矢量分块矩阵
问题描述:
我给出了一个向量,并希望将其转换为某个块矩阵。考虑一个简单的例子:矢量分块矩阵
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
这个载体应形式的分块矩阵排列:
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,我想做到这一点的更大的矢量/矩阵。因此,我的问题:是否有一种通用的方法来对齐某个形式的块矩阵中的矢量(没有循环)?
答
不用手动操作的方式分割,我们使用克罗内克积上的稍微不同的向量如下可以使用%/%
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
答
一种替代。
# create initial vector
aNew <- rep(1:3, 3) * rep(1:3, each=3)
aNew
[1] 1 2 3 2 4 6 3 6 9
注意,重新处于相同的顺序向量a的唯一值,也就是说,它等同于unique(a)
。将aNew转换为3X3矩阵,然后针对它和3X3单位矩阵执行Kronecker产品。
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
也许相关的帖子:https://*.com/questions/17495841/block-diagonal-binding-of-matrices – zx8754