arrays - R reshape 数组以包含矩阵列表中的数据两次作为 keras 的准备

Question

我正在尝试创建一个 3D 数组，该数组复制元素(数据的副本)，并且两者的行保持相同。这最终是一个时间序列的 keras/tensorflow lstm 输入数组(样本、时间步长、特征)，其中，出于不同的原因，我确定我的时间步长也是我的特征，并且我的样本是固定的(行)。

一些数据:

perhaps <- matrix(sample(c(0:1), 20, replace = TRUE), nrow = 4, 
ncol = 5)
print(perhaps)
perhaps
     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    1    0    1
[2,]    0    0    0    0    1
[3,]    0    1    1    0    1
[4,]    1    0    0    1    1

perhaps_2_lst <- list(perhaps, perhaps)
all.equal(perhaps_2_lst[[1]], perhaps_2_lst[[2]])
[1] TRUE

#construct array from SOF inquestion:15213463
perhaps_arr <- array(
data = do.call(rbind, lapply(perhaps_2_lst, as.vector)),
dim = c(dim = c(dim(perhaps_2_lst[[1]])[1], 
dim(perhaps_2_lst[[1]])[2], dim(perhaps_2_lst[[1]])[2]))

dim(perhaps_arr)
[1] 4 5 5

令人鼓舞。

print(perhaps_arr)
, , 1

     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    0    1    1
[2,]    0    0    0    1    1
[3,]    0    1    0    0    0
[4,]    0    1    0    0    0

, , 2

     [,1] [,2] [,3] [,4] [,5]
[1,]    1    0    0    1    1
[2,]    1    0    0    1    1
[3,]    0    0    1    1    1
[4,]    0    0    1    1    1

, , 3

     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    0    1    1
[2,]    0    0    0    1    1
[3,]    0    1    0    0    0
[4,]    0    1    0    0    0

, , 4

     [,1] [,2] [,3] [,4] [,5]
[1,]    1    0    0    1    1
[2,]    1    0    0    1    1
[3,]    0    0    1    1    1
[4,]    0    0    1    1    1

, , 5

     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    0    1    1
[2,]    0    0    0    1    1
[3,]    0    1    0    0    0
[4,]    0    1    0    0    0

哦。我猜对数组不太熟悉。我期待这样的事情:

, , 1
#perhaps
     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    1    0    1
[2,]    0    0    0    0    1
[3,]    0    1    1    0    1
[4,]    1    0    0    1    1
, , 2
#perhaps
     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    1    0    1
[2,]    0    0    0    0    1
[3,]    0    1    1    0    1
[4,]    1    0    0    1    1

即。重复相同的数据。现在被难住了。对于理解这里发生的事情的建议和澄清非常感谢。

Answer 1

我们可以使用复制

n <- 2
replicate(n, perhaps)
#, , 1

#     [,1] [,2] [,3] [,4] [,5]
#[1,]    0    0    1    0    1
#[2,]    0    0    0    0    1
#[3,]    0    1    1    0    1
#[4,]    1    0    0    1    1

#, , 2

#     [,1] [,2] [,3] [,4] [,5]
#[1,]    0    0    1    0    1
#[2,]    0    0    0    0    1
#[3,]    0    1    1    0    1
#[4,]    1    0    0    1    1