
>



[Rd] Quiz: How to get a "named column" from a data frame
Duncan Murdoch