From josh.m.ulrich at gmail.com  Wed Aug 22 16:15:00 2012
From: josh.m.ulrich at gmail.com (Joshua Ulrich)
Date: Wed, 22 Aug 2012 09:15:00 -0500
Subject: [Rd] Quiz: How to get a "named column" from a data frame
In-Reply-To: <5033E2B5.8000308@gmail.com>
References: <20527.44736.486888.854733@stat.math.ethz.ch>
	<CAPPM_gRmc+VE3ZGJ3TzTn6m4qb4ToUq30X0USATmmEjxq3u4dA@mail.gmail.com>
	<20527.50144.797572.795413@stat.math.ethz.ch>
	<5033E2B5.8000308@gmail.com>
Message-ID: <CAPPM_gSOvpTxvCU3n7C_+TedAkG0j-_tMtZgwbg64Sk1qhFq0w@mail.gmail.com>

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
>>
>> ______________________________________________
>> 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
Joshua Ulrich