我有个审判的想法
定义一个函数来链接矩阵列表中的一些kron调用：

In [50]: def foo(alist):
    ...:     M = alist[0]
    ...:     for a in alist[1:]:
    ...:         M = sparse.kron(M,a)
    ...:     return M
    ...:

你的两个矩阵：

In [51]: A=sparse.csr_matrix([[0., 0.],
    ...:    [1., 0.]])
    ...: Id=sparse.identity(2)

做一个列表，在这个例子中，3个id，2个在A之后：

In [52]: alist = [Id]*3+[A]+[Id]*2

In [53]: alist
Out[53]: 
[<2x2 sparse matrix of type '<class 'numpy.float64'>'
    with 2 stored elements (1 diagonals) in DIAgonal format>,
 <2x2 sparse matrix of type '<class 'numpy.float64'>'
    with 2 stored elements (1 diagonals) in DIAgonal format>,
 <2x2 sparse matrix of type '<class 'numpy.float64'>'
    with 2 stored elements (1 diagonals) in DIAgonal format>,
 <2x2 sparse matrix of type '<class 'numpy.float64'>'
    with 1 stored elements in Compressed Sparse Row format>,
 <2x2 sparse matrix of type '<class 'numpy.float64'>'
    with 2 stored elements (1 diagonals) in DIAgonal format>,
 <2x2 sparse matrix of type '<class 'numpy.float64'>'
    with 2 stored elements (1 diagonals) in DIAgonal format>]

并将其传递给foo：

In [54]: M = foo(alist)

In [55]: M
Out[55]: 
<64x64 sparse matrix of type '<class 'numpy.float64'>'
    with 512 stored elements (blocksize = 2x2) in Block Sparse Row format>

然后，只需迭代可能的Id/A混合列表，使用不同数量的before和after。
它节省打字，如果不是速度。