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
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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
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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)?
我们可以使用而不是手动进行拆分 %/%
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
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