是否可以在Julia中重塑稀疏数组？

我试图在Julia中重塑稀疏数组，但这样做会出错。有可能吗？如果是，怎么做

一些挫折的快照：

julia> A = sparse([1, 2, 3], [1, 2, 3], [1, 2, 3])
3×3 SparseMatrixCSC{Int64,Int64} with 3 stored entries:
  [1, 1]  =  1
  [2, 2]  =  2
  [3, 3]  =  3

julia> reshape(A, (5, 5))
ERROR: DimensionMismatch("parent has 9 elements, which is incompatible with size (5, 5)")
Stacktrace:

我还尝试了调整大小！（A，4*4），但似乎没有任何效果。同样，我们非常感谢您的帮助

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编辑：我应该更清楚，我想专门调整数组的大小，以便在3x3范围之外添加更多元素（在示例代码中）。例如，

A[4，4]=1

当前不起作用。

您可以使用创建一个新的稀疏数组

julia> i, j, v = findnz(A)
([1, 2, 3], [1, 2, 3], [1, 2, 3])

julia> sparse(i, j, v, 5, 5)
5×5 SparseMatrixCSC{Int64, Int64} with 3 stored entries:
 1  ⋅  ⋅  ⋅  ⋅
 ⋅  2  ⋅  ⋅  ⋅
 ⋅  ⋅  3  ⋅  ⋅
 ⋅  ⋅  ⋅  ⋅  ⋅
 ⋅  ⋅  ⋅  ⋅  ⋅

sparse

的反方向是

findnz

，它检索用于创建稀疏数组的输入。看

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（编辑，只是要注意，我的输出与您的不同，因为我在撰写本文时使用了Julia v1.6（发布候选1或

rc1

），它有一个非常酷的新

show

稀疏矩阵方法。）

当您将包含

n个
元素的矩阵重塑为包含不同数量元素的矩阵时，您希望得到什么结果？
reformate
的思想不是改变而是移动元素，例如
reformate（A，9，1）
工作我更新了我的问题，以展示我真正想要实现的目标。你看过这个吗：？
julia> J = Base.ReshapedArray(A, (4, 4), ())
4×4 reshape(::SparseMatrixCSC{Int64,Int64}, 4, 4) with eltype Int64:
 1  2    3     #undef
 0  0  #undef  #undef
 0  0  #undef  #undef
 0  0  #undef  #undef

julia> J[4, 4] = 4
ERROR: BoundsError: attempt to access 3×3 SparseMatrixCSC{Int64,Int64} at index [1, 6]
Stacktrace:

julia> i, j, v = findnz(A)
([1, 2, 3], [1, 2, 3], [1, 2, 3])

julia> sparse(i, j, v, 5, 5)
5×5 SparseMatrixCSC{Int64, Int64} with 3 stored entries:
 1  ⋅  ⋅  ⋅  ⋅
 ⋅  2  ⋅  ⋅  ⋅
 ⋅  ⋅  3  ⋅  ⋅
 ⋅  ⋅  ⋅  ⋅  ⋅
 ⋅  ⋅  ⋅  ⋅  ⋅