Running sum

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

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch]On Behalf Of Sean Davis
Sent: Friday, November 19, 2004 1:09 PM
To: r-help
Subject: [R] Running sum


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

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

cumsum(1:10)

max(cumsum(1:10))

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?

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

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

-----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

?cumsum

On Fri, 19 Nov 2004, 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?

?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

Y

runSum2(1:10)

running(1:10, width = 2, fun = sum)

Perhaps I'm missing something, but isn't the maximum of the cumulative
sum simply the last value, ie. sum(x)?

Hadley

On Fri, 2004-11-19 at 16:44 -0600, hadley wickham wrote:
 

Perhaps I'm missing something, but isn't the maximum of the cumulative
sum simply the last value, ie. sum(x)?

Hadley
   

Indeed!

And...going back to Sean's original post he did write:

"I just need max(Y)."

Thus, a whole group of us (independently) managed to over-engineer the
 

solution. 

Presuming of course, that Sean does not require the entire cumulative
sum vector for other purposes.

I wonder what that says about human cognition....

Thanks,

Marc
<Cleaning his bi-focals...>

On Fri, 2004-11-19 at 16:44 -0600, hadley wickham wrote:

Perhaps I'm missing something, but isn't the maximum of the cumulative
sum simply the last value, ie. sum(x)?

Hadley

Indeed!

And...going back to Sean's original post he did write:

"I just need max(Y)."

Thus, a whole group of us (independently) managed to over-engineer the
solution. 

Presuming of course, that Sean does not require the entire cumulative
sum vector for other purposes.

I wonder what that says about human cognition....

Thanks,

Marc
<Cleaning his bi-focals...>

I have forgotten the details of the original question but it seems to me 
that if the elements of x could be both positive and negative then the 
maximum of the cumulative sum doesn't have to be the last sum.

Please, Please, please, ... was it ever said that the numbers was only 
positive?

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch 
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of 
Philippe Grosjean
Sent: Friday, November 19, 2004 9:11 PM
To: MSchwartz at MedAnalytics.com; 'Sean Davis'
Cc: 'R-Help'
Subject: RE: [R] 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

Perhaps I'm missing something, but isn't the maximum of the cumulative
sum simply the last value, ie. sum(x)?