Calculate cumsum() while ignoring NA values

Question

Consider the following named vector x.

( x <- setNames(c(1, 2, 0, NA, 4, NA, NA, 6), letters[1:8]) )
# a  b  c  d  e  f  g  h 
# 1  2  0 NA           


        

Answer 1

Do you want something like this:

x2 <- x
x2[!is.na(x)] <- cumsum(x2[!is.na(x)])

x2

[edit] Alternatively, as suggested by a comment above, you can change NA's to 0's -

miss <- is.na(x)
x[miss] <- 0
cs <- cumsum(x)
cs[miss] <- NA
# cs is the requested cumsum

Answer 2

Here's a function I came up from the answers to this question. Thought I'd share it, since it seems to work well so far. It calculates the cumulative FUNC of x while ignoring NA. FUNC can be any one of sum(), prod(), min(), or max(), and x is a numeric vector.

cumSkipNA <- function(x, FUNC)
{
    d <- deparse(substitute(FUNC))
    funs <- c("max", "min", "prod", "sum")
    stopifnot(is.vector(x), is.numeric(x), d %in% funs)
    FUNC <- match.fun(paste0("cum", d))
    x[!is.na(x)] <- FUNC(x[!is.na(x)])
    x
}

set.seed(1)
x <- sample(15, 10, TRUE)
x[c(2,7,5)] <- NA
x
# [1]  4 NA  9 14 NA 14 NA 10 10  1
cumSkipNA(x, sum)
# [1]  4 NA 13 27 NA 41 NA 51 61 62
cumSkipNA(x, prod)
# [1]      4     NA     36    504     NA   7056     NA
# [8]  70560 705600 705600
cumSkipNA(x, min)
# [1]  4 NA  4  4 NA  4 NA  4  4  1
cumSkipNA(x, max)
# [1]  4 NA  9 14 NA 14 NA 14 14 14 

Definitely nothing new, but maybe useful to someone.

Answer 3

It's an old question but tidyr gives a new solution. Based on the idea of replacing NA with zero.

require(tidyr)

cumsum(replace_na(x, 0))

 a  b  c  d  e  f  g  h 
 1  3  3  3  7  7  7 13 

Answer 4

You can do this in one line with:

cumsum(ifelse(is.na(x), 0, x)) + x*0
#  a  b  c  d  e  f  g  h 
#  1  3  3 NA  7 NA NA 13

Or, similarly:

library(dplyr)
cumsum(coalesce(x, 0)) + x*0
#  a  b  c  d  e  f  g  h 
#  1  3  3 NA  7 NA NA 13