Converting Unsymmetric Vector-List into Matrix - r

Question

How is it possible to convert a list object (with different length) into a matrix object in an efficient way! Following example clarify the afore-mentioned goal:
imagine you have a list object of structure:
l <- list(c(1,2), c(5,7,3,11))
print(l)
# [[1]]
# [1] 1 2
# [[2]]
# [1] 5 7 3 11
The aim is to get a matrix or data.frame in form of:
[,1] [,2] [,3] [,4]
[1,] 1 2 NA NA
[2,] 5 7 3 11
It's very easy to tackle the problem with for-loop. Do you have any idea, how is it possible to make this kind of transformation easily? Thank you in advance!

Answer 1

You could also try
t(sapply(l, `length<-`, max(sapply(l, length))))
# [,1] [,2] [,3] [,4]
#[1,] 1 2 NA NA
#[2,] 5 7 3 11

Answer 2

Here's one way to do it:
n <- max(sapply(l, length))
t(sapply(l, function(x) if(length(x) < n) c(x, rep(NA, n - length(x))) else x))
[,1] [,2] [,3] [,4]
[1,] 1 2 NA NA
[2,] 5 7 3 11
First we find out the maximum vector length per list element and store it in n (which is 4 in this case).
Then, we sapply over the list and check if the length of the list element is equal to n and if it is, return it, if it's shorter than n, return the list element + NA repeated as often as the difference in length. This returns a matrix. We use t() on that matrix to transpose it and get the desired result.

Answer 3

If you're open to using a package, you could also consider stri_list2matrix from the "stringi" package:
library(stringi)
l <- list(c(1,2), c(5,7,3,11))
stri_list2matrix(l, byrow = TRUE)
# [,1] [,2] [,3] [,4]
# [1,] "1" "2" NA NA
# [2,] "5" "7" "3" "11"
Regarding your question about doing this efficiently, #akrun's answer is already pretty efficient, but could be made more efficient by using vapply instead of sapply. The "stringi" approach is also pretty efficient (and has the benefit of not resorting to cryptic code like length<-).
funDD <- function() {
n <- max(sapply(l, length))
t(sapply(l, function(x) if(length(x) < n) c(x, rep(NA, n - length(x))) else x))
}
funAK <- function() t(sapply(l, `length<-`, max(sapply(l, length))))
funAM <- function() {
x <- max(vapply(l, length, 1L))
t(vapply(l, `length<-`, numeric(x), x))
}
funStringi <- function() stri_list2matrix(l, byrow = TRUE)
## Make a big list to test on
set.seed(1)
l <- lapply(sample(3:10, 1000000, TRUE), function(x) sample(10, x, TRUE))
system.time(out1 <- funDD())
# user system elapsed
# 5.81 0.33 7.02
library(microbenchmark)
microbenchmark(funAK(), funAM(), funStringi(), times = 10)
# Unit: seconds
# expr min lq mean median uq max neval
# funAK() 2.350877 2.499963 2.974141 3.123008 3.200545 3.418648 10
# funAM() 1.154151 1.238235 1.337607 1.287610 1.494964 1.508884 10
# funStringi() 2.080901 2.168248 2.352030 2.344763 2.462959 2.716910 10