R - fill na values sequentially from last nonzero value

One option:

library(dplyr)

df %>%
  group_by(idx = cumsum(!(is.na(a) | a == 0)), is.na(a)) %>%
  mutate(rn = row_number()) %>%
  group_by(idx) %>%
  mutate(a = coalesce(a, first(a) + rn)) %>%
  ungroup() %>%
  select(a)

Output:

# A tibble: 12 x 1
       a
   <int>
 1 11127
 2     0
 3     0
 4 11128
 5     0
 6     0
 7 11129
 8     0
 9 11580
10     0
11 11581
12     0

If speed is an issue, perhaps the data.table equivalent will be slightly faster:

library(data.table)

setDT(df)[, rn := rowid(a), .(cumsum(!(is.na(a) | a == 0)), is.na(a))][
  , a := fcoalesce(a, first(a) + rn), by = cumsum(!(is.na(a) | a == 0))][
    , rn := NULL]

EDIT

IMO grouping and then getting the row index for NAs is not really elegant; it's far better what you can see in other solutions (e.g. using cumsum).

Using fcoalesce, the issue could then be solved in a single data.table step:

library(data.table)

setDT(df)[, a := fcoalesce(a, first(a) + cumsum(is.na(a))), by = cumsum(!(is.na(a) | a == 0))]

One solution utilizing dplyr could be:

df %>%
 group_by(id = cumsum(!is.na(a) & a != 0)) %>%
 mutate(a = ifelse(is.na(a), first(a) + cumsum(is.na(a)), a))

       a    id
   <int> <int>
 1 11127     1
 2     0     1
 3     0     1
 4 11128     1
 5     0     1
 6     0     1
 7 11129     1
 8     0     1
 9 11580     2
10     0     2
11 11581     2
12     0     2

A base R approach with cumsum(logical) and ave.

nze <- df1$a != 0 & !is.na(df1$a)
ave(df1$a, cumsum(nze), FUN = function(x){
  na <- is.na(x)
  x[na] <- x[!na][1] + seq_along(which(na))
  x
})
# [1] 11127     0     0 11128     0     0 11129     0 11580     0 11581     0

Then assign this result.

df1$a <- ave(df1$a, cumsum(nze), FUN = function(x){
  na <- is.na(x)
  x[na] <- x[!na][1] + seq_along(which(na))
  x
})