Fill NAs in R with zero if the next valid data point is more than 2 intervals away

Question

I have multiple vectors with NAs and my intention to fill NA which are more than 2 intervals from a valid data point with 0. for example:

x <- c(3, 4, NA, NA, NA, 3, 3)

Expected output is,

3, 4, NA, 0, NA, 3, 3 

Answer 1

Update -

Here's probably one of the simplest and fastest solutions (Thanks to answer from G. Grothendieck). Simply knowing whether the value is NA on either side of any NA is sufficient information. Therefore, using lead and lag from dplyr package -

na2zero <- function(x) {
  x[is.na(lag(x, 1, 0)) & is.na(lead(x, 1, 0)) & is.na(x)] <- 0
  x
}

na2zero(x = c(3, 4, NA, NA, NA, 3, 3))
[1]  3  4 NA  0 NA  3  3

na2zero(x = c(3, 4, NA, NA, NA, NA, NA, 3, 3))
[1]  3  4 NA  0  0  0 NA  3  3

na2zero(x = c(3, 4, NA, NA, NA, 3, 3, NA, NA, 1, NA, 0, 0, rep(NA, 4L)))
[1]  3  4 NA  0 NA  3  3 NA NA  1 NA  0  0 NA  0  0 NA

---

Previous Answer (also fast) -

Here's one way using rle and replace from base R. This method turns every NA, that is not an endpoint in the running length, into a 0 -

na2zero <- function(x) {
  run_lengths <- rle(is.na(x))$lengths
  replace(x, 
    sequence(run_lengths) != 1 &
    sequence(run_lengths) != rep(run_lengths, run_lengths) &
    is.na(x),
  0)
}

na2zero(x = c(3, 4, NA, NA, NA, 3, 3))
[1]  3  4 NA  0 NA  3  3

na2zero(x = c(3, 4, NA, NA, NA, NA, NA, 3, 3))
[1]  3  4 NA  0  0  0 NA  3  3

---

Updated Benchmarks -

set.seed(2)
x <- c(3, 4, NA, NA, NA, 3, 3)
x <- sample(x, 1e5, T)

microbenchmark(
  Rui(x),
  Shree_old(x), Shree_new(x),
  markus(x),
  IceCreamT(x),
  Uwe1(x), Uwe2(x), Uwe_Reduce(x),
  Grothendieck(x),
  times = 50
)

all.equal(Shree_dplyr(x), Rui(x)) # [1] TRUE
all.equal(Shree_dplyr(x), Shree_rle(x)) # [1] TRUE
all.equal(Shree_dplyr(x), markus(x)) # [1] TRUE
all.equal(Shree_dplyr(x), Uwe1(x)) # [1] TRUE
all.equal(Shree_dplyr(x), Uwe2(x)) # [1] TRUE
all.equal(Shree_dplyr(x), Uwe_Reduce(x)) # [1] TRUE
all.equal(Shree_dplyr(x), Grothendieck(x)) # [1] TRUE


Unit: milliseconds
           expr        min         lq        mean     median          uq        max neval
         Rui(x) 286.026540 307.586604  342.620266 318.404731  363.844258  518.03330    50
   Shree_rle(x)  51.556489  62.038875   85.348031  65.012384   81.882141  327.57514    50
 Shree_dplyr(x)   3.996918   4.258248   17.210709   6.298946   10.335142  207.14732    50
      markus(x) 853.513854 885.419719 1001.450726 919.930389 1018.353847 1642.25435    50
   IceCreamT(x)  12.162079  13.773873   22.555446  15.021700   21.271498  199.08993    50
        Uwe1(x) 162.536980 183.566490  225.801038 196.882049  269.020395  439.17737    50
        Uwe2(x)  83.582360  93.136277  115.608342  99.165997  115.376903  309.67290    50
  Uwe_Reduce(x)   1.732195   1.871940    4.215195   2.016815    4.842883   25.91542    50
Grothendieck(x) 620.814291 688.107779  767.749387 746.699435  850.442643  982.49094    50

PS: Do check out TiredSquirell's answer which seems like a base version of Uwe's lead-lag answer but is somewhat faster (not benchmarked above).

Answer 2

Maybe there are simpler solutions but this one works.

na2zero <- function(x){
  ave(x, cumsum(abs(c(0, diff(is.na(x))))), FUN = function(y){
    if(anyNA(y)){
      if(length(y) > 2) y[-c(1, length(y))] <- 0
    }
    y
  })
}

na2zero(x)
#[1]  3  4 NA  0 NA  3  3

X <- list(x, c(x, x), c(3, 4, NA, NA, NA, NA, 3, 3))
lapply(X, na2zero)