Hi
not knowing dsn do you by chance looking for cumsum?
Or maybe something like this?
> x
[1] 1 2 3 4 5 6 7 8 9 10
> y
[1] 2 3 4 5 6 7 8 9 10 11
> x[1:9]+y[2:10]
[1] 4 6 8 10 12 14 16 18 20
>
HTH
Petr
On 10 Sep 2006 at 23:45, Rick Bischoff wrote:
To: R-Help <r-help@stat.math.ethz.ch>
From: Rick Bischoff <[EMAIL PROTECTED]>
Date sent: Sun, 10 Sep 2006 23:45:38 -0400
Subject: [R] faster way?
> Hi,
>
> Is there a faster way to do this? It takes forever, even on a
> moderately sized dataset.
>
>
> n <- dim(dsn)[1]
> dsn2 <- dsn[order(-dsn$xhat),]
> dsn2[1, "cumx"] <- dsn2[1, "xhat"]
>
> for (i in 2:n) {
> dsn2[i, "cumx"] <- dsn2[i - 1, "cumx"] + dsn2[i, "xhat"]
> }
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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> PLEASE do read the posting guide
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> minimal, self-contained, reproducible code.
Petr Pikal
[EMAIL PROTECTED]
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