faster way?

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

To:             	R-Help <r-help at stat.math.ethz.ch>
From:           	Rick Bischoff <rdbisch at gmail.com>
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"]
}


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Petr Pikal
petr.pikal at precheza.cz