On Thu, Jan 20, 2011 at 10:12 AM, Ortiz, John 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 +
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.
Try this:
replace(x + ave(x, x, FUN = seq) * .01, !(duplicated(x) | duplicated(x,
fromLast = TRUE)), x)
On Thu, Jan 20, 2011 at 1:12 PM, Ortiz, John 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
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ortiz, John
> Sent: Thursday, January 20, 2011 7:13 AM
> To: r-help@r-project.org
> Subject: [R] Identify duplicate numbers and to increase a value
>
> H
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
ards
Moritz
Moritz Grenke
http://www.360mix.de
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von Ortiz, John
Gesendet: Donnerstag, 20. Januar 2011 16:13
An: r-help@r-project.org
Betreff: [R] Identify duplicate numbers and
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
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