将矩阵转换为列向量列表

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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

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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

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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 ;-)

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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

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最佳答案

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)))

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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"

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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

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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

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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

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convertRowsToList {BBmisc}

Convert rows (columns) of data.frame or matrix to lists.

BBmisc::convertColsToList(x)

ref: http://berndbischl.github.io/BBmisc/man/convertRowsToList.html

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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

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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)

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Use asplit to convert a matrix into a list of vectors. Use the MARGIN argument to give the margins to split by. For a matrix 1 indicates rows, 2 indicates columns.

asplit(x, MARGIN = 1) # split into list of row vectors
asplit(x, MARGIN = 2) # split into list of column vectors

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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)

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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