Re: [Rd] Quiz: How to get a named column from a data frame
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])) 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
Re: [Rd] Quiz: How to get a named column from a data frame
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
[Rd] Suggestion for Writing R Extensions
Hi, Early in Chapter 1, the manual states: A computing environment including a number of tools is assumed; the R Installation and Administration manual describes what is needed. Under a Unix-alike most of the tools are likely to be present by default, but Microsoft Windows may require careful setup. Would it make sense to add links and/or mention the relevant appendices (A and D) of R Installation and Administration? Best, -- Joshua Ulrich | about.me/joshuaulrich FOSS Trading | www.fosstrading.com __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] Suggestion for Writing R Extensions
Hello, Zitat von Joshua Ulrich josh.m.ulr...@gmail.com (Wed, 22 Aug 2012 12:35:51 -0500) [...] Would it make sense to add links and/or mention the relevant appendices (A and D) of R Installation and Administration? [...] I do not understad, what your question is all about. Maybe you should be more verbose. Do you want to make the document better, or do you have questions on how to understand what it talks about? Ciao, Oliver __ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
Re: [Rd] Suggestion for Writing R Extensions
On Wed, Aug 22, 2012 at 2:58 PM, Oliver Bandel oli...@first.in-berlin.de wrote: Hello, Zitat von Joshua Ulrich josh.m.ulr...@gmail.com (Wed, 22 Aug 2012 12:35:51 -0500) [...] Would it make sense to add links and/or mention the relevant appendices (A and D) of R Installation and Administration? [...] I do not understad, what your question is all about. Maybe you should be more verbose. Do you want to make the document better, or do you have questions on how to understand what it talks about? The sentences I quoted, which you removed from the thread, point the reader to R Installation and Administration. Rather than point the reader to the entire manual, I suggest pointing them to the relevant appendices (A and D) of R Installation and Administration. Ciao, Oliver __ 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
