cumsum along row of data.frame with NA in R

Question

The hypothetical case is that there exist NA in a data.frame

> a <- c(1:5, NA, 7:10)
> b <- 1:10
> c <- 1:10
> 
> data <-         


        

Answer 1

Given your desired result (where you don't mind NA becoming 0), I guess the easiest thing is to first remove the NA values using is.na and then carry on as before.

data[ is.na(data) ] <- 0
data.frame(t(apply(data,1,cumsum)))

Answer 2

Simon's is definitely the simplest. I was surprised to learn a few things from this exercise: 1. cumsum doesn't have a na.rm argument 2. sum(NA, na.rm=TRUE) equals 0

Here is the code that brought me to the same solution:

cumsum.alt <- function(x){
    res <- NaN*seq(x)
    for(i in seq(x)){
        res[i] <- sum(x[1:i], na.rm=TRUE)
    }
    res
}

t(apply(data, 1, cumsum.alt))

To return the NA´s, a slight modification can be used:

cumsum.alt <- function(x){
    res <- NaN*seq(x)
    for(i in seq(x)){
        if(sum(is.na(x[1])) == i){
            res[i] <- NaN
        } else {
            res[i] <- sum(x[1:i], na.rm=TRUE)
        }
    }
    res
}

t(apply(data, 1, cumsum.alt))