Peter Dalgaard wrote: > Duncan Murdoch <[EMAIL PROTECTED]> writes: > > >> Suppose I have two columns, x,y. I can use order(x,y) to calculate a >> permutation that puts them into increasing order of x, >> with ties broken by y. >> >> I'd like instead to calculate the rank of each pair under the same >> ordering, but the rank() function doesn't take multiple values >> as input. Is there a simple way to get what I want? >> >> E.g. >> >> > x <- c(1,2,3,4,1,2,3,4) >> > y <- c(1,2,3,1,2,3,1,2) >> > rank(x+y/10) >> [1] 1 3 6 7 2 4 5 8 >> >> gives me the answer I want, but only because I know the range of y and >> the size of gaps in the x values. What do I do in general? >> > > Still not quite general, but in the absence of ties: > > >> z[order(x,y)]<-1:8 >> z >> > [1] 1 3 6 7 2 4 5 8 > >
Thanks to all who have replied. Unfortunately for me, ties do exist, and I'd like them to get identical ranks. John Fox's suggestion would handle ties properly, but I'm worried about rounding error giving spurious ties. Duncan Murdoch ______________________________________________ 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
