在原处修改scipy稀疏矩阵

3 投票
1 回答
3580 浏览
提问于 2025-04-17 17:50

基本上,我只是想做一个简单的矩阵乘法,具体来说,就是提取矩阵的每一列,然后通过除以它的长度来进行归一化处理。

    #csc sparse matrix
    self.__WeightMatrix__ = self.__WeightMatrix__.tocsc()
    #iterate through columns
    for Col in xrange(self.__WeightMatrix__.shape[1]):
       Column = self.__WeightMatrix__[:,Col].data
       List = [x**2 for x in Column]
       #get the column length
       Len = math.sqrt(sum(List))
       #here I assumed dot(number,Column) would do a basic scalar product
       dot((1/Len),Column)
       #now what? how do I update the original column of the matrix, everything that have been returned are copies, which drove me nuts and missed pointers so much

我查阅了scipy稀疏矩阵的文档,但没有找到有用的信息。我希望能有一个函数返回矩阵的指针或引用,这样我就可以直接修改它的值。谢谢。

1 个回答

6

在CSC格式中,你有两个可以写入的属性,dataindices。它们分别存储了矩阵中非零的数值和对应的行索引。你可以利用这些属性来实现你的需求,方法如下:

def sparse_row_normalize(sps_mat) :
    if sps_mat.format != 'csc' :
        msg = 'Can only row-normalize in place with csc format, not {0}.'
        msg = msg.format(sps_mat.format)
        raise ValueError(msg)
    row_norm = np.sqrt(np.bincount(sps_mat.indices, weights=mat.data * mat_data))
    sps_mat.data /= np.take(row_norm, sps_mat.indices)

为了验证这个方法确实有效:

>>> mat = scipy.sparse.rand(4, 4, density=0.5, format='csc')
>>> mat.toarray()
array([[ 0.        ,  0.        ,  0.58931687,  0.31070526],
       [ 0.24024639,  0.02767106,  0.22635696,  0.85971295],
       [ 0.        ,  0.        ,  0.13613897,  0.        ],
       [ 0.        ,  0.13766507,  0.        ,  0.        ]])
>>> mat.toarray() / np.sqrt(np.sum(mat.toarray()**2, axis=1))[:, None]
array([[ 0.        ,  0.        ,  0.88458487,  0.46637926],
       [ 0.26076366,  0.03003419,  0.24568806,  0.93313324],
       [ 0.        ,  0.        ,  1.        ,  0.        ],
       [ 0.        ,  1.        ,  0.        ,  0.        ]])
>>> sparse_row_normalize(mat)
>>> mat.toarray()
array([[ 0.        ,  0.        ,  0.88458487,  0.46637926],
       [ 0.26076366,  0.03003419,  0.24568806,  0.93313324],
       [ 0.        ,  0.        ,  1.        ,  0.        ],
       [ 0.        ,  1.        ,  0.        ,  0.        ]])

而且这个方法也很快,使用numpy可以避免Python中的循环,让操作更加高效:

In [2]: mat = scipy.sparse.rand(10000, 10000, density=0.005, format='csc')

In [3]: mat
Out[3]: 
<10000x10000 sparse matrix of type '<type 'numpy.float64'>'
    with 500000 stored elements in Compressed Sparse Column format>

In [4]: %timeit sparse_row_normalize(mat)
100 loops, best of 3: 14.1 ms per loop

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