matrix - 如何在 Julia 中以对角线形式连接矩阵

Question

我有一个矩阵 A我想建立一个块对角矩阵 M作为

A 0 ⋯ 0
0 A ⋯ 0
⋮ ⋮ ⋱ ⋮
0 0 ⋯ A

哪里 0 s 是与 A 大小相同的矩阵充满零。有没有方便的方法来做到这一点？
换句话说，是否有等效于 Python 的 linalg.block_diag (来自 scipy)？

Answer 1

您可以使用 blockdiag来自 SparseArrays.jl 标准库。例如:

julia> using SparseArrays

julia> A = sparse(rand(3,2)) # A must be sparse for blockdiag
3×2 SparseMatrixCSC{Float64, Int64} with 6 stored entries:
 0.465226  0.473656
 0.230678  0.391924
 0.928409  0.772551

julia> M = blockdiag(A, A, A)
9×6 SparseMatrixCSC{Float64, Int64} with 18 stored entries:
 0.465226  0.473656   ⋅         ⋅         ⋅         ⋅
 0.230678  0.391924   ⋅         ⋅         ⋅         ⋅
 0.928409  0.772551   ⋅         ⋅         ⋅         ⋅
  ⋅         ⋅        0.465226  0.473656   ⋅         ⋅
  ⋅         ⋅        0.230678  0.391924   ⋅         ⋅
  ⋅         ⋅        0.928409  0.772551   ⋅         ⋅
  ⋅         ⋅         ⋅         ⋅        0.465226  0.473656
  ⋅         ⋅         ⋅         ⋅        0.230678  0.391924
  ⋅         ⋅         ⋅         ⋅        0.928409  0.772551

如果你真的需要 M为了密集，你可以把它转换回来

julia> Matrix(M)
9×6 Matrix{Float64}:
 0.465226  0.473656  0.0       0.0       0.0       0.0
 0.230678  0.391924  0.0       0.0       0.0       0.0
 0.928409  0.772551  0.0       0.0       0.0       0.0
 0.0       0.0       0.465226  0.473656  0.0       0.0
 0.0       0.0       0.230678  0.391924  0.0       0.0
 0.0       0.0       0.928409  0.772551  0.0       0.0
 0.0       0.0       0.0       0.0       0.465226  0.473656
 0.0       0.0       0.0       0.0       0.230678  0.391924
 0.0       0.0       0.0       0.0       0.928409  0.772551