thanks to all for your replies. very neat. On Wed, 1 Nov 2006, Gabor Grothendieck wrote:
> Try: > > rep(1:3, length = 17) > > or > > as.numeric(gl(3, 1, 17)) > > or > > y <- 1:17 > replace(y, TRUE, 1:3) # ignore warning > > > On 11/1/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote: > > > > > > i need a strange algorithm that i can easily do in a loop but need > > o do without looping. > > > > suppose i have a vector of length y filled with zeros and a number x. > > > > then, i want a new vector which is (1,......x,1......x,1.....x,1.....y mod > > x ) > > > > so if y was of length 17 and x was 3, the resultant vector should be > > > > (1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2) > > > > so basically , the algorithm should repeat seq(1,x) as many times as > > needed to fill y but > > then , if the last one doesn;'t fit perfectly, it should stop whenever it > > hits the end of y ? > > > > rep(seq(1:x,length(y)/x) will work when there is a perfect fit but > > it can't handle the non fit cases. it just cuts > > off the non integer part of length(y)/x and then fills the vector > > do that the resultant vector is less than the length of y. > > > > thanks. > > > > ______________________________________________ > > R-help@stat.math.ethz.ch 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. > > > ______________________________________________ R-help@stat.math.ethz.ch 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.