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

From: Martin Maechler <maechler_at_stat.math.ethz.ch>
Date: Sat, 18 Aug 2012 18:33:36 +0200

>>>>> 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).

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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