Question

给定scipy.sparse.crs_matrix，我想提取Numpy密集代数中的子矩阵表示为

 A[0::2, 0::2]

即，A_{new}(i,j) = A(2*i,2*j)（“棋盘黑方矩阵”）。

Answer 1

如果您首先将矩阵转换为COO格式，那么这是一块蛋糕：

def sps_black_squares(a):
    a = a.tocoo()
    idx = (a.row % 2 == 0) & (a.col % 2 == 0)
    new_shape = tuple((j-1) // 2 + 1 for j in a.shape)
    return sps.csr_matrix((a.data[idx], (a.row[idx]//2, a.col[idx]//2)),
                          shape=new_shape)

%timeit sps_black_squares(a)
1000 loops, best of 3: 315 us per loop

%timeit sps.csr_matrix(a.toarray()[::2, ::2])
100 loops, best of 3: 6.55 ms per loop

np.allclose(sps_black_squares(a).toarray(), a.toarray()[::2, ::2])
Out[119]: True

用于scipy :: crs_matrix的棋盘子矩阵

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