可以使用lapply：

lapply(seq(dim(MyArray)[3]), function(x) MyArray[ , , x])

# [[1]]
#           [,1]       [,2]      [,3]
# [1,] 0.2050745 0.21410846 0.2433970
# [2,] 0.9662453 0.93294504 0.1466763
# [3,] 0.5775559 0.86977616 0.6950287
# [4,] 0.4626039 0.04009952 0.5197830
# 
# [[2]]
#           [,1]      [,2]      [,3]
# [1,] 0.6323070 0.2684788 0.7232186
# [2,] 0.1986486 0.2096121 0.2878846
# [3,] 0.3064698 0.7326781 0.8339690
# [4,] 0.3068035 0.4559094 0.8783581
# 
# [[3]]
#           [,1]      [,2]      [,3]
# [1,] 0.9557156 0.9069851 0.3415961
# [2,] 0.5287296 0.6292590 0.8291184
# [3,] 0.4023539 0.8106378 0.4257489
# [4,] 0.7199638 0.2708597 0.6327383

在plyr中有一个方便的函数：

alply(MyArray,3)
$`1`
          [,1]       [,2]       [,3]
[1,] 0.7643427 0.27546113 0.31131581
[2,] 0.6254926 0.19449191 0.04617286
[3,] 0.5879341 0.10484810 0.08056612
[4,] 0.4423744 0.09046864 0.82333646

$`2`
          [,1]      [,2]      [,3]
[1,] 0.3726026 0.3063512 0.4997664
[2,] 0.8757070 0.2309768 0.9274503
[3,] 0.9269987 0.5751226 0.9347077
[4,] 0.4063655 0.4593746 0.4830263

$`3`
          [,1]       [,2]      [,3]
[1,] 0.7538325 0.18824996 0.3679285
[2,] 0.4985409 0.61026876 0.4134485
[3,] 0.3209792 0.60056130 0.8887652
[4,] 0.0160972 0.06534362 0.2618056

只需添加.dims参数即可保留维名称：

dimnames(MyArray) <- Map(paste0, letters[seq_along(dim(MyArray))],
                                  lapply(dim(MyArray), seq))

alply(MyArray,3,.dims = TRUE)
$c1
    b
a           b1         b2        b3
  a1 0.4752803 0.01728003 0.1744352
  a2 0.7144411 0.13353980 0.1069188
  a3 0.2429445 0.60039428 0.8610824
  a4 0.9757289 0.71712288 0.5941202

$c2
    b
a            b1         b2        b3
  a1 0.07118296 0.43761119 0.3174442
  a2 0.16458581 0.65040897 0.5654846
  a3 0.88711374 0.07655825 0.7163768
  a4 0.07117881 0.79314705 0.9054457

$c3
    b
a            b1        b2         b3
  a1 0.04761279 0.5668479 0.04145537
  a2 0.72320804 0.2692747 0.74700930
  a3 0.82138686 0.3604211 0.57163369
  a4 0.53325169 0.8831302 0.71119421

为了好玩（因为我迟到了），这里有另一个只使用R基的。像@joran一样，它是可编程的，你可以很容易地沿着任何给定的维度n进行分割：

split.along.dim <- function(a, n)
  setNames(lapply(split(a, arrayInd(seq_along(a), dim(a))[, n]),
                  array, dim = dim(a)[-n], dimnames(a)[-n]),
           dimnames(a)[[n]])

identical(split.along.dim(MyArray, n = 3), MyList)
# [1] TRUE

它还将保留您的所有dimname（如果有的话），请参见例如：

dimnames(MyArray) <- Map(paste0, letters[seq_along(dim(MyArray))],
                                 lapply(dim(MyArray), seq))
split.along.dim(MyArray, n = 3)

在tidyverse的purrr包中有一个函数array_tree()，它可以最小化地做到这一点：

A1 <- matrix(runif(12),4,3)
A2 <- matrix(runif(12),4,3)
A3 <- matrix(runif(12),4,3)

MyList <- list(a1=A1, a2=A2, a3=A3)

MyArray <- purrr::array_tree(MyList)

$`a1`
           [,1]       [,2]       [,3]
[1,] 0.18895576 0.16225488 0.09941778
[2,] 0.69737985 0.01757565 0.84838836
[3,] 0.06849385 0.71726810 0.52981969
[4,] 0.83352338 0.90922401 0.55946707

$a2
          [,1]      [,2]       [,3]
[1,] 0.6498039 0.5015537 0.48840965
[2,] 0.7745612 0.4346254 0.40873822
[3,] 0.4169687 0.3634961 0.01878936
[4,] 0.3753315 0.3008145 0.94580448

$a3
          [,1]      [,2]      [,3]
[1,] 0.3967292 0.8117389 0.9184360
[2,] 0.5729680 0.9466026 0.4086640
[3,] 0.5744074 0.5913192 0.6038389
[4,] 0.4507735 0.2285655 0.8815671

如图所示，默认情况下，它会保留名称。一般来说，这应该优于plyr::alply()（@joran的答案），因为plyr不再处于积极的开发中。

在R的最新版本（>= 4.1.0，2021年5月）中，apply现在有一个simplify=参数，这可以通过在第三维上循环来实现，如：

apply(MyArray, 3, identity, simplify=FALSE)

它还将保留每个矩阵的列表names和dimnames。