python - 访问 coo_matrix 中的元素

Question

这是一个非常简单的问题。对于像 coo_matrix 这样的 SciPy 稀疏矩阵，如何访问单个元素？

类比本征线性代数库。可以使用 coeffRef 访问元素 (i,j)，如下所示:

myMatrix.coeffRef(i,j)

Answer 1

来自 coo_matrix 的文档:

 |  Intended Usage
 |      - COO is a fast format for constructing sparse matrices
 |      - Once a matrix has been constructed, convert to CSR or
 |        CSC format for fast arithmetic and matrix vector operations
 |      - By default when converting to CSR or CSC format, duplicate (i,j)
 |        entries will be summed together.  This facilitates efficient
 |        construction of finite element matrices and the like. (see example)

事实上，csr_matrix 以预期的方式支持索引:

>>> from scipy.sparse import coo_matrix
>>> m = coo_matrix([[1, 2, 3], [4, 5, 6]])
>>> m1 = m.tocsr()
>>> m1[1, 2]
6
>>> m1
<2x3 sparse matrix of type '<type 'numpy.int64'>'
    with 6 stored elements in Compressed Sparse Row format>

(我从文档中找到上述引用的方式是 >>> help(m) ，相当于 the online docs )。