Hi, the cut function might be helpful.
vec=1: 163863 fcut=cut(vec,seq(1, 163863+1,by= 6069),include.lowest = T,right=F) aggregate(vec,by=list(fcut),min) aggregate(vec,by=list(fcut),max) cheers Peter On 25. Apr 2017, at 14:33, PIKAL Petr <petr.pi...@precheza.cz<mailto:petr.pi...@precheza.cz>> wrote: Hi -----Original Message----- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Thomas Mailund Sent: Tuesday, April 25, 2017 11:20 AM To: r-help@r-project.org<mailto:r-help@r-project.org>; Saifuddin, Miah Mohammad <miah.mohammadsaif...@mavs.uta.edu<mailto:miah.mohammadsaif...@mavs.uta.edu>> Subject: Re: [R] asking for help If you write something like indices <- rep(1:(163863/6069), each = 6069) You can get similar result by indices2 <- 0:163862%/%6069 but starting with zero. or the same with indices2 <- (0:163862%/%6069)+1 Cheers Petr you can get the i’th block of rows with table[indices == i,] It looks like a little more work than a loop would be since you have to run through all rows for each block, but the implicit loop in this approach is likely to be faster than an explicit for-loop. Cheers Thomas On 25 Apr 2017, 07.01 +0200, Saifuddin, Miah Mohammad <miah.mohammadsaif...@mavs.uta.edu<mailto:miah.mohammadsaif...@mavs.uta.edu>>, wrote: I have a data frame having 163863 values. I want to subset it so that each set has 6069 values in it. for example 1:6069 as first, 6070: 6070+6068 as second. how can I do that, preferably in a loop. TIA [[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. 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