On Tue, Aug 21, 2012 at 2:34 PM, Duncan Murdoch <murdoch.dun...@gmail.com> wrote: > On 12-08-18 12:33 PM, Martin Maechler wrote: >>>>>>> >>>>>>> Joshua Ulrich <josh.m.ulr...@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 >> > <maech...@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@r-project.org mailing list >> >> https://stat.ethz.ch/mailman/listinfo/r-devel >> >> ______________________________________________ >> R-devel@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-devel >> > ______________________________________________ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
