Using sample() to sample from nested lists in R - r

Question

I'm looking for a way to use sample() to sample values from different lists based on a value in another column of a data.table - at the moment I'm getting a recursive indexing failed error - code and more explanation below:
First set up example data:
library(stats)
library(data.table)
# list of three different nest survival rates
survival<-list(0.91,0.95,0.99)
# incubation period
inc.period<-28
# then set up function to use the geometric distribution to generate 3 lists of incubation outcomes based on the nest survivals and incubation period above.
# e.g. less than 28 is a nest failure, 28 is a successful nest.
create.sample <- function(survival){
outcome<-rgeom(100,1-survival)
fifelse(outcome > inc.period, inc.period, outcome)
}
# then create list of 100 nest outcomes with 3 different survival values using lapply
inc.outcomes <- lapply(survival,create.sample)
# set up a data.table - each row of data will be a nest.
index<-c(1:3)
iteration<-1:20
dt = CJ(index,iteration)
Then I want to make a new column 'inc.period' which samples from the 'inc.outcomes' list using the index column of the dt to select which of the three 'inc.outcomes' lists to sample from (with a different sample for each row of data).
So e.g. when index = 1, the sampled value comes from inc.outcomes[[1]] - which is the low nest survival list, when index = 2 I sample from inc.outcomes[[2]] etc.
The code would look something like this but this doesn't work (I get a recursive indexing failed error):
dt[,inc.period:= sample(inc.outcomes[[index]],nrow(dt),replace = TRUE)]
Any help or advice gratefully received, also suggestions for different approaches to this problem - this is for an update to code that runs in a shiny simulation so quicker options preferred!

Answer 1

Two problems:
inc.outcomes[[index]] is a problem since index is 60-long here, meaning you are ultimately trying inc.outcomes[[ c(1,1,...,2,2,...,3,3) ]], which is incorrect. [[-indexing is either length-1 (for most uses) or a vector as long as its list is nested. For example, in list(list(1,2),list(3,4))[[ c(1,2) ]] the [[c(1,2)]] with length-2 works because the have 2-deep nested lists. Since inc.outcomes is only 1-deep, we can only have length-1 in the [[ indexing.
This means we need to do this by-index. (An from this, we need to change from nrow(dt) to .N, but frankly we should be using that anyway even without by=.)
dt[, inc.period := sample(inc.outcomes[[ index[1] ]], .N, replace = TRUE), by = index]
# index iteration inc.period
# <int> <int> <num>
# 1: 1 1 17
# 2: 1 2 17
# 3: 1 3 21
# 4: 1 4 24
# 5: 1 5 3
# 6: 1 6 1
# 7: 1 7 17
# 8: 1 8 0
# 9: 1 9 1
# 10: 1 10 0
# ---
# 51: 3 11 0
# 52: 3 12 0
# 53: 3 13 28
# 54: 3 14 28
# 55: 3 15 9
# 56: 3 16 28
# 57: 3 17 7
# 58: 3 18 28
# 59: 3 19 28
# 60: 3 20 28
My data:
dt <- setDT(structure(list(index = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), iteration = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L)), row.names = c(NA, -60L), class = c("data.table", "data.frame"), sorted = c("index", "iteration")))