我有两个矢量t和delta - 两个长度n。我想用该向量创建一个稀疏的DIA矩阵A，然后调整列：对于所有i，我想通过ith列向左移动A中的delta[i]条目。
控制列的简单方法是COO格式。这是我认为会起作用的：
from scipy.sparse import diags
A = diags([t], offsets=[-1]).tocoo()
A.col = A.col - delta
然而，在我的例子中，A.nnz == len(A.col)只是216，而t和delta的长度是239。鉴于nnz存储"Number of stored values, including explicit zeros."，我不明白这是怎么回事。
我该如何解决这个问题？这是我的示例数据：
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])
In [29]: t = np.array([1.2, 3.2, 4, 0, 0])
In [30]: A = sparse.diags([t], offsets=[-1])
In [31]: A
Out[31]:
<6x6 sparse matrix of type '<class 'numpy.float64'>'
with 5 stored elements (1 diagonals) in DIAgonal format>
转换为coo剥离了0。
In [32]: Ac = A.tocoo()
In [33]: Ac
Out[33]:
<6x6 sparse matrix of type '<class 'numpy.float64'>'
with 3 stored elements in COOrdinate format>
看看A.tocoo的代码（dia到coo版本）。它有(self.data != 0)面具。
如果我直接制作coo矩阵，它会保留零，至少是暂时的：
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. ]])
内部零删除：
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>