# Speed up computation on lists of lists in R (scalability issues)

My goal is to take a source list of vectors and apply a mapping to each vector. The mapping creates a list of lists. When the vector size in each list element is large (~20000), and the number of elements in the source list is large (~5000), my program does not seem to be efficient. How can I optimize my code?

I've tried two things: implementation in Rcpp and mclapply. The Rcpp implementation is comparable as expected, and the mclapply implementation may not be that efficient because I'm running my function in a larger function that is already parallelized over multiple cores.

Basic example

source.list = rep(list(seq(6)),3)
target.list = list(c(1,2),c(3,4),c(5,6))

result = map_partition(source.list,target.list)

> result
[[1]]
[[1]][[1]]
[1] 1 2

[[1]][[2]]
[1] 3 4

[[1]][[3]]
[1] 5 6

[[2]]
[[2]][[1]]
[1] 1 2

[[2]][[2]]
[1] 3 4

[[2]][[3]]
[1] 5 6

[[3]]
[[3]][[1]]
[1] 1 2

[[3]][[2]]
[1] 3 4

[[3]][[3]]
[1] 5 6


R implementation

map.partition.R <- function(invec, partitionlist) {
lst <- lapply(partitionlist, function(x) invec[invec %in% x])
return(lst)
}

result = lapply(X=source.list, FUN=map.partition,
partitionlist=target.list)

• Is function(x) invec[invec %in% x] what you really intend to apply or is it just here for illustration purposes? If it is really what you intend to do, I might have an idea using a data.table merge. – flodel Nov 17 '18 at 1:55
• @flodel - it is indeed what I want to use. I want to "match" the elements of each vector in source.list and and each vector in target.list – stats134711 Nov 17 '18 at 13:03

You have two lapply calls in there, so essentially two hidden for loops contributing to the slow execution.

Here, one way to tap into faster compiled functions is to rely on the data.table package for merging (a.k.a. a database join) your two data sets.

First, we turn your two lists into two data.tables:

idx1 <- rep(seq_along(source.list), lengths(source.list))
idx2 <- rep(seq_along(target.list), lengths(target.list))

library(data.table)
X <- data.table(id = unlist(source.list), idx1 = idx1,
pos = seq_along(idx1), key = "id")
Y <- data.table(id = unlist(target.list), idx2 = idx2, key = "id")


where idx1 and idx2 are the variables telling us in which element of source.list and target.list a specific item (id) belongs. Also pos is just a temporary row number we will use later to sort the data back, as the next statement below (the merge) shuffles things around:

Z <- Y[X, allow.cartesian=TRUE]
Z <- Z[order(Z$pos)]  At this point, if it's possible for you, I would recommend you stop here. By that, I mean to make the next steps of your analysis use this Z data.table rather than a nested list, as to avoid further slow processing loops (for/lapply/Map/etc.). However, if you really need the nested list output, you can do: no_name_split <- function(...) unname(split(...)) gp1 <- factor(Z$idx1, seq_along(source.list))
gp2 <- factor(Z$idx2, seq_along(target.list)) res <- Map(no_name_split, no_name_split(Z$id, gp1),
no_name_split(gp2,  gp1))


Below is a simulation with a larger dataset, comparing computation times and checking that the results are identical:

set.seed(632)
source.list <- replicate(500, sample(100000, 1000), simplify = FALSE)
target.list <- replicate(555, sample(100000, 1055), simplify = FALSE)

system.time({
idx1 <- rep(seq_along(source.list), lengths(source.list))
idx2 <- rep(seq_along(target.list), lengths(target.list))

library(data.table)
X <- data.table(id = unlist(source.list), idx1 = idx1,
pos = seq_along(idx1), key = "id")
Y <- data.table(id = unlist(target.list), idx2 = idx2, key = "id")
Z <- Y[X, allow.cartesian = TRUE]
Z <- Z[order(Z$pos)] no_name_split <- function(...) unname(split(...)) gp1 <- factor(Z$idx1, seq_along(source.list))
gp2 <- factor(Z$idx2, seq_along(target.list)) res <- Map(no_name_split, no_name_split(Z$id, gp1),
no_name_split(gp2,  gp1))
})
#    user  system elapsed
#   3.394   0.382   3.646

system.time({
result <- lapply(X = source.list, FUN = map.partition.R,
partitionlist = target.list)
})
#    user  system elapsed
#  23.943   5.329  36.999

identical(res, result)
# [1] TRUE