Running sums for every row of the previous 25 rows

Question

I've been trying for a while now to produce a code that brings me a new vector of the sum of the 25 previous rows for an original vector.

So if we say I have a variable Y with 500 rows and I would like a running sum, in a new vector, which contains the sum of rows [1:25] then [2:26] for the length of Y, such as this:

y<-1:500
runsum<-function(x){
   cumsum(x)-cumsum(x[26:length(x)])
}

new<-runsum(y)

I've tried using some different functions here and then even using the apply functions on top but none seem to produce the right answers....

Would anyone be able to help? I realise it's probably very easy for many of the community here but any help would be appreciated

Thanks

[possible duplicate](http://stackoverflow.com/questions/10930810/cumulative-sum-for-n-rows) — Roland, Jun 14 '12 at 14:13

score 3 · Accepted Answer · answered Jun 14 '12 at 12:56

3

This function calculates the sum of the 24 preceding values and the actual value:

 movsum <- function(x,n=25){filter(x,rep(1,n), sides=1)}

It is easy to adapt to sum only preceding values, if this is what you really want.

answered Jun 14 '12 at 12:56

Roland

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Jase_ · Answer 2 · 2017-04-24T16:25:44.697

3

In addition to Roland's answer you could use the zoo library

library ( zoo )
y <- 1:500
rollapply ( zoo ( y ), 25, sum )

HTH

edited Apr 24 '17 at 16:25

answered Jun 14 '12 at 13:01

Jase_

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can the 25 here by dynamic ? i.e. a vector of values – deltascience Nov 15 '17 at 15:46

score 1 · Answer 3 · answered Jun 14 '12 at 13:03

I like Roland's answer better as it relies on a time series function and will probably be pretty quick. Since you mentioned you started going down the path of using apply() and friends, here's one approach to do that:

y<-1:500
#How many to sum at a time?
n <- 25
#Create a matrix of the appropriate start and end points
mat <- cbind(start = head(y, -(n-1)), end = tail(y, -(n-1)))
#Check output
rbind(head(mat,3), tail(mat,3))
#-----
       start end
           1  25
           2  26
           3  27
[474,]   474 498
[475,]   475 499
[476,]   476 500

#add together
apply(mat, 1, function(x) sum(y[x[1]]:y[x[2]]))

#Is it the same as Roland's answer after removing the NA values it returns?
all.equal(apply(mat, 1, function(x) sum(y[x[1]]:y[x[2]])),
          movsum(y)[-c(1:n-1)])
#-----
[1] TRUE

Thanks - it helpful to see the apply function being used in a manner related to my work - I'll try this way out too so I at least know how it all works! — Ash, Jun 15 '12 at 13:29

Running sums for every row of the previous 25 rows

3 Answers3