Efficiently remove all NULL values in a list and all sublists

Question

Consider the following list:

> l1 <- list(NULL,1,2,list(NULL,3,list(NULL,4)))
> str(l1)
List of 4
 $ : NULL
 $ : num 1
 $ : num 2
 $ :List of 3
  ..$ : NULL
  ..$ : num 3
  ..$ :List of 2
  .. ..$ : NULL
  .. ..$ : num 4

To remove NULL values from the first level, simply call

l1[vapply(l1,is.null,logical(1L))] <- NULL

Now I want to remove all NULL values at all levels, I come up with the following code.

list.clean <- function(.data, fun = is.null, recursive = FALSE) {
  if(recursive) {
    .data <- lapply(.data, function(.item) {
      if(is.list(.item)) list.clean(.item, fun, TRUE)
      else .item
    })
  }
  .data[vapply(.data,fun,logical(1L))] <- NULL
  .data
}

And calling

> list.clean(l1, recursive = TRUE)
[[1]]
[1] 1

[[2]]
[1] 2

[[3]]
[[3]][[1]]
[1] 3

[[3]][[2]]
[[3]][[2]][[1]]
[1] 4

Although it works right now, is there a better or faster way to do so?

Answer 1

Using an external package, this can now also be done with rrapply in the rrapply-package (a revised version of base rapply). Set how = "prune" to prune all list elements that do not satisfy a function defined in the condition argument:

library(rrapply)

l1 <- list(NULL,1,2,list(NULL,3,list(NULL,4)))

rrapply(l1, condition = Negate(is.null), how = "prune")
#> [[1]]
#> [1] 1
#> 
#> [[2]]
#> [1] 2
#> 
#> [[3]]
#> [[3]][[1]]
#> [1] 3
#> 
#> [[3]][[2]]
#> [[3]][[2]][[1]]
#> [1] 4

We can benchmark computation times for large list objects against OP's list.clean function and G. Grothendieck's rmNull function:

## benchmark recursion functions
rmNull <- function(x) {
  x <- Filter(Negate(is.null), x)
  lapply(x, function(x) if (is.list(x)) rmNull(x) else x)
}

list.clean <- function(.data, fun = is.null, recursive = FALSE) {
  if(recursive) {
    .data <- lapply(.data, function(.item) {
      if(is.list(.item)) list.clean(.item, fun, TRUE)
      else .item
    })
  }
  .data[vapply(.data,fun,logical(1L))] <- NULL
  .data
}

## recursively create nested list with dmax layers and 50% NULL elements
f <- function(len, d, dmax) {
  x <- vector(mode = "list", length = len)
  for(i in seq_along(x)) {
    if(d + 1 < dmax) {
      x[[i]] <- Recall(len, d + 1, dmax)
    } else {
      x[[i]] <- list(1, NULL)
    }
  }
  return(x)
}

## long shallow list (3 layers, total 5e5 nodes)
x_long <- f(len = 500, d = 1, dmax = 3)

microbenchmark::microbenchmark(
  rmNull = rmNull(x_long),
  list.clean = list.clean(x_long, recursive = TRUE),
  rrapply = rrapply(x_long, condition = Negate(is.null), how = "prune"),
  check = "equal",
  times = 5L
)
#> Unit: milliseconds
#>        expr       min        lq      mean    median        uq       max
#>      rmNull 2381.5536 2535.6871 2559.4045 2546.0375 2571.9462 2761.7982
#>  list.clean 1954.4046 1973.7983 2012.2158 2010.7334 2049.8020 2072.3409
#>     rrapply  288.5784  297.9041  382.3111  301.3147  460.5107  563.2475

## deeply nested list (18 layers, total 2^18 nodes)
x_deep <- f(len = 2, d = 1, dmax = 18)

microbenchmark::microbenchmark(
  rmNull = rmNull(x_deep),
  list.clean = list.clean(x_deep, recursive = TRUE),
  rrapply = rrapply(x_deep, condition = Negate(is.null), how = "prune"),
  check = "equal",
  times = 5L
)
#> Unit: milliseconds
#>        expr       min        lq      mean    median       uq       max
#>      rmNull 2306.5788 2360.2663 2422.2578 2367.9296 2530.201 2546.3135
#>  list.clean 1708.1192 1829.1303 2014.2162 2157.2148 2180.023 2196.5937
#>     rrapply  174.5385  187.9491  271.4967  200.9263  206.739  587.3306

Answer 2

This can be done recursively:

rmNull <- function(x) {
   x <- Filter(Negate(is.null), x)
   lapply(x, function(x) if (is.list(x)) rmNull(x) else x)
}
l2 <- rmNull(l1)

giving:

> str(l2)
List of 3
 $ : num 1
 $ : num 2
 $ :List of 2
  ..$ : num 3
  ..$ :List of 1
  .. ..$ : num 4