Hello John,
If many numbers are duplicated, then one way is to coerce to a factor
and use the levels() function. For instance:
x <- c(1,1,2,2,2,3,3,4,1,1,2,4)
X <- factor(x)
for (i in levels(X))
{
loc <- (X==i); len = length(loc)
x[loc] <- x[loc] + 0.01 * (1:len)
}
x
[1] 1.01 1.02 2.01 2.02 2.03 3.01 3.02 4.01 1.03 1.04 2.04 4.02
Hope that helps
James Lawrence
On Thu, 2011-01-20 at 08:00 -0800, Joshua Wiley wrote:
> 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.
> >
>
>
>
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