[Rd] Quiz: How to get a "named column" from a data frame
ruipbarradas at sapo.pt
Sat Aug 18 18:01:18 CEST 2012
A bit more general
nv <- c(a=1, d=17, e=101); nv
nv2 <- c(a="a", d="d", e="e")
df2 <- data.frame(VAR = nv, CHAR = nv2); df2
identical( nv, drop(t( df2 )) ) # TRUE
identical( nv, drop(t( df2[] )) ) # FALSE
Em 18-08-2012 16:16, Joshua Ulrich escreveu:
> 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)))
>  TRUE
> 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
>> a 1
>> d 17
>> e 101
>> Now how, can I get 'nv' back from 'df' ? I.e., how to get
>>> identical(nv, .......)
>>  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.
>> R-devel at r-project.org mailing list
> R-devel at r-project.org mailing list
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