CODEWITHSUNDEEP
rrollapply
89_Simple
89_Simple

Reputation: 3805

Rolling sum in R

df <- data.frame(x = seq(1:10))

I want this:

df$y <- c(1, 2, 3, 4, 5, 15, 20 , 25, 30, 35)

i.e. each y is the sum of previous five x values. This implies the first five y will be same as x

What I get is this:

df$y1 <- c(df$x[1:4], RcppRoll::roll_sum(df$x, 5)) 

  x  y y1
  1  1  1
  2  2  2
  3  3  3
  4  4  4
  5  5 15
  6 15 20
  7 20 25
  8 25 30
  9 30 35
  10 35 40

In summary, I need y but I am only able to achieve y1

Upvotes: 1

Views: 2234

Answers (3)

G. Grothendieck
G. Grothendieck

Reputation: 269654

1) enhanced sum function Define a function Sum which sums its first 5 values if it receives 6 values and returns the last value otherwise. Then use it with partial=TRUE in rollapplyr:

Sum <- function(x) if (length(x) < 6) tail(x, 1) else sum(head(x, -1))
rollapplyr(x, 6, Sum, partial = TRUE)
##  [1]  1  2  3  4  5 15 20 25 30 35

2) sum 6 and subtract off original Another possibility is to take the running sum of 6 elements filling in the first 5 elements with NA and subtracting off the original vector. Finally fill in the first 5.

replace(rollsumr(x, 6, fill = NA) - x, 1:5, head(x, 5))
##  [1]  1  2  3  4  5 15 20 25 30 35

3) specify offsets A third possibility is to use the offset form of width to specify the prior 5 elements:

c(head(x, 5), rollapplyr(x, list(-(1:5)), sum))
## [1]  1  2  3  4  5 15 20 25 30 35

4) alternative specification of offsets In this alternative we specify an offset of 0 for each of the first 5 elements and offsets of -(1:5) for the rest.

width <- replace(rep(list(-(1:5)), length(x)), 1:5, list(0))
rollapply(x, width, sum)
## [1]  1  2  3  4  5 15 20 25 30 35

Note

The scheme for filling in the first 5 elements seems quite unusual and you might consider using partial sums for the first 5 with NA or 0 for the first one since there are no prior elements fir that one:

rollapplyr(x, list(-(1:5)), sum, partial = TRUE, fill = NA)
## [1] NA  1  3  6 10 15 20 25 30 35

rollapplyr(x, list(-(1:5)), sum, partial = TRUE, fill = 0)
## [1]  0  1  3  6 10 15 20 25 30 35

rollapplyr(x, 6, sum, partial = TRUE) - x
## [1]  0  1  3  6 10 15 20 25 30 35

Upvotes: 4

Jim Chen
Jim Chen

Reputation: 3729

mysum <- function(x, k = 5) {
  res <- x[1:k]
  append<-sapply(2:(len(x)+1-k),function(i) sum(x[i:(i+k-1)]))
  return(c(res,append))
}
mysum(df$x)

Upvotes: 0

r.user.05apr
r.user.05apr

Reputation: 5456

A simple approach would be:

df <- data.frame(x = seq(1:10))
mysum <- function(x, k = 5) {
  res <- rep(NA, length(x))
  for (i in seq_along(x)) {
    if (i <= k) { # edited ;-)
      res[i] <- x[i]
    } else {
      res[i] <- sum(x[(i-k):(i-1)])
    }
  }
  res
}
mysum(df$x)
# [1]  1  2  3  4  5 15 20 25 30 35

Upvotes: 0

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