Replace NA with the previous value with the limit

Question

I would like to replace up to an nNA value in a vector with a last value other than NA.

For example, if:

a <- c(1,NA,NA,NA,NA,NA,2,NA,1,NA,NA,NA)
n <- 2

I would like to get:

c(1,1,1,NA,NA,NA,2,2,1,1,1,NA)

n - the maximum number of NA values ​​that can be replaced by this element).

I know the function na.locf(), but I do not know how to set the limit n. Can this be done?

Answer 1

Here the option is used na.locfandrle

library(zoo)
r <- rle(is.na(a))
a <- na.locf(a)
is.na(a) <- sequence(r$lengths) > n & rep(r$values, r$lengths)
a
# [1]  1  1  1 NA NA NA  2  2  1  1  1 NA

, a ( NA), NA na.locf , , NA, , n NA.

Answer 2

"a" NA s:

is = seq_along(a)
i = cummax((!is.na(a)) * is)
i
# [1] 1 1 1 1 1 1 7 7 9 9 9 9

NA , NA "n":

wh = (is - i) > n
i[wh] = is[wh]
i
# [1]  1  1  1  4  5  6  7  7  9  9  9 12

"a":

a[i]
# [1]  1  1  1 NA NA NA  2  2  1  1  1 NA

Answer 3

base R, cumsum diff, ave we replace NA , 'n'

ave(a, cumsum(c(TRUE, diff(is.na(a)) < 0)), 
      FUN = function(x) replace(x, is.na(x) & seq_along(x) <= n + 1, x[1]))
#[1]  1  1  1 NA NA NA  2  2  1  1  1 NA

---

ave(a, cumsum(!is.na(a)), FUN = function(x) replace(x, pmin(length(x), seq(n+1)), x[1]))
#[1]  1  1  1 NA NA NA  2  2  1  1  1 NA

Answer 4

, split replace R

f <- function(a, n) {
  # split the vector based on the position of non-NA values
  l <- split(a, cumsum(seq_along(a) %in% which(!is.na(a))))
  unlist(lapply(l, function(r) replace(r, 1:(n+1), r[1])[seq_along(r)]),use.names = FALSE)
}

f(a, n = 2)
#[1]  1  1  1 NA NA NA  2  2  1  1  1 NA

f(a, n = 3)
#[1]  1  1  1  1 NA NA  2  2  1  1  1  1

Benchmarking (randomly generated vector size 7467)

library(microbenchmark)
library(dplyr)
library(zoo)
set.seed(123)
a <- unlist(replicate(1000, c(sample(10, 2), rep(NA, sample.int(10, 1)))))
length(a)
# [1] 7467
n <- 3
f_989 <- function(a, n) {
  # split the vector based on the position of non-NA values
  l <- split(a, cumsum(seq_along(a) %in% which(!is.na(a))))
  unlist(lapply(l, function(r) replace(r, 1:(n+1), r[1])[seq_along(r)]),use.names = FALSE)
}
f_zx8754 <- function(a, n)
data.frame(a) %>% mutate(gr = cumsum(!is.na(a))) %>% 
  group_by(gr) %>% 
  mutate(res = if_else(row_number() <= n + 1, na.locf(a), a)) %>% 
  .$res
f_docendo_discimus <- function(a, n){
    r <- rle(is.na(a))
    a <- na.locf(a)
    is.na(a) <- sequence(r$lengths) > n & rep(r$values, r$lengths)
    a
}
f_akrun <- function(a,n) 
ave(a, cumsum(!is.na(a)), FUN = function(x) replace(x, pmin(length(x), seq(n+1)), x[1]))

f_alexis_laz=function(a,n){
    is = seq_along(a)
    i = cummax((!is.na(a)) * is)
    wh = (is - i) > n
    i[wh] = is[wh]
    a[i]
}
r <- f_989(a,n)
identical(r, f_zx8754(a,n))
# [1] TRUE
identical(r, f_docendo_discimus(a,n))
# [1] TRUE
identical(r, f_akrun(a,n))
# [1] TRUE
identical(r, f_alexis_laz(a,n))
# [1] TRUE
res <- microbenchmark("f1"=f_989(a,n), "f2"=f_zx8754(a,n), 
"f3"=f_docendo_discimus(a,n), "f4"=f_akrun(a,n), "f5"=f_alexis_laz(a,n))

---

print(res, order="mean")

# Unit: microseconds
 # expr        min         lq       mean      median          uq        max neval
   # f5    129.804    137.014    161.106    141.6715    151.7375   1031.511   100
   # f3   1249.351   1354.215   1471.478   1392.9750   1482.2140   2553.086   100
   # f1   4736.895   5093.852   5630.367   5345.3450   6069.9260   8848.513   100
   # f4  22165.601  23936.866  24660.990  24485.6725  24883.6440  29453.177   100
   # f2 205854.339 215582.174 221524.448 218643.9540 224211.0435 261512.922   100

Answer 5

Using dplyr::group_byand zoo::na.locf:

library(dplyr)
library(zoo)

data.frame(a) %>% 
  mutate(gr = cumsum(!is.na(a))) %>% 
  group_by(gr) %>% 
  mutate(res = if_else(row_number() <= n + 1, na.locf(a), a)) %>% 
  .$res

# [1]  1  1  1 NA NA NA  2  2  1  1  1 NA