Yes, I think unlist() is still better. One caution (for all): make sure the columns are all of the same type/class/mode or you may be in for nasty surprises.
Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 "Data is not information. Information is not knowledge. And knowledge is certainly not wisdom." Clifford Stoll On Thu, Apr 30, 2015 at 9:32 AM, Marc Schwartz <marc_schwa...@me.com> wrote: > Hi, > > Given that a data frame is a list: > > unlist(mydata[, 1:3]) > > For example: > >> all(unlist(iris[, 1:3]) == do.call(c, iris[, 1:3])) > [1] TRUE > > > Also, note that the returned result in both cases above retains names: > >> unlist(iris[, 1:3]) > Sepal.Length1 Sepal.Length2 Sepal.Length3 Sepal.Length4 > 5.1 4.9 4.7 4.6 > Sepal.Length5 Sepal.Length6 Sepal.Length7 Sepal.Length8 > 5.0 5.4 4.6 5.0 > Sepal.Length9 Sepal.Length10 Sepal.Length11 Sepal.Length12 > 4.4 4.9 5.4 4.8 > ... > > You can just get the plain vector by using: > >> unlist(iris[, 1:3], use.names = FALSE) > [1] 5.1 4.9 4.7 4.6 5.0 5.4 4.6 5.0 4.4 4.9 5.4 4.8 4.8 4.3 5.8 5.7 > [17] 5.4 5.1 5.7 5.1 5.4 5.1 4.6 5.1 4.8 5.0 5.0 5.2 5.2 4.7 4.8 5.4 > [33] 5.2 5.5 4.9 5.0 5.5 4.9 4.4 5.1 5.0 4.5 4.4 5.0 5.1 4.8 5.1 4.6 > [49] 5.3 5.0 7.0 6.4 6.9 5.5 6.5 5.7 6.3 4.9 6.6 5.2 5.0 5.9 6.0 6.1 > ... > > > As Bert notes, the indices would need to be adjusted for the actual data > frame in question. > > Regards, > > Marc > > > >> On Apr 30, 2015, at 10:52 AM, Bert Gunter <gunter.ber...@gene.com> wrote: >> >> ... and if this is what is wanted, somewhat cleaner and more >> generalizable for programming would be: >> >> do.call(c, mydata[,1:3]) >> >> ## where the column indices might have to be adjusted to get the >> desired columns. >> >> Cheers, >> >> Bert >> >> Bert Gunter >> Genentech Nonclinical Biostatistics >> (650) 467-7374 >> >> "Data is not information. Information is not knowledge. And knowledge >> is certainly not wisdom." >> Clifford Stoll >> >> >> >> >> On Wed, Apr 29, 2015 at 11:51 PM, Jim Lemon <drjimle...@gmail.com> wrote: >>> Hi Olufemi, >>> I sounds like you have a data frame (let's call it "mydata") with at >>> least three elements (columns). You may be trying to use c() in this >>> way: >>> >>> y1to3<-c(y1,y2,y3) >>> >>> in which case it won't work. However: >>> >>> y1to3<-c(mydata$y1,mydata$y2,mydata$y3) >>> >>> might do what you want, substituting whatever the name of your data >>> frame is for "mydata". >>> >>> Jim >>> >>> >>> On Thu, Apr 30, 2015 at 1:20 PM, Jeff Newmiller >>> <jdnew...@dcn.davis.ca.us> wrote: >>>> I am sure you can use c() because columns may be vectors even though >>>> vectors are not columns, but you really need to follow the posting guide >>>> and provide a reproducible example for us to show you how. You might find >>>> [1] helpful, in particular as it describes the use of the dput function to >>>> give us a few rows of your data. >>>> >>>> [1] >>>> http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example >>>> --------------------------------------------------------------------------- >>>> Jeff Newmiller The ..... ..... Go Live... >>>> DCN:<jdnew...@dcn.davis.ca.us> Basics: ##.#. ##.#. Live Go... >>>> Live: OO#.. Dead: OO#.. Playing >>>> Research Engineer (Solar/Batteries O.O#. #.O#. with >>>> /Software/Embedded Controllers) .OO#. .OO#. rocks...1k >>>> --------------------------------------------------------------------------- >>>> Sent from my phone. Please excuse my brevity. >>>> >>>> On April 29, 2015 7:48:24 PM PDT, Olufemi Bolarinwa >>>> <dafemli...@yahoo.co.uk> wrote: >>>>> Thank you Jeff for your response. >>>>> My y1, y2, y3 are actually 3 columns in the data so I cannot use the >>>>> c() function to concatenate them. I am confusing the "columns" with >>>>> vectors. I actually meant columns. >>>>> Any help will be much appreciated >>>>> Olufemi >>>>> >>>>> >>>>> >>>>> On Wednesday, 29 April 2015, 22:31, Jeff Newmiller >>>>> <jdnew...@dcn.davis.ca.us> wrote: >>>>> >>>>> >>>>> Vectors are not "columns" or "rows". Use the c() function to >>>>> concatenate vectors. >>>>> --------------------------------------------------------------------------- >>>>> Jeff Newmiller The ..... ..... Go >>>>> Live... >>>>> DCN:<jdnew...@dcn.davis.ca.us> Basics: ##.#. ##.#. Live >>>>> Go... >>>>> Live: OO#.. Dead: OO#.. Playing >>>>> Research Engineer (Solar/Batteries O.O#. #.O#. with >>>>> /Software/Embedded Controllers) .OO#. .OO#. >>>>> rocks...1k >>>>> --------------------------------------------------------------------------- >>>>> >>>>> Sent from my phone. Please excuse my brevity. >>>>> >>>>> On April 29, 2015 6:56:46 PM PDT, Olufemi Bolarinwa >>>>> <dafemli...@yahoo.co.uk> wrote: >>>>>> Hello,I am estimating a system of nonlinear equation where I need to >>>>>> stack my vector of y. I have data of about 6000units. I tried using >>>>> the >>>>>> rbind but instead of having a vector of 1 by 18000, it is giving me a >>>>> 3 >>>>>> by 6000 so that my matrix multiplication is non-conformable. The stack >>>>>> command requires an identifier but in this case, I do not have a >>>>> unique >>>>>> identifier. >>>>>> I would like to stack the the first 6000 units of y1 on the 2nd 6000 >>>>>> units of y2 and 6000 units of y3. >>>>>> Any help will be greatly appreciated. >>>>>> ThanksOlufemi > ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
