Running sum

?cumsum is not exactly the answer (as I understand it), but a part of it.
I propose:

runSum2 <- function(x)
	cumsum(x)[-1] - c(0, cumsum(x[1:(length(x) - 2)]))

# Example
a <- round(runif(10, 0, 10))
a
runSum2(a)
max(runSum2(a)) # To get only the max

Best,

Philippe

..............................................<??}))><........
 ) ) ) ) )
( ( ( ( (    Prof. Philippe Grosjean
 ) ) ) ) )
( ( ( ( (    Numerical Ecology of Aquatic Systems
 ) ) ) ) )   Mons-Hainaut University, Pentagone
( ( ( ( (    Academie Universitaire Wallonie-Bruxelles
 ) ) ) ) )   6, av du Champ de Mars, 7000 Mons, Belgium  
( ( ( ( (       
 ) ) ) ) )   phone: + 32.65.37.34.97, fax: + 32.65.37.33.12
( ( ( ( (    email: Philippe.Grosjean at umh.ac.be
 ) ) ) ) )      
( ( ( ( (    web:   http://www.umh.ac.be/~econum
 ) ) ) ) )
..............................................................

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch 
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Marc Schwartz
Sent: Friday, November 19, 2004 7:57 PM
To: Sean Davis
Cc: R-Help
Subject: Re: [R] Running sum

On Fri, 2004-11-19 at 13:08 -0500, Sean Davis wrote:

I have vector X of length N that I want to have a running sum for 
(called Y).  I just need max(Y).  I do this with a "for" 

loop like so:

     Y <- vector(length=N)
     Y[1] <- X[1]
     for (i in 2:N) {
       Y[i] <- Y[i-1]+X[i]
     }
     return(max(Y))

Is there a faster way to do this?

Thanks,
Sean

Something like:

cumsum(1:10)

 [1]  1  3  6 10 15 21 28 36 45 55

max(cumsum(1:10))

[1] 55

Does that help?

Marc Schwartz

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