[Rd] Quiz: How to get a "named column" from a data frame
Bert Gunter
gunter.berton at gene.com
Sat Aug 18 18:39:02 CEST 2012
Sorry! -- Change that to drop = FALSE !
drop(t(df[,1,drop=FALSE]))
t(df[,1,drop=FALSE])[1,]
-- Bert
On Sat, Aug 18, 2012 at 9:37 AM, Bert Gunter <bgunter at gene.com> wrote:
> Yes, but either
>
> drop(t(df[,1,drop=TRUE]))
>
> or
>
> t(df[,1,drop=TRUE])[1,]
>
> does work. My minimal effort to check timings found that the first
> version was a hair faster.
>
> -- Bert
>
> On Sat, Aug 18, 2012 at 9:01 AM, Rui Barradas <ruipbarradas at sapo.pt> wrote:
>> Hello,
>>
>> 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[1] )) ) # TRUE
>> identical( nv, drop(t( df2[[1]] )) ) # FALSE
>>
>> Rui Barradas
>>
>> 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)))
>>> [1] TRUE
>>>
>>> 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
>>
>>
>> ______________________________________________
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>
>
>
> --
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
>
> Internal Contact Info:
> Phone: 467-7374
> Website:
> http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
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