On 08-Nov-11 08:59:38, Ted Harding wrote: > On 08-Nov-11 07:46:15, flokke wrote: >> Sorry, but I dont think that I get what you mean (I am a >> quite new user though) >> I hae a data file of a sample, so I cannot use random numbers, >> the only thing I want to do is to randomly reassign the order >> of the items. > > The simplest method of randomly re-ordering is to use > sample() to re-arrange (1:N) randomly, where N is the > number of items in the data. Then use the result to > access the items. Example: > > D <- data.frame(X1=c(1.1,2.1,3.1,4.1), > X2=c(1.2,2.2,3.2,4.2)) > N <- nrow(D) > ix <- sample((1:N)) > D > # X1 X2 > # 1 1.1 1.2 > # 2 2.1 2.2 > # 3 3.1 3.2 > # 4 4.1 4.2 > N > # [1] 4 > ix > # [1] 3 2 4 1 > D[ix,] > # X1 X2 > # 3 3.1 3.2 > # 2 2.1 2.2 > # 4 4.1 4.2 > # 1 1.1 1.2 > > Note that the defaults for sample() are (see ?sample): > > For 'sample' the default for 'size' is the number > of items inferred from the first argument, so that > 'sample(x)' generates a random permutation of the > elements of 'x' (or '1:x'). > > and the default for the 'replace' option is "FALSE", > so sample((1:N)) samples N from (1:N) without replacement, > i.e. a random permutation. > > Ted.
While I am at it, an alternative to this use of sample() is to use order() to find the permutation which re-arranges a set of random numbers into increasing order. This in effect returns a random permutation of (1:N). Hence, instead of "ix <- sample(1:N))" in the above, you could use: ix <- order(runif(N)) Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <[email protected]> Fax-to-email: +44 (0)870 094 0861 Date: 08-Nov-11 Time: 09:30:33 ------------------------------ XFMail ------------------------------ ______________________________________________ [email protected] 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.
