Dear Pele,
Probably not the best way to proceed but it works:
X<-read.table(textConnection("ID X2
1.00 1.00
2.00 0.00
3.00 1.00
4.00 3058
5.00 0.00
6.00 6.00"),header=TRUE)
closeAllConnections()
X
x3<-0
for(i in 2:(nrow(X)+1)) x3<-c(x3, X$X2[i-1]+0.24*x3[i-1])
X$x3<-x3[-1]
X
HTH,
Jorge
On Mon, Feb 23, 2009 at 3:59 PM, Pele <drdi...@yahoo.com> wrote:
>
> Hi R users,
>
> Is there an easy way in R to generate the results table below using table 1
> and the formula (simplified version of the real problem)? It would be easy
> if I knew the R equivalent of SAS's retain function, but could not find
> one.
>
> Thanks in Advance for any help!
>
> table1:
>
> ID X2 X3
> 1.00 1.00 0
> 2.00 0.00
> 3.00 1.00
> 4.00 3058
> 5.00 0.00
> 6.00 6.00
>
> Formula: X3 = x2 + (.24 * x3)
>
> where the values in the x3 column of the result table are retained from
> previous x3 rows.. Also the first x3 value is initialized to 0 to start
>
> e.g.
> for ID=1 we have 1 + .24(0) = 1.00
> for ID=2 we have 0 + .24(1) = 0.24
> for ID=3 we have 1 + .24(.24) = 1.06
> for ID=4 we have 3058 + .24(1.06) = 3058.25
> etc.............
>
> Results:
> ID X2 x3
> 1.00 1.00 1.00
> 2.00 0.00 0.24
> 3.00 1.00 1.06
> 4.00 3058 3058.25
> 5.00 0.00 733.98
> 6.00 6.00 182.16
> --
> View this message in context:
> http://www.nabble.com/Formula-that-includes-previous-row-values-tp22170010p22170010.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
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