使用偏移量i(矢量化方法)拆分大小为n的块中的矩阵

Question

我想将大小的矩阵分成大小的k x l块,n x n考虑到一个问题o(就像Mathematica的分区函数那样).

例如,给定矩阵之A类的

A <- matrix(seq(1:16), nrow = 4, ncol = 4)

     [,1] [,2] [,3] [,4]
[1,]    1    5    9   13
[2,]    2    6   10   14
[3,]    3    7   11   15
[4,]    4    8   12   16

和块大小= 3,偏移= 1,我想要输出我得到的四个子矩阵

A[1:3, 1:3]
A[1:3, 2:4]
A[2:4, 1:3]
A[2:4, 2:4]

如果offset等于2或3,则此示例的输出应该只是我得到的子矩阵

A[1:3, 1:3]

我该如何对此进行矢量化？

Answer 1

可能有更优雅的方式。myPartition以下是我通过编写一个模拟 mathematica 函数的函数来实现的方法Partition。首先使用 use沿我们要考虑的Map行和列轴构建可能的索引，然后使用from构建子集索引的所有可能组合的列表。最后用于对矩阵进行子集化并返回子集矩阵列表；seqoffsetcross2purrrlapply

offset和的测试结果如下1，似乎表现符合预期：23

library(purrr)
ind <- function(k, n, o) Map(`:`, seq(1, k-n+1, by = o), seq(n, k, by = o))

# this is a little helper function that generates subset index according to dimension of the 
# matrix, the first sequence construct the starting point of the subset index with an interval 
# of o which is the offset while the second sequence construct the ending point of the subset index
# use Map to construct vector from start to end which in OP's case will be 1:3 and 2:4. 

myPartition <- function(mat, n, o) {
    lapply(cross2(ind(nrow(mat),n,o), ind(ncol(mat),n,o)), function(i) mat[i[[1]], i[[2]]])
}

# This is basically an lapply. we use cross2 to construct combinations of all subset index
# which will be 1:3 and 1:3, 1:3 and 2:4, 2:4 and 1:3 and 2:4 and 2:4 in OP's case. Use lapply
# to loop through the index and subset.

# Testing case for offset = 1
myPartition(A, 3, 1)

# [[1]]
#      [,1] [,2] [,3]
# [1,]    1    5    9
# [2,]    2    6   10
# [3,]    3    7   11

# [[2]]
#      [,1] [,2] [,3]
# [1,]    2    6   10
# [2,]    3    7   11
# [3,]    4    8   12

# [[3]]
#      [,1] [,2] [,3]
# [1,]    5    9   13
# [2,]    6   10   14
# [3,]    7   11   15

# [[4]]
#      [,1] [,2] [,3]
# [1,]    6   10   14
# [2,]    7   11   15
# [3,]    8   12   16

# Testing case for offset = 2
myPartition(A, 3, 2)
# [[1]]
#      [,1] [,2] [,3]
# [1,]    1    5    9
# [2,]    2    6   10
# [3,]    3    7   11

# Testing case for offset = 3
myPartition(A, 3, 3)
# [[1]]
#      [,1] [,2] [,3]
# [1,]    1    5    9
# [2,]    2    6   10
# [3,]    3    7   11