How to implement coalesce efficiently in R

On my machine, using Reduce gets a 5x performance improvement:

coalesce2 <- function(...) {
  Reduce(function(x, y) {
    i <- which(is.na(x))
    x[i] <- y[i]
    x},
  list(...))
}

> microbenchmark(coalesce(a,b,c),coalesce2(a,b,c))
Unit: microseconds
               expr    min       lq   median       uq     max neval
  coalesce(a, b, c) 97.669 100.7950 102.0120 103.0505 243.438   100
 coalesce2(a, b, c) 19.601  21.4055  22.8835  23.8315  45.419   100

Looks like coalesce1 is still available

coalesce1 <- function(...) {
    ans <- ..1
    for (elt in list(...)[-1]) {
        i <- is.na(ans)
        ans[i] <- elt[i]
    }
    ans
}

which is faster still (but more-or-less a hand re-write of Reduce, so less general)

> identical(coalesce(a, b, c), coalesce1(a, b, c))
[1] TRUE
> microbenchmark(coalesce(a,b,c), coalesce1(a, b, c), coalesce2(a,b,c))
Unit: microseconds
               expr     min       lq   median       uq     max neval
  coalesce(a, b, c) 336.266 341.6385 344.7320 355.4935 538.348   100
 coalesce1(a, b, c)   8.287   9.4110  10.9515  12.1295  20.940   100
 coalesce2(a, b, c)  37.711  40.1615  42.0885  45.1705  67.258   100

Or for larger data compare

coalesce1a <- function(...) {
    ans <- ..1
    for (elt in list(...)[-1]) {
        i <- which(is.na(ans))
        ans[i] <- elt[i]
    }
    ans
}

showing that which() can sometimes be effective, even though it implies a second pass through the index.

> aa <- sample(a, 100000, TRUE)
> bb <- sample(b, 100000, TRUE)
> cc <- sample(c, 100000, TRUE)
> microbenchmark(coalesce1(aa, bb, cc),
+                coalesce1a(aa, bb, cc),
+                coalesce2(aa,bb,cc), times=10)
Unit: milliseconds
                   expr       min        lq    median        uq       max neval
  coalesce1(aa, bb, cc) 11.110024 11.137963 11.145723 11.212907 11.270533    10
 coalesce1a(aa, bb, cc)  2.906067  2.953266  2.962729  2.971761  3.452251    10
  coalesce2(aa, bb, cc)  3.080842  3.115607  3.139484  3.166642  3.198977    10

Using dplyr package:

library(dplyr)
coalesce(a, b, c)
# [1]  1  2 NA  4  6

Benchamark, not as fast as accepted solution:

coalesce2 <- function(...) {
  Reduce(function(x, y) {
    i <- which(is.na(x))
    x[i] <- y[i]
    x},
    list(...))
}

microbenchmark::microbenchmark(
  coalesce(a, b, c),
  coalesce2(a, b, c)
)

# Unit: microseconds
#                expr    min     lq     mean median      uq     max neval cld
#   coalesce(a, b, c) 21.951 24.518 27.28264 25.515 26.9405 126.293   100   b
#  coalesce2(a, b, c)  7.127  8.553  9.68731  9.123  9.6930  27.368   100  a 

But on a larger dataset, it is comparable:

aa <- sample(a, 100000, TRUE)
bb <- sample(b, 100000, TRUE)
cc <- sample(c, 100000, TRUE)

microbenchmark::microbenchmark(
  coalesce(aa, bb, cc),
  coalesce2(aa, bb, cc))

# Unit: milliseconds
#                   expr      min       lq     mean   median       uq      max neval cld
#   coalesce(aa, bb, cc) 1.708511 1.837368 5.468123 3.268492 3.511241 96.99766   100   a
#  coalesce2(aa, bb, cc) 1.474171 1.516506 3.312153 1.957104 3.253240 91.05223   100   a

From data.table >= 1.12.3 you can use fcoalesce.

library(data.table)
fcoalesce(a, b, c)
# [1]  1  2 NA  4  6

---

fcoalesce can also take "a single plain list, data.table or data.frame". Thus, if the vectors above were columns in a data.frame (or a data.table), we could simply supply the name of the data set:

d = data.frame(a, b, c)
# or d = data.table(a, b, c) 
fcoalesce(d)
# [1]  1  2 NA  4  6

---

For more info, including a benchmark, see NEWS item #18 for development version 1.12.3.