Ari*_*man 15 functional-programming r
聪明之后lapply
,我留下了一个二维矩阵列表.
例如:
set.seed(1)
test <- replicate( 5, matrix(runif(25),ncol=5), simplify=FALSE )
> test
[[1]]
[,1] [,2] [,3] [,4] [,5]
[1,] 0.8357088 0.29589546 0.9994045 0.2862853 0.6973738
[2,] 0.2377494 0.14704832 0.0348748 0.7377974 0.6414624
[3,] 0.3539861 0.70399206 0.3383913 0.8340543 0.6439229
[4,] 0.8568854 0.10380669 0.9150638 0.3142708 0.9778534
[5,] 0.8537634 0.03372777 0.6172353 0.4925665 0.4147353
[[2]]
[,1] [,2] [,3] [,4] [,5]
[1,] 0.1194048 0.9833502 0.9674695 0.6687715 0.1928159
[2,] 0.5260297 0.3883191 0.5150718 0.4189159 0.8967387
[3,] 0.2250734 0.2292448 0.1630703 0.3233450 0.3081196
[4,] 0.4864118 0.6232975 0.6219023 0.8352553 0.3633005
[5,] 0.3702148 0.1365402 0.9859542 0.1438170 0.7839465
[[3]]
...
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我想把它变成一个三维数组:
set.seed(1)
replicate( 5, matrix(runif(25),ncol=5) )
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显然,如果我使用复制,我可以打开simplify
,但sapply
不能正确简化结果,并stack
完全失败. do.call(rbind,mylist)
把它变成2d矩阵而不是3d数组.
我可以用循环来做到这一点,但我正在寻找一种简洁实用的方法来处理它.
我提出的最接近的方式是:
array( do.call( c, test ), dim=c(dim(test[[1]]),length(test)) )
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但我觉得它不够优雅(因为它反汇编然后重新组装向量的数组属性,并且需要大量测试才能确保安全(例如每个元素的尺寸相同).
mne*_*nel 10
您可以使用该abind
包然后使用abind(test, along = 3)
library(abind)
testArray <- abind(test, along = 3)
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或者您可以simplify = 'array'
在呼叫中使用sapply
,(而不是lapply
).simplify = 'array'
是不一样的simplify = TRUE
,因为它会改变说法higher
在simplify2array
例如
foo <- function(x) matrix(1:10, ncol = 5)
# the default is simplify = TRUE
sapply(1:5, foo)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 1 1
[2,] 2 2 2 2 2
[3,] 3 3 3 3 3
[4,] 4 4 4 4 4
[5,] 5 5 5 5 5
[6,] 6 6 6 6 6
[7,] 7 7 7 7 7
[8,] 8 8 8 8 8
[9,] 9 9 9 9 9
[10,] 10 10 10 10 10
# which is *not* what you want
# so set `simplify = 'array'
sapply(1:5, foo, simplify = 'array')
, , 1
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 5 7 9
[2,] 2 4 6 8 10
, , 2
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 5 7 9
[2,] 2 4 6 8 10
, , 3
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 5 7 9
[2,] 2 4 6 8 10
, , 4
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 5 7 9
[2,] 2 4 6 8 10
, , 5
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 5 7 9
[2,] 2 4 6 8 10
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