将scipy稀疏矩阵转换为基于索引的numpy数组

我有一个scipy稀疏矩阵，其中N个值为非零值，我希望返回一个带有形状（N，3）的numpy数组，其中第一列包含非零值的索引，最后一列包含相应的非零值

例如：

我想要

mymatrix.toarray()
matrix([[0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.83885831, 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.        , 0.        , 1.13395003, 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.        , 0.        , 0.57979727, 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.        , 0.75500017, 0.        , 0.81459546, 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.87997548, 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        , 0.        ]])

成为

np.array([[3, 2, 0.83885831], [4,5,1.13395003], [6,5,0.57979727], [7,4,0.75500017], [7,6,0.81459546], [8,9,0.87997548]])

array([[3.        , 2.        , 0.83885831],
       [4.        , 5.        , 1.13395003],
       [6.        , 5.        , 0.57979727],
       [7.        , 4.        , 0.75500017],
       [7.        , 6.        , 0.81459546],
       [8.        , 9.        , 0.87997548]])

我如何有效地做到这一点

转换后，我将迭代行-因此，如果有一个有效的选项可以在不进行转换的情况下迭代行，我也希望：

for index_i, index_j, value in mymatrix.iterator():
     do_something(index_i, index_j, value)

对于迭代，dok（dictionary of keys）格式看起来像是自然匹配；你可以做：

for (i,j), v in your_sparse_matrix.todok().items():
    etc.

Nx3坐标值记录列表可通过coo格式轻松获取：

 coo = your_sparse_matrix.tocoo()
 np.column_stack((coo.row,coo.col,coo.data))

显然，这也可以用于迭代；您必须测试在您的用例中哪个更快