我的问题对你们中的许多人来说可能听起来微不足道,但经过长时间的互联网搜索,我仍然没有得到以下问题的答案:
如何将三维数组转换为“三维”列表?
假设我有以下内容:
A1 <- matrix(runif(12),4,3)
A2 <- matrix(runif(12),4,3)
A3 <- matrix(runif(12),4,3)
MyList <- list(A1,A2,A3)
MyArray <- array(NA,c(4,3,3))
MyArray[,,1] <- A1
MyArray[,,2] <- A2
MyArray[,,3] <- A3
有没有一种方法可以将MyArray
转换成与MyList
具有“相同结构”的列表?
非常感谢您的帮助!最好的,罗曼
发布于 2013-11-26 03:12:39
为了好玩(因为我来晚了),这里是另一个只使用基数R的n
,就像@joran的一样,它是可编程的,您可以很容易地沿着任何给定的维度拆分
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,如果您有任何,请参见示例:
dimnames(MyArray) <- Map(paste0, letters[seq_along(dim(MyArray))],
lapply(dim(MyArray), seq))
split.along.dim(MyArray, n = 3)
发布于 2013-11-25 17:05:25
您可以使用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
发布于 2019-02-11 19:31:13
在tidyverse的purrr
包中有一个函数array_tree()
,它可以以最小的麻烦完成这项工作:
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,这应该比plyr::alply()
(@joran的答案)更好,因为plyr
不再处于活跃的开发中。
https://stackoverflow.com/questions/20198751
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