Reputation: 3815
How can I set all vector elements to NA in a list of vectors?
Essentially, I'd like to keep an existing list's structure and names but empty all values, to fill them in later. I provide a minimal example with a couple solutions below. I prefer base and tidyverse (esp. purrr) solutions, but can get on board with any approach which is better than what I have below.
my_list <- list(A = c('a' = 1, 'b' = 2, 'c' = 3), B = c('x' = 10, 'y' = 20))
ret_list <- my_list
# Approach 1
for (element_name in names(my_list)) {
ret_list[[element_name]][] <- NA
}
ret_list
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
# Approach 2
lapply(my_list, function(x) {x[] <- NA; return(x)})
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
Upvotes: 9
Views: 1799
Reputation: 48211
Here's another one for numeric vectors:
lapply(my_list, `*`, NA) # Instead of * it could also be +, -, etc.
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
More generally,
lapply(my_list, replace, TRUE, NA)
and
lapply(ret_list, ifelse, NA, NA)
Upvotes: 8
Reputation: 269526
If the list is not restricted to one level then use rapply
.
# test data modified from question
my_list2 <- list(list(A = c(a = 1, b = 2, c = 3)), B = c(x = 10, y = 20))
rapply(my_list2, function(x) replace(x, TRUE, NA), how = "list")
which can also be written as:
rapply(my_list2, replace, list = TRUE, values = NA, how = "list")
Upvotes: 4
Reputation: 2208
another alternative with dplyr
:
lapply(my_list, function(x) dplyr::na_if(x,x))
Upvotes: 4
Reputation: 12937
Another way around
relist(replace( unlist(my_list), TRUE, NA ), skeleton = my_list)
#$A
# a b c
#NA NA NA
#$B
# x y
#NA NA
Upvotes: 4
Reputation: 76402
You can use function is.na<-
in a lapply
loop.
ret_list <- lapply(my_list, `is.na<-`)
ret_list
#$A
# a b c
#NA NA NA
#
#$B
# x y
#NA NA
Upvotes: 8