Keith, I am simply baffled! Didn't think a second about doing it this way, tsss - Great!
Thanks also for Daniel, Jim's and Bart's proposals! R is cool, I realise it every day again :-) Thanks!! On Wed, Jul 9, 2008 at 12:33 PM, Jim Lemon <[EMAIL PROTECTED]> wrote: > On Wed, 2008-07-09 at 11:40 +0200, Anne-Marie Ternes wrote: >> Hi, >> >> if given the value of, say, 15000, I would like to be able to divide >> that value recursively by, say, 5, and to get a vector of a determined >> length, say 9, the last value being (set to) zero- i.e. like this: >> >> 15000 3000 600 120 24 4.8 0.96 0.192 0 >> >> These are in fact concentration values from an experiment. For my >> script, I get only the starting value (here 15000), and the factor by >> which concentration is divided for each well, the last one having, by >> definition, no antagonist at all. >> >> I have tried to use "seq", but it can "only" do positive or negative >> increment. I didn't either find a way with "rep", "sweep" etc. These >> function normally start from an existing vector, which is not the case >> here, I have only got a single value to start with. >> >> I suppose I could do something "loopy", but I'm sure there is a better >> way to do it. >> > Well, if you really want to do it recursively (and maybe loopy as well) > > recursivdiv<-function(x,denom,lendiv,firstpass=TRUE) { > if(firstpass) lendiv<-lendiv-1 > if(lendiv > 1) { > divvec<-c(x/denom,recursivdiv(x/denom,denom,lendiv-1,FALSE)) > cat(divvec,ndiv,"\n") > } > else divvec<-0 > if(firstpass) divvec<-c(x,divvec) > return(divvec) > } > > 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.