R: Removing Null Elements from a List

R: removing NULL elements from a list

The closest you'll be able to get is to first name the list elements and then remove the NULLs.

names(x) <- seq_along(x)

## Using some higher-order convenience functions
Filter(Negate(is.null), x)
# $`11`
# [1] 123
# 
# $`13`
# [1] 456

# Or, using a slightly more standard R idiom
x[sapply(x, is.null)] <- NULL
x
# $`11`
# [1] 123
# 
# $`13`
# [1] 456

Recursive remove NULL elements of list of lists

One approach is to use rrapply in the rrapply-package (extension of base rapply):

library(rrapply)

x <- list(a=1, b=2, c=list(ca=1, cb=2, cc=NULL), d=NULL)

rrapply(x, condition = Negate(is.null), how = "prune")
#> $a
#> [1] 1
#> 
#> $b
#> [1] 2
#> 
#> $c
#> $c$ca
#> [1] 1
#> 
#> $c$cb
#> [1] 2

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Benchmark timings

Benchmarking the computation time of rrapply against rlist's list.clean function for some large nested lists, I get the following results:

## 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(
  rlist = rlist::list.clean(x_long, recursive = TRUE),
  rrapply = rrapply::rrapply(x_long, condition = Negate(is.null), how = "prune"),
  check = "equal",
  times = 5L
)
#> Unit: milliseconds
#>     expr       min        lq      mean    median        uq       max neval
#>    rlist 2331.4914 2343.3001 2438.9674 2441.3850 2512.3484 2566.3121     5
#>  rrapply  353.7169  393.0646  400.8198  399.7971  417.7235  439.7972     5

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

microbenchmark::microbenchmark(
  rlist = rlist::list.clean(x_deep, recursive = TRUE),
  rrapply = rrapply::rrapply(x_deep, condition = Negate(is.null), how = "prune"),
  check = "equal",
  times = 5L
)
#> Unit: milliseconds
#>     expr       min        lq      mean    median        uq       max neval
#>    rlist 2167.2946 2251.5203 2279.9963 2292.5045 2332.4432 2356.2188     5
#>  rrapply  268.9463  274.7437  325.9585  292.4559  354.1607  439.4857     5

Extract non null elements from a list in R

Here's another option:

Filter(Negate(is.null), x)

Extract non null elements from a list in R

Here's another option:

Filter(Negate(is.null), x)

Remove null list-columns

We could use select_if. NULL doesn't exist within a vector. So, the condition should be to check if the column is a list and if all the elements are NULL (is.null), negate (!) to select all the other columns

library(dplyr)
library(purrr)
null_list %>% 
    select_if(~ !(is.list(.) && all(map_lgl(., is.null))))
# A tibble: 3 x 1
#   name
#  <int>
#1     1
#2     2
#3     3

NOTE: This check if all the values in the list are NULL and then it removes those columns

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If it is deeply nested,

example %>%     
    select(where(~ !(is.list(.) && is.null(unlist(.)))))
# A tibble: 3 x 3
#      a     b c         
#  <dbl> <dbl> <list>    
#1     1     4 <NULL>    
#2     2     5 <NULL>    
#3     3     6 <list [3]>

Or to select only the list column with any non-NULL elements

example %>%     
    select(where(~ is.list(.) && !is.null(unlist(.))))
# A tibble: 3 x 1
#   c         
#  <list>    
#1 <NULL>    
#2 <NULL>    
#3 <list [3]>

data

 example <- tibble(a = c(1, 2, 3), b = c(4, 5, 6),
   c = list(NULL, NULL, list(1,2,3)),
   d = list(NULL, NULL, list(x = NULL, y = NULL, z = NULL)))

How to Remove Null Values in a List

This would be much easier if you just used Filter() and Map() For example

df.numeric.summary <- function(data.frame1){
    my_summary <- function(x) c(
      "Mean"=mean(x, na.rm = TRUE),
      "Median"=median(x, na.rm=TRUE),
      "IQR"=IQR(x, na.rm=TRUE))

    Map(my_summary, Filter(is.numeric, data.frame1))
}

You can test with

df.numeric.summary(iris)

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