Well, you can have exactly 70:30%, i.e. 70% 1s and 30% 0s, but in random order. For example:
Popn <- c(rep(1,70),rep(0,30)) Samp <- sample(Pop) (see '?sample' for this usage -- the result of sample(x) is a random permutation of the elements of x). In probabilistic terms, this is a "conditional" sample, i.e. what you would get by using sample(...,replace=TRUE) but rejecting samples which do not have 70% 1s and 30% 0s until you get a sample which does. Ted. On 05-Jul-11 18:25:48, Joshua Wiley wrote: > Hi Ana, > > Look at the documentation for ?sample, specifically, the "prob" > argument. In your case this should work: > > sample(c(0,1), 100, replace = TRUE, prob = c(.3, .7)) > > note that you may not have *exactly* 70% 1 and 30%, in any given > sample. > > HTH, > > Josh > > On Tue, Jul 5, 2011 at 11:21 AM, Ana Kolar <annako...@yahoo.com> wrote: >> Hi there, >> >> I guess this is an easy one, but still: >> >> I would like to randomly sample 0s and 1s but in a way that >> I end up having for example 70% of 1s and the rest of 0s and >> not 50:50 as this function does: sample(c(0,1), 100, replace = TRUE) >> >> Any recommendations? >> >> Many thanks! >> Ana -------------------------------------------------------------------- E-Mail: (Ted Harding) <ted.hard...@wlandres.net> Fax-to-email: +44 (0)870 094 0861 Date: 05-Jul-11 Time: 19:37:06 ------------------------------ XFMail ------------------------------ ______________________________________________ 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.
