如何将核矩阵放在矩阵的对角线上？答案

【问题标题】：how put kernel matrix in a diagonal of a matrix?如何将核矩阵放在矩阵的对角线上？
【发布时间】：2019-08-13 22:43:34
【问题描述】：

我有一个包含 9200 行和 6 列的数据集。我找到了这个数据框的内核，代码如下：

  #kernel
  library("kernlab", "v0.9-27")
  D<-as.matrix(X1)
  rbf <- rbfdot(sigma = 0.05)
  kernel<-kernelMatrix(rbf, D)

现在我想把它作为一个方阵的对角线。例如假设核矩阵是

    K11  k12
    k21  k22

我需要一个矩阵

    K11  k12   0    0    0    0    0    0
    k21  k22   0    0    0    0    0    0
    0     0   K11  k12   0    0    0    0
    0     0   k21  k22   0    0    0    0
    0     0    0    0   K11  k12   0    0
    0     0    0    0   K21  k22   0    0
    0     0    0    0    0    0    K11  k12
    0     0    0    0    0    0    K21  k22

【问题讨论】：

Matrix::bdiag(replicate(2, kernel, simple = FALSE))
我收到此错误：没有将“函数”强制为“CsparseMatrix”的方法或默认值
sherek_66;好吧，似乎它使用的是 stats::kernel 函数，而不是您问题中的 kernel 对象。运行您问题中的代码，然后运行评论中的代码^^，它将起作用。

标签： r dataframe matrix

【解决方案1】：

您可能需要根据实际需要调整尺寸 -

# this is your kernel matrix
m1 <- matrix(c("K11","k12","k21","k22"), nrow = 2, ncol = 2, byrow = T)

# this is output matrix; change nrow and ncol as per needs
m2 <- matrix("0", nrow = 8, ncol = 8)

for(i in seq(1, nrow(m2), nrow(m1))) {
  m2[i:(i+nrow(m1)-1), i:(i+nrow(m1)-1)] <- m1
}

m2
     [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8] 
[1,] "K11" "k12" "0"   "0"   "0"   "0"   "0"   "0"  
[2,] "k21" "k22" "0"   "0"   "0"   "0"   "0"   "0"  
[3,] "0"   "0"   "K11" "k12" "0"   "0"   "0"   "0"  
[4,] "0"   "0"   "k21" "k22" "0"   "0"   "0"   "0"  
[5,] "0"   "0"   "0"   "0"   "K11" "k12" "0"   "0"  
[6,] "0"   "0"   "0"   "0"   "k21" "k22" "0"   "0"  
[7,] "0"   "0"   "0"   "0"   "0"   "0"   "K11" "k12"
[8,] "0"   "0"   "0"   "0"   "0"   "0"   "k21" "k22"

【讨论】：

【解决方案2】：

一般来说，要回答这个问题，我最好建议使用 R base 的 kronecker 乘积。它所做的只是将第一个矩阵的所有元素（作为标量）乘以整个第二个矩阵，然后转换为一个组合维度的矩阵，就像你的情况一样。

例如：

D <- diag(1,4) # Identity matrix, in your case you should change 4 by 6
M <- matrix(1:4, 2, 2) # Matrix 2x2
kronecker(D,M) # Matrix you are looking for
#output

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

【讨论】：