Question

我有一个9200行和6列的数据集。我使用以下代码找到了该数据框的内核：

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

现在，我想将其作为方矩阵的对角线。例如，假设内核矩阵为

    K11  k12
    k21  k22

我需要一个像这样的矩阵

    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

Answer 1

您可能必须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"

Answer 2

一般来说，要回答这个问题，我建议使用R kronecker的{{1}}乘积。它所做的只是将第一个矩阵的所有元素（按标量）乘以整个第二个矩阵，然后转换为组合维的矩阵（如您的情况）。

例如：

base

如何将内核矩阵放在矩阵的对角线上？

2 个答案: