Python 创建稀疏直径矩阵，然后更改列

我有两个向量t和delta，都是长度n。我想用这个向量创建一个稀疏的DIA矩阵a，然后调整列：对于所有I，我想将a by delta[I]列中的第I个条目向左移动

控制列的一种简单方法是采用COO格式。以下是我认为有效的方法：

from scipy.sparse import diags
A = diags([t], offsets=[-1]).tocoo()
A.col = A.col - delta

然而，在我的示例中，A.nnz==lenA.col仅为216，而t和delta的长度为239。考虑到nnz的商店，我不明白这是怎么发生的

我如何解决这个问题？以下是我的示例数据：

from numpy import np
t = np.array([ 2.655,  2.655,  2.655,  2.655,  2.655,  2.655,  2.655,  2.655,
        2.655,  2.655,  2.655,  2.655,  2.655,  2.655,  2.655,  2.655,
        2.655,  2.655,  2.655,  2.655,  2.655,  2.655,  2.655,  2.155,
        2.155,  2.155,  2.155,  2.155,  2.155,  2.155,  2.155,  2.155,
        2.155,  2.155,  2.155,  2.155,  2.155,  2.155,  2.155,  2.155,
        2.155,  2.155,  2.155,  2.155,  2.155,  2.155,  2.155,  1.655,
        1.655,  1.655,  1.655,  1.655,  1.655,  1.655,  1.655,  1.655,
        1.655,  1.655,  1.655,  1.655,  1.655,  1.655,  1.655,  1.655,
        1.655,  1.655,  1.655,  1.655,  1.655,  1.655,  1.655,  1.155,
        1.155,  1.155,  1.155,  1.155,  1.155,  1.155,  1.155,  1.155,
        1.155,  1.155,  1.155,  1.155,  1.155,  1.155,  1.155,  1.155,
        1.155,  1.155,  1.155,  1.155,  1.155,  1.155,  1.155,  0.655,
        0.655,  0.655,  0.655,  0.655,  0.655,  0.655,  0.655,  0.655,
        0.655,  0.655,  0.655,  0.655,  0.655,  0.655,  0.655,  0.655,
        0.655,  0.655,  0.655,  0.655,  0.655,  0.655,  0.655,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,
        0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.405,  0.   ,
        0.   ,  0.   ,  0.   ,  0.   ,  0.   ,  0.   ,  0.   ,  0.   ,
        0.   ,  0.   ,  0.   ,  0.   ,  0.   ,  0.   ,  0.   ,  0.   ,
        0.   ,  0.   ,  0.   ,  0.   ,  0.   ,  0.   ,  0.   ])
delta = np.array([0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5,
   5, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5,
   5, 5, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5,
   5, 5, 5, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4,
   5, 5, 5, 5, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4,
   4, 5, 5, 5, 5, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4,
   4, 4, 5, 5, 5, 5, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4,
   4, 4, 4, 5, 5, 5, 5, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3,
   4, 4, 4, 4, 5, 5, 5, 5, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3,
   3, 4, 4, 4, 4, 5, 5, 5, 5, 0, 0, 0, 0, 1, 1, 1, 1, 2, 0, 2, 2, 3, 2,
   1, 3, 4, 3, 0, 3, 5, 4, 1, 0])

查看A.tocoo的代码，即dia-to-coo版本。它有自己的数据！=0面具

如果我直接生成coo矩阵，它将保留零，至少暂时保留：

In [58]: A.A
Out[58]: 
array([[ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 1.2,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  3.2,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  4. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ]])
In [59]: M = sparse.coo_matrix((t, (np.arange(1,6),np.arange(5))),shape=(6,6))
In [60]: M
Out[60]: 
<6x6 sparse matrix of type '<class 'numpy.float64'>'
    with 5 stored elements in COOrdinate format>
In [61]: M.A
Out[61]: 
array([[ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 1.2,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  3.2,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  4. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ]])

就地零位消除：

In [64]: M.eliminate_zeros()
In [65]: M
Out[65]: 
<6x6 sparse matrix of type '<class 'numpy.float64'>'
    with 3 stored elements in COOrdinate format>

In [58]: A.A
Out[58]: 
array([[ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 1.2,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  3.2,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  4. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ]])
In [59]: M = sparse.coo_matrix((t, (np.arange(1,6),np.arange(5))),shape=(6,6))
In [60]: M
Out[60]: 
<6x6 sparse matrix of type '<class 'numpy.float64'>'
    with 5 stored elements in COOrdinate format>
In [61]: M.A
Out[61]: 
array([[ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 1.2,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  3.2,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  4. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ]])

In [64]: M.eliminate_zeros()
In [65]: M
Out[65]: 
<6x6 sparse matrix of type '<class 'numpy.float64'>'
    with 3 stored elements in COOrdinate format>