For all possible pairs, you'll have 20^2 pairs. This is a way to do it: expand.grid(x[1:20],x[1:20])
HTH, Erin On Wed, Oct 8, 2008 at 4:43 AM, Jim Lemon <[EMAIL PROTECTED]> wrote: > Stephen Cole wrote: >> >> ... >> I have a vector of 20 values >> >> x <- c(20,18, 45, 16, 47, 47, 15, 26, 14,14,12,16,35,27,18,94,16,26,26,30) >> >> 1. >> I want to select random pairs from this data set but do it without >> replacement exhaustively >> > > matrix(x[sample(1:20,20)],nrow=2) > > then step through the columns of the resulting matrix >> >> I know i can select random pairs without replacement using >> >> sample(N,n,replace=F) >> However i am wondering if there is any way to get 10 random pairs from >> this >> data set without repeating any of the data points >> that is to say if i got a (20, 94) for one pair, i would like to get 9 >> other >> pairs from the data without again getting 20 or 94? >> >> 2. >> The second thing i would like to do is be able to select all possible >> pairs >> of numbers and calculate each pairs variance. > > I think you want to use the combn function, but you are going to get a lot > of pairs... > > Jim > > ______________________________________________ > R-help@r-project.org mailing list > 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. > -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] ______________________________________________ R-help@r-project.org mailing list 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.
