python - 将 n 个最大元素保留在矩阵的行或列中的可靠方法

Question

我想从密集矩阵中创建一个稀疏矩阵，这样在每一行或每一列中仅保留n最大的元素。我执行以下操作:

def sparsify(K, min_nnz = 5):
'''
This function eliminates the elements which are smaller that the maximal element in the matrix,

Parameters
----------
K : ndarray
    K - the input matrix
min_nnz:
    the minimal number of elements in row or column to be preserved

'''
cond = np.bitwise_or(K >= -np.partition(-K, min_nnz - 1, axis = 1)[:, min_nnz - 1][:, None], 
                     K >= -np.partition(-K, min_nnz - 1, axis = 0)[min_nnz - 1, :][None, :])

return spsp.csr_matrix(np.where(cond, K, 0))

这种方法按预期工作，但似乎不是最有效、最稳健的方法。您建议采取什么更好的方法？

使用示例:

A = np.random.rand(10, 10)
A_sp = sparsify(A, min_nnz = 3)

Answer 1

您可以使用 coo_matrix 仅使用您需要的值来构建，而不是创建另一个密集矩阵:

return spsp.coo_matrix((K[cond], np.where(cond)), shape = K.shape)

至于其余的，您可以短路第二个维度，但您节省的时间将完全取决于您的输入

def sparsify(K, min_nnz = 5):

    '''
    This function eliminates the elements which are smaller that the maximal element in the matrix,
    
    Parameters
    ----------
    K : ndarray
        K - the input matrix
    min_nnz:
        the minimal number of elements in row or column to be preserved
    
    '''
    cond = K >= -np.partition(-K, min_nnz - 1, axis = 0)[min_nnz - 1, :]
    mask = cond.sum(1) < min_nnz
    cond[mask] = np.bitwise_or(cond[mask], 
                               K[mask] >= -np.partition(-K[mask], 
                                                        min_nnz - 1, 
                                                        axis = 1)[:, min_nnz - 1][:, None])
    
    return spsp.coo_matrix((K[cond], np.where(cond)), shape = K.shape)

测试:

sparsify(A)
Out[]: 
<10x10 sparse matrix of type '<class 'numpy.float64'>'
    with 58 stored elements in COOrdinate format>

sparsify(A).A
Out[]: 
array([[0.        , 0.        , 0.61362248, 0.        , 0.73648987,
        0.64561856, 0.40727807, 0.61674005, 0.53533315, 0.        ],
       [0.8888361 , 0.64548039, 0.94659603, 0.78474203, 0.        ,
        0.        , 0.78809603, 0.88938798, 0.        , 0.37631541],
       [0.69356682, 0.        , 0.        , 0.        , 0.        ,
        0.7386594 , 0.71687659, 0.67750768, 0.58002451, 0.        ],
       [0.67241433, 0.71923718, 0.95888737, 0.        , 0.        ,
        0.        , 0.82773085, 0.69788448, 0.63736915, 0.4263064 ],
       [0.        , 0.65831794, 0.        , 0.        , 0.59850093,
        0.        , 0.        , 0.61913869, 0.65024867, 0.50860294],
       [0.75522891, 0.        , 0.93342402, 0.8284258 , 0.64471939,
        0.6990814 , 0.        , 0.        , 0.        , 0.32940821],
       [0.        , 0.88458635, 0.62460096, 0.60412265, 0.66969674,
        0.        , 0.40318741, 0.        , 0.        , 0.44116059],
       [0.        , 0.        , 0.500971  , 0.92291245, 0.        ,
        0.8862903 , 0.        , 0.375885  , 0.49473635, 0.        ],
       [0.86920647, 0.85157893, 0.89883006, 0.        , 0.68427193,
        0.91195162, 0.        , 0.        , 0.94762875, 0.        ],
       [0.        , 0.6435456 , 0.        , 0.70551006, 0.        ,
        0.8075527 , 0.        , 0.9421039 , 0.91096934, 0.        ]])

sparsify(A).A.astype(bool).sum(0)
Out[]: array([5, 6, 7, 5, 5, 6, 5, 7, 7, 5])

sparsify(A).A.astype(bool).sum(1)
Out[]: array([6, 7, 5, 7, 5, 6, 6, 5, 6, 5])