Thank you. Using ‘each’ is shorter code, so better if the requested order was important versus being an initial solution.
blocC <- c(rep(x=c(1:13), times=84)) #it makes 84 copies of the numbers 1:13 blocC <- arrange(.data = data.frame(blocC), blocC) #sort to get requested order. blocD <- c(rep(x=c(1:13), each=84)) #it makes 84 copies of 1, then 84 copies of 2 … blocD <- data.frame(blocC=blocD) #give the column the same name. identical(blocC, blocD) all.equal(blocC, blocD) and both identical() and all.equal() return TRUE. Either way gives the same end product. Tim From: Bert Gunter <bgunter.4...@gmail.com> Sent: Thursday, June 13, 2024 9:13 PM To: Ebert,Timothy Aaron <teb...@ufl.edu> Cc: Francesca PANCOTTO <francesca.panco...@unimore.it>; r-help@r-project.org Subject: Re: [R] Create a numeric series in an efficient way [External Email] Nope. She would have wanted the 'each' argument = 84. See ?rep. -- Bert On Thu, Jun 13, 2024 at 3:54 PM Ebert,Timothy Aaron <teb...@ufl.edu<mailto:teb...@ufl.edu>> wrote: Maybe this was your solution? blocC <- c(rep(x=c(1:13), times=84)) blocC <- arrange(.data = data.frame(blocC), blocC) The second line sorts, but that may not be needed depending on application. The object class is also different in the sorted solution. Tim -----Original Message----- From: R-help <r-help-boun...@r-project.org<mailto:r-help-boun...@r-project.org>> On Behalf Of Francesca PANCOTTO via R-help Sent: Thursday, June 13, 2024 2:22 PM To: r-help@r-project.org<mailto:r-help@r-project.org> Subject: Re: [R] Create a numeric series in an efficient way [External Email] I apologize, I solved the problem, sorry for that. f. Il giorno gio 13 giu 2024 alle ore 16:42 Francesca PANCOTTO < francesca.panco...@unimore.it<mailto:francesca.panco...@unimore.it>> ha scritto: > Dear Contributors > I am trying to create a numeric series with repeated numbers, not > difficult task, but I do not seem to find an efficient way. > > This is my solution > > blocB <- c(rep(x = 1, times = 84), rep(x = 2, times = 84), rep(x = 3, > times = 84), rep(x = 4, times = 84), rep(x = 5, times = 84), rep(x = > 6, times = 84), rep(x = 7, times = 84), rep(x = 8, times = 84), rep(x > = 9, times = 84), rep(x = 10, times = 84), rep(x = 11, times = 84), > rep(x = 12, times = 84), rep(x = 13, times = 84)) > > which works but it is super silly and I need to create different > variables similar to this, changing the value of the repetition, 84 in this > case. > Thanks for any help. > > > F. > [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org<mailto: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. ______________________________________________ R-help@r-project.org<mailto: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<http://www.r-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] ______________________________________________ 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.
