[R] Running sum

[Philippe Grosjean]

?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

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( ( ( ( (    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
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..............................................................

 

> -----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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