Is this what you want:
> x
cola colb
1 1 c
2 1 i
3 1 i
4 1 c
5 2 i
6 2 c
7 2 c
8 2 i
9 2 i
> # generate run lengths
> z <- rle(x$colb)
> # calculate offset of each starting sequence
> offsets <- head(cumsum(c(1, z$lengths)), -1)
> # now determine where the 'i' is more than 1
> long <- which(z$lengths > 1 & z$values == 'i')
> # now replace all 'i' with 'I'
> x$colb[x$colb == 'i'] <- 'I'
> # replace lon runs with 'h'
> for (i in long){
+ x$colb[(offsets[i]+1):(offsets[i] + z$lengths[i] - 1)] <-
rep('h', z$lengths[i] - 1)
+ }
> x
cola colb
1 1 c
2 1 I
3 1 h
4 1 c
5 2 I
6 2 c
7 2 c
8 2 I
9 2 h
>
>
On Fri, Jun 27, 2008 at 5:17 AM, poul <[EMAIL PROTECTED]> wrote:
>
> Hello every body,
> I am quite a new user of R so I beg your pardon for this naive question and
> the lake of syntax with wich I ask it:
>
> I have a data frames like this:
>
> cola colb
> 1 c
> 1 i
> 1 i
> 1 c
> 2 i
> 2 c
> 2 c
> 2 i
> 2 i
> ...
> 10000
>
> I would like ,for each level of cola and for x in colb:
>
> -if colb[x]=="i" and colb[x-1] does not exist (first row in dataframes),
> then replace colb[x] by "l"
> -if colb[x]=="i" and colb[x-1]=="c" (previous row in data frmes), then
> replace colb[x] by "l"
> -if colb[x]=="i" and colb[x-1]=="i" (previous row in data frmes), then
> replace colb[x] by "h"
>
> Thanks in advance
>
> Paul
>
> --
> View this message in context:
> http://www.nabble.com/change-values-in-data-frames-tp18151205p18151205.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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