Question

我有一个给定的向量，并希望将其转换为某个块矩阵。考虑这个简单的例子：

k <- c(1,2,3)
a <- rep(apply(expand.grid(k, k), 1, prod), each=3)
a
[1] 1 1 1 2 2 2 3 3 3 2 2 2 4 4 4 6 6 6 3 3 3 6 6 6 9 9 9

此向量应在以下形式的块矩阵中对齐：

rbind(
cbind(diag(a[1:3]), diag(a[4:6]), diag(a[7:9])),
cbind(diag(a[10:12]), diag(a[13:15]), diag(a[16:18]) ),
cbind(diag(a[19:21]), diag(a[22:24]), diag(a[25:27]) ) 
)

       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
 [1,]    1    0    0    2    0    0    3    0    0
 [2,]    0    1    0    0    2    0    0    3    0
 [3,]    0    0    1    0    0    2    0    0    3
 [4,]    2    0    0    4    0    0    6    0    0
 [5,]    0    2    0    0    4    0    0    6    0
 [6,]    0    0    2    0    0    4    0    0    6
 [7,]    3    0    0    6    0    0    9    0    0
 [8,]    0    3    0    0    6    0    0    9    0
 [9,]    0    0    3    0    0    6    0    0    9

现在这当然是一个小而简单的例子，我想为更大的矢量/矩阵做这个。因此我的问题是：是否有一种通用的方法来对齐某种形式的块矩阵中的向量（没有循环）？

Answer 1

我们可以使用%/%

而不是手动执行拆分

k <- 3
lst <- split(a, (seq_along(a)-1)%/%k + 1)
do.call(rbind, lapply(split(lst, (seq_along(lst)-1) %/% k + 1), 
       function(x) do.call(cbind,  lapply(x, function(y) diag(y)))))
#       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
# [1,]    1    0    0    2    0    0    3    0    0
# [2,]    0    1    0    0    2    0    0    3    0
# [3,]    0    0    1    0    0    2    0    0    3
# [4,]    2    0    0    4    0    0    6    0    0
# [5,]    0    2    0    0    4    0    0    6    0
# [6,]    0    0    2    0    0    4    0    0    6
# [7,]    3    0    0    6    0    0    9    0    0
# [8,]    0    3    0    0    6    0    0    9    0
# [9,]    0    0    3    0    0    6    0    0    9

Answer 2

在稍微不同的向量上使用Kronecker产品的替代方案如下。

# create initial vector
aNew <- rep(1:3, 3) * rep(1:3, each=3)
aNew
[1] 1 2 3 2 4 6 3 6 9

请注意，aNew是矢量a的唯一值，顺序相同，即相当于unique(a)。将aNew转换为3X3矩阵，然后针对它执行Kronecker产品和3X3单位矩阵。

matrix(aNew, 3, 3) %x% diag(3)
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
 [1,]    1    0    0    2    0    0    3    0    0
 [2,]    0    1    0    0    2    0    0    3    0
 [3,]    0    0    1    0    0    2    0    0    3
 [4,]    2    0    0    4    0    0    6    0    0
 [5,]    0    2    0    0    4    0    0    6    0
 [6,]    0    0    2    0    0    4    0    0    6
 [7,]    3    0    0    6    0    0    9    0    0
 [8,]    0    3    0    0    6    0    0    9    0
 [9,]    0    0    3    0    0    6    0    0    9

矢量到块矩阵

2 个答案: