rlistsapply

replace numeric(0) with 0 in R list

I have a list:

myList
$`0`
$`0`$item1
numeric(0)

$`0`$item2
[1] 350

$`0`$item3
numeric(0)


$`1`
$`1`$item1
numeric(0)

$`1`$item2
[1] 56

$`1`$item3
numeric(0)

I am using an sapply function on this list, but get an error:

invalid 'type' (list) of argument

how can i convert all numeric(0) to 0, like the other items?

Solution

  • In case the depth is not known or not equal you can use a recursive function:

    f <- function(x) {
  if(is.list(x)) lapply(x, f)
  else ifelse(length(x) == 0, 0, x)
}

f(myList)
#$`0`
#$`0`$item1
#[1] 0
#
#$`0`$item2
#[1] 350
#
#$`0`$item3
#[1] 0
#
#
#$`1`
#$`1`$item1
#[1] 0
#
#$`1`$item2
#[1] 56
#
#$`1`$item3
#[1] 0
#
#
#$`2`
#[1] 4
#
#$`3`
#[1] 0

    Another option is to use rapply

    rapply(myList, \(x) {if(length(x)==0) 0 else x}, how="replace")

    Data:

    myList <- list(`0` = list(item1 = numeric(0), item2 = 350, item3 = numeric(0)),
               `1` = list(item1 = numeric(0), item2 = 56, item3 = numeric(0)),
               `2` = 4, `3` = numeric(0))