How to fill in the preceding numbers whenever there is a 0 in R?

Question

I have a string of numbers:

n1 = c(1, 1, 0, 6, 0, 0, 10, 10, 11, 12, 0, 0, 19, 23, 0, 0)

I need to replace 0 with the corresponding number righ

Answer 1

Here's a solution using data.table:

require(data.table) ## >= 1.9.2
idx = which(!n1 %in% 0L)
DT <- data.table(val=n1[idx], idx=idx)
setattr(DT, 'sorted', "idx")
n1 = DT[J(seq_along(n1)), roll=Inf]$val
#  [1]  1  1  1  6  6  6 10 10 11 12 12 12 19 23 23 23

---

Benchmarks on bigger data:

require(zoo)
require(data.table)

set.seed(1L)
n1 = sample(c(0:10), 1e6, TRUE)

## data.table
dt_fun <- function(n1) {
    idx = which(!n1 %in% 0L)
    DT <- data.table(val=n1[idx], idx=idx)
    setattr(DT, 'sorted', "idx")
    DT[J(seq_along(n1)), roll=Inf]$val
}

# na.locf from zoo - gagolews
zoo_fun <- function(n1) {
    wh_na <- which(is.na(n1))
    n1[n1 == 0] <- NA
    n2 <- na.locf(n1)
    n2[wh_na] <- NA
    n2
}

## rle - thelatemail
rle_fun <- function(n1) {
    r <- rle(n1)
    r$values[which(r$values==0)] <- r$values[which(r$values==0)-1]
    inverse.rle(r)
}

flodel_fun <- function(n1) n1[cummax(seq_along(n1) * (n1 != 0))]

require(microbenchmark)
microbenchmark(a1 <- dt_fun(n1), 
               a2 <- zoo_fun(n1), 
               a3 <- rle_fun(n1), 
               a4 <- flodel_fun(n1), times=10L)

Here's the benchmarking result:

# Unit: milliseconds
#                  expr       min        lq    median        uq       max neval
#      a1 <- dt_fun(n1) 155.49495 164.04133 199.39133 243.22995 289.80908    10
#     a2 <- zoo_fun(n1) 596.33039 632.07841 671.51439 682.85950 697.33500    10
#     a3 <- rle_fun(n1) 356.95103 377.61284 383.63109 406.79794 495.09942    10
#  a4 <- flodel_fun(n1)  51.52259  55.54499  56.20325  56.39517  60.15248    10