How to convert a matrix to a list of column-vectors in R?
ListRMatrixList Problem Overview
Say you want to convert a matrix to a list, where each element of the list contains one column. list() or as.list() obviously won't work, and until now I use a hack using the behaviour of tapply :
x <- matrix(1:10,ncol=2)
tapply(x,rep(1:ncol(x),each=nrow(x)),function(i)i)
I'm not completely happy with this. Anybody knows a cleaner method I'm overlooking?
(for making a list filled with the rows, the code can obviously be changed to :
tapply(x,rep(1:nrow(x),ncol(x)),function(i)i)
)
List Solutions
Solution 1 - List
Gavin's answer is simple and elegant. But if there are many columns, a much faster solution would be:
lapply(seq_len(ncol(x)), function(i) x[,i])
The speed difference is 6x in the example below:
> x <- matrix(1:1e6, 10)
> system.time( as.list(data.frame(x)) )
user system elapsed
1.24 0.00 1.22
> system.time( lapply(seq_len(ncol(x)), function(i) x[,i]) )
user system elapsed
0.2 0.0 0.2
Solution 2 - List
In the interests of skinning the cat, treat the array as a vector as if it had no dim attribute:
split(x, rep(1:ncol(x), each = nrow(x)))
Solution 3 - List
data.frames are stored as lists, I believe. Therefore coercion seems best:
as.list(as.data.frame(x))
> as.list(as.data.frame(x))
$V1
[1] 1 2 3 4 5
$V2
[1] 6 7 8 9 10
Benchmarking results are interesting. as.data.frame is faster than data.frame, either because data.frame has to create a whole new object, or because keeping track of the column names is somehow costly (witness the c(unname()) vs c() comparison)? The lapply solution provided by @Tommy is faster by an order of magnitude. The as.data.frame() results can be somewhat improved by coercing manually.
manual.coerce <- function(x) {
x <- as.data.frame(x)
class(x) <- "list"
x
}
library(microbenchmark)
x <- matrix(1:10,ncol=2)
microbenchmark(
tapply(x,rep(1:ncol(x),each=nrow(x)),function(i)i) ,
as.list(data.frame(x)),
as.list(as.data.frame(x)),
lapply(seq_len(ncol(x)), function(i) x[,i]),
c(unname(as.data.frame(x))),
c(data.frame(x)),
manual.coerce(x),
times=1000
)
expr min lq
1 as.list(as.data.frame(x)) 176221 183064
2 as.list(data.frame(x)) 444827 454237
3 c(data.frame(x)) 434562 443117
4 c(unname(as.data.frame(x))) 257487 266897
5 lapply(seq_len(ncol(x)), function(i) x[, i]) 28231 35929
6 manual.coerce(x) 160823 167667
7 tapply(x, rep(1:ncol(x), each = nrow(x)), function(i) i) 1020536 1036790
median uq max
1 186486 190763 2768193
2 460225 471346 2854592
3 449960 460226 2895653
4 271174 277162 2827218
5 36784 37640 1165105
6 171088 176221 457659
7 1052188 1080417 3939286
is.list(manual.coerce(x))
[1] TRUE
Solution 4 - List
Converting to a data frame thence to a list seems to work:
> as.list(data.frame(x))
$X1
[1] 1 2 3 4 5
$X2
[1] 6 7 8 9 10
> str(as.list(data.frame(x)))
List of 2
$ X1: int [1:5] 1 2 3 4 5
$ X2: int [1:5] 6 7 8 9 10
Solution 5 - List
Using plyrcan be really useful for things like this:
library("plyr")
alply(x,2)
$`1`
[1] 1 2 3 4 5
$`2`
[1] 6 7 8 9 10
attr(,"class")
[1] "split" "list"
Solution 6 - List
Use asplit to convert a matrix into a list of vectors
asplit(x, 1) # split into list of row vectors
asplit(x, 2) # split into list of column vectors
Solution 7 - List
I know this is anathema in R, and I don't really have a lot of reputation to back this up, but I'm finding a for loop to be rather more efficient. I'm using the following function to convert matrix mat to a list of its columns:
mat2list <- function(mat)
{
list_length <- ncol(mat)
out_list <- vector("list", list_length)
for(i in 1:list_length) out_list[[i]] <- mat[,i]
out_list
}
Quick benchmark comparing with mdsummer's and the original solution:
x <- matrix(1:1e7, ncol=1e6)
system.time(mat2list(x))
user system elapsed
2.728 0.023 2.720
system.time(split(x, rep(1:ncol(x), each = nrow(x))))
user system elapsed
4.812 0.194 4.978
system.time(tapply(x,rep(1:ncol(x),each=nrow(x)),function(i)i))
user system elapsed
11.471 0.413 11.817
Solution 8 - List
The new function asplit() will be coming to base R in v3.6. Up until then and in similar spirit to the answer of @mdsumner we can also do
split(x, slice.index(x, MARGIN))
as per the docs of asplit(). As previously shown however, all split() based solutions are much slower than @Tommy's lapply/`[` . This also holds for the new asplit(), at least in its current form.
split_1 <- function(x) asplit(x, 2L)
split_2 <- function(x) split(x, rep(seq_len(ncol(x)), each = nrow(x)))
split_3 <- function(x) split(x, col(x))
split_4 <- function(x) split(x, slice.index(x, 2L))
split_5 <- function(x) lapply(seq_len(ncol(x)), function(i) x[, i])
dat <- matrix(rnorm(n = 1e6), ncol = 100)
#> Unit: milliseconds
#> expr min lq mean median uq max neval
#> split_1(dat) 16.250842 17.271092 20.26428 18.18286 20.185513 55.851237 100
#> split_2(dat) 52.975819 54.600901 60.94911 56.05520 60.249629 105.791117 100
#> split_3(dat) 32.793112 33.665121 40.98491 34.97580 39.409883 74.406772 100
#> split_4(dat) 37.998140 39.669480 46.85295 40.82559 45.342010 80.830705 100
#> split_5(dat) 2.622944 2.841834 3.47998 2.88914 4.422262 8.286883 100
dat <- matrix(rnorm(n = 1e6), ncol = 1e5)
#> Unit: milliseconds
#> expr min lq mean median uq max neval
#> split_1(dat) 204.69803 231.3023 261.6907 246.4927 289.5218 413.5386 100
#> split_2(dat) 229.38132 235.3153 253.3027 242.0433 259.2280 339.0016 100
#> split_3(dat) 208.29162 216.5506 234.2354 221.7152 235.3539 342.5918 100
#> split_4(dat) 214.43064 221.9247 240.7921 231.0895 246.2457 323.3709 100
#> split_5(dat) 89.83764 105.8272 127.1187 114.3563 143.8771 209.0670 100
Solution 9 - List
There's a function array_tree() in the tidyverse's purrr package that does this with minimum fuss:
x <- matrix(1:10,ncol=2)
xlist <- purrr::array_tree(x, margin=2)
xlist
#> [[1]]
#> [1] 1 2 3 4 5
#>
#> [[2]]
#> [1] 6 7 8 9 10
Use margin=1 to list by row instead. Works for n-dimensional arrays. It preserves names by default:
x <- matrix(1:10,ncol=2)
colnames(x) <- letters[1:2]
xlist <- purrr::array_tree(x, margin=2)
xlist
#> $a
#> [1] 1 2 3 4 5
#>
#> $b
#> [1] 6 7 8 9 10
(this is a near word-for-word copy of my answer to a similar question here)
Solution 10 - List
Under Some R Help site accessible via nabble.com I find:
c(unname(as.data.frame(x)))
as a valid solution and in my R v2.13.0 install this looks ok:
> y <- c(unname(as.data.frame(x)))
> y
[[1]]
[1] 1 2 3 4 5
[[2]]
[1] 6 7 8 9 10
Can't say anythng about performance comparisons or how clean it is ;-)
Solution 11 - List
You could use apply and then c with do.call
x <- matrix(1:10,ncol=2)
do.call(c, apply(x, 2, list))
#[[1]]
#[1] 1 2 3 4 5
#
#[[2]]
#[1] 6 7 8 9 10
And it looks like it will preserve the column names, when added to the matrix.
colnames(x) <- c("a", "b")
do.call(c, apply(x, 2, list))
#$a
#[1] 1 2 3 4 5
#
#$b
#[1] 6 7 8 9 10
Solution 12 - List
convertRowsToList {BBmisc}
Convert rows (columns) of data.frame or matrix to lists.
BBmisc::convertColsToList(x)
ref: http://berndbischl.github.io/BBmisc/man/convertRowsToList.html
Solution 13 - List
In the trivial case where the number of columns is small and constant, then I've found that the fastest option is to simply hard-code the conversion:
mat2list <- function (mat) lapply(1:2, function (i) mat[, i])
mat2list2 <- function (mat) list(mat[, 1], mat[, 2])
## Microbenchmark results; unit: microseconds
# expr min lq mean median uq max neval
## mat2list(x) 7.464 7.932 8.77091 8.398 8.864 29.390 100
## mat2list2(x) 1.400 1.867 2.48702 2.333 2.333 27.525 100
Solution 14 - List
The simplest way to create a list that has the columns of a matrix mat as its elements is to use the fact that a data.frame object in R is internally represented as a list of the columns. Thus all that is needed is the following line
mat.list <- as.data.frame(mat)
Solution 15 - List
A dplyr readable renewed approach for the same thing:
x <- matrix(1:10,ncol=2)
library(dplyr)
x %>% as_tibble() %>%
as.list()
$V1
[1] 1 2 3 4 5
$V2
[1] 6 7 8 9 10