Hi John, If you only have one duplicated number (e.g., just 2), then this will work:
x <- c(1,2,3,5,6,2,8,9,2,2) xd <- duplicated(x) x[xd] <- x[xd] + seq(sum(xd))/100 x otherwise, I think a different framework than duplicated() will be necessary, because it will matter not just if the number is duplicated but which one how many times and where. Cheers, Josh On Thu, Jan 20, 2011 at 7:12 AM, Ortiz, John <ort...@si.edu> wrote: > Hi everybody. > > I want to identify duplicate numbers and to increase a value of 0.01 for each > time that it is duplicated. > > Example: > x=c(1,2,3,5,6,2,8,9,2,2) > > I want to do this: > > 1 > 2 + 0.01 > 3 > 5 > 6 > 2 + 0.02 > 8 > 9 > 2 + 0.03 > 2 + 0.04 > > I am trying to get something like this: > > 1 > 2.01 > 3 > 5 > 6 > 2.02 > 8 > 9 > 2.03 > 2.04 > > Actually I just know the way to identify the duplicated numbers > > rbind(x, duplicated(x) | duplicated(x, fromLast=TRUE)) > > [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] > x 1 2 3 5 6 2 8 9 2 2 > 0 1 0 0 0 1 0 0 1 1 > > Some advice? > > Thanks and regards > John Ortiz > > ______________________________________________ > 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. > -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ ______________________________________________ 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.
