Unlist the second to last list of a nested list - r

Question

I have a deeply nested list of lists. In the "center" of the nested list is a vector containing n integers. I need to count how many integers are in each nested list, then unlist one level above to have a vector of these counts (i.e., instead of list(0, 1:5, 0, 0, 1:3) at the center of the nest, I want c(0, 5, 0, 0, 3).
This seems relatively simple - I was able to use rapply to accomplish the first part, i.e. convert list(0, 1:5, 0, 0, 1:3) to list(0, 5, 0, 0, 3). My specific question I need help with is how to unlist the innermost lists to a vector (instead of list(0, 5, 0, 0, 3) I want c(0, 5, 0, 0, 3)
I have searched and tried various apply, lapply, unlist approaches but none of them are quite right as they target the very innermost list. Since the list I want to unlist is the second to last element, I am struggling finding a way to accomplish this elegantly.
In the sample data below, I can get the desired outcome 2 ways: either multiple lapply functions or a for loop. However, my actual data contain many more lists and millions of datapoints, so these are likely not effective options.
Below is (1) sample data, (2) what I have tried, and (3) sample data having the desired structure.
Sample Data
have_list <- list(scenario1 = list(method1 = list(place1 = list(0, 1:5, 0, 0, 1:3),
place2 = list(1:2, 0, 1:10, 0, 0),
place3 = list(0:19, 0, 0, 0, 0),
place4 = list(1:100, 0, 0, 1:4, 0)),
method2 = list(place1 = list(1:5, 1:5, 0, 0, 1:3),
place2 = list(0, 0, 1:5, 0, 0),
place3 = list(0:19, 0, 1:7, 0, 0),
place4 = list(1:22, 0, 0, 1:4, 0)),
method3 = list(place1 = list(0, 1:2, 1:6, 0, 1:3),
place2 = list(1:2, 0, 1:6, 1:4, 0),
place3 = list(0:19, 0, 0, 0, 1:2),
place4 = list(1:12, 0, 0, 1:12, 0))),
scenario2 = list(method1 = list(place1 = list(0, 1:5, 0, 0, 1:3),
place2 = list(1:2, 0, 1:10, 0, 0),
place3 = list(0:19, 0, 0, 0, 0),
place4 = list(1:100, 0, 0, 1:4, 0)),
method2 = list(place1 = list(1:5, 1:5, 0, 0, 1:3),
place2 = list(0, 0, 1:5, 0, 0),
place3 = list(0:19, 0, 1:7, 0, 0),
place4 = list(1:22, 0, 0, 1:4, 0)),
method3 = list(place1 = list(0, 1:2, 1:6, 0, 1:3),
place2 = list(1:2, 0, 1:6, 1:4, 0),
place3 = list(0:19, 0, 0, 0, 1:2),
place4 = list(1:12, 0, 0, 1:12, 0))))
What I have tried
And SO questions I have visited:
Unlist LAST level of a list in R
R - unlist nested list of dates
Issues unlisting a nested-list
# Get number of integers in each nested list
lengths <- rapply(have_list, function(x) unlist(length(x)), how = "list") # this works fine
#' Each count is currently still in its own list of length 1,
#' Convert each count to vector
#' In the "middle" the nested list:
# I have list(0, 5, 0, 0, 3)
# I want c(0, 5, 0, 0, 3)
# Attempts to unlist the counts
# Unlist the counts
test1 <- rapply(lengths, unlist, how = "list") # doesn't work
test2 <- unlist(lengths, recursive = FALSE) # doesn't work
test3 <- lapply(lengths, function(x) lapply(x, unlist)) # doesnt work
test4 <- lapply(lengths, function(x) lapply(x, unlist, recursive = FALSE)) # doesnt work
test5 <- rapply(have_list, function(x) unlist(length(x)), how = "list") #doesnt work
test6 <- rapply(have_list, function(x) unlist(length(x)), how = "unlist") #doesnt work
Data structure I want
# This works on test data but is impractical for real data
want_list <- lapply(lengths, function(w) lapply(w, function(x) lapply(x, unlist)))
# or
want_list <- lengths
## for loops work but is not practical
for (i in 1:length(lengths)){
for (j in 1:length(lengths[[i]])){
for (k in 1:length(lengths[[i]][[j]])){
want_list[[i]][[j]][[k]] <- unlist(lengths[[i]][[j]][[k]])
}
}
}

Answer 1

An option is to melt the nested list with rrapply, replace the 'value' column with the lengths and then use the recursive split (rsplit) from collapse
library(rrapply)
library(collapse)
dat <- transform(rrapply(have_list, how = "melt"), value= lengths(value))
out <- rsplit(dat$value, dat[1:3])
-testing with OP' expected
identical(out, want_list)
[1] TRUE

Answer 2

Another solution with rrapply() could be to apply lengths() only to the lists of vectors using a condition function:
library(rrapply)
out <- rrapply(have_list, classes = "list", condition = \(x) is.numeric(x[[1]]), f = lengths)
identical(want_list, out)
#> [1] TRUE

Answer 3

This can be done by using recursion. A simple recursion will be:
my_fun <- function(x) if(is.list(x[[1]])) lapply(x, my_fun) else lengths(x)
out <- my_fun(have_list)
identical(out, want_list)
[1] TRUE