Question

我一直试图将python scipy稀疏矩阵除以其行的向量和。这是我的代码

sparse_mat = bsr_matrix((l_data, (l_row, l_col)), dtype=float)
sparse_mat = sparse_mat / (sparse_mat.sum(axis = 1)[:,None])

然而，无论我如何尝试，它都会抛出错误

sparse_mat = sparse_mat / (sparse_mat.sum(axis = 1)[:,None])
File "/usr/lib/python2.7/dist-packages/scipy/sparse/base.py", line 381, in __div__
return self.__truediv__(other)
File "/usr/lib/python2.7/dist-packages/scipy/sparse/compressed.py", line 427, in __truediv__
raise NotImplementedError
NotImplementedError

任何人都知道我哪里出错了？

Answer 1

您可以通过从行和的倒数创建稀疏对角矩阵然后将其与矩阵相乘来避开问题。在产品中，对角矩阵向左移动，矩阵向右移动。

示例：

>>> a
array([[0, 9, 0, 0, 1, 0],
       [2, 0, 5, 0, 0, 9],
       [0, 2, 0, 0, 0, 0],
       [2, 0, 0, 0, 0, 0],
       [0, 9, 5, 3, 0, 7],
       [1, 0, 0, 8, 9, 0]])
>>> b = sparse.bsr_matrix(a)
>>> 
>>> c = sparse.diags(1/b.sum(axis=1).A.ravel())
>>> # on older scipy versions the offsets parameter (default 0)
... # is a required argument, thus
... # c = sparse.diags(1/b.sum(axis=1).A.ravel(), 0)
...
>>> a/a.sum(axis=1, keepdims=True)
array([[ 0.        ,  0.9       ,  0.        ,  0.        ,  0.1       ,  0.        ],
       [ 0.125     ,  0.        ,  0.3125    ,  0.        ,  0.        ,  0.5625    ],
       [ 0.        ,  1.        ,  0.        ,  0.        ,  0.        ,  0.        ],
       [ 1.        ,  0.        ,  0.        ,  0.        ,  0.        ,  0.        ],
       [ 0.        ,  0.375     ,  0.20833333,  0.125     ,  0.        ,  0.29166667],
       [ 0.05555556,  0.        ,  0.        ,  0.44444444,  0.5       ,  0.        ]])
>>> (c @ b).todense() # on Python < 3.5 replace c @ b with c.dot(b)
matrix([[ 0.        ,  0.9       ,  0.        ,  0.        ,  0.1       ,  0.        ],
        [ 0.125     ,  0.        ,  0.3125    ,  0.        ,  0.        ,  0.5625    ],
        [ 0.        ,  1.        ,  0.        ,  0.        ,  0.        ,  0.        ],
        [ 1.        ,  0.        ,  0.        ,  0.        ,  0.        ,  0.        ],
        [ 0.        ,  0.375     ,  0.20833333,  0.125     ,  0.        ,  0.29166667],
        [ 0.05555556,  0.        ,  0.        ,  0.44444444,  0.5       ,  0.        ]])

Answer 2

有趣的事情正在发生。我执行元素划分没有问题。我想知道这是否是Py2问题。我正在使用Py3。

In [1022]: A=sparse.bsr_matrix([[2,4],[1,2]])
In [1023]: A
Out[1023]: 
<2x2 sparse matrix of type '<class 'numpy.int32'>'
    with 4 stored elements (blocksize = 2x2) in Block Sparse Row format>
In [1024]: A.A
Out[1024]: 
array([[2, 4],
       [1, 2]], dtype=int32)
In [1025]: A.sum(axis=1)
Out[1025]: 
matrix([[6],
        [3]], dtype=int32)
In [1026]: A/A.sum(axis=1)
Out[1026]: 
matrix([[ 0.33333333,  0.66666667],
        [ 0.33333333,  0.66666667]])

或尝试另一个例子：

In [1027]: b=sparse.bsr_matrix([[0, 9, 0, 0, 1, 0],
      ...:        [2, 0, 5, 0, 0, 9],
      ...:        [0, 2, 0, 0, 0, 0],
      ...:        [2, 0, 0, 0, 0, 0],
      ...:        [0, 9, 5, 3, 0, 7],
      ...:        [1, 0, 0, 8, 9, 0]])
In [1028]: b
Out[1028]: 
<6x6 sparse matrix of type '<class 'numpy.int32'>'
    with 14 stored elements (blocksize = 1x1) in Block Sparse Row format>
In [1029]: b.sum(axis=1)
Out[1029]: 
matrix([[10],
        [16],
        [ 2],
        [ 2],
        [24],
        [18]], dtype=int32)
In [1030]: b/b.sum(axis=1)
Out[1030]: 
matrix([[ 0.        ,  0.9       ,  0.        ,  0.        ,  0.1       , 0.        ],
        [ 0.125     ,  0.        ,  0.3125    ,  0.        ,  0.        , 0.5625    ],
 ....
        [ 0.05555556,  0.        ,  0.        ,  0.44444444,  0.5       ,     0.        ]])

这种稀疏/密集的结果也是密集的，其中c*b（c是稀疏对角线）是稀疏的。

In [1039]: c*b
Out[1039]: 
<6x6 sparse matrix of type '<class 'numpy.float64'>'
    with 14 stored elements in Compressed Sparse Row format>

稀疏和是一个密集矩阵。它是2d，所以不需要扩展它的尺寸。事实上，如果我尝试我得到一个错误：

In [1031]: A/(A.sum(axis=1)[:,None])
....
ValueError: shape too large to be a matrix.

Answer 3

根据 this message，为了保持矩阵稀疏，您可以访问 data 值并使用（非零）索引：

sums = np.asarray(A.sum(axis=1)).squeeze()  # this is dense
A.data /= sums[A.nonzero()[0]]

如果除以非零行平均值而不是总和，则可以

nnz = A.getnnz(axis=1)  # this is also dense
means = sums / nnz
A.data /= means[A.nonzero()[0]]

scipy稀疏矩阵除法

3 个答案: