用对角线归一化 sparse.csc_matrix

Question

我有一个 scipy.sparse.csc_matrix，dtype = np.int32。我想用该列中的对角线元素有效地划分矩阵的每一列（或行，以更快的 csc_matrix 为准）。所以 mnew[:,i] = m[:,i]/m[i,i] 。请注意，我需要将我的矩阵转换为 np.double（因为 mnew 元素将在 [0,1] 中）并且由于矩阵很大且非常稀疏，我想知道我是否可以在某些 efficient/no 中做到这一点loop/never 走密集路。

最佳，

伊利亚

Answer 1

制作一个稀疏矩阵：

In [379]: M = sparse.random(5,5,.2, format='csr')
In [380]: M
Out[380]: 
<5x5 sparse matrix of type '<class 'numpy.float64'>'
    with 5 stored elements in Compressed Sparse Row format>
In [381]: M.diagonal()
Out[381]: array([ 0.,  0.,  0.,  0.,  0.])

对角线上有太多 0 - 让我们添加一个非零对角线：

In [382]: D=sparse.dia_matrix((np.random.rand(5),0),shape=(5,5))
In [383]: D
Out[383]: 
<5x5 sparse matrix of type '<class 'numpy.float64'>'
    with 5 stored elements (1 diagonals) in DIAgonal format>
In [384]: M1 = M+D


In [385]: M1
Out[385]: 
<5x5 sparse matrix of type '<class 'numpy.float64'>'
    with 10 stored elements in Compressed Sparse Row format>

In [387]: M1.A
Out[387]: 
array([[ 0.35786668,  0.81754484,  0.        ,  0.        ,  0.        ],
       [ 0.        ,  0.41928992,  0.        ,  0.01371273,  0.        ],
       [ 0.        ,  0.        ,  0.4685924 ,  0.        ,  0.35724102],
       [ 0.        ,  0.        ,  0.77591294,  0.95008721,  0.16917791],
       [ 0.        ,  0.        ,  0.        ,  0.        ,  0.16659141]])

现在用对角线划分每一列是微不足道的（这是一个矩阵'product'）

In [388]: M1/M1.diagonal()
Out[388]: 
matrix([[ 1.        ,  1.94983185,  0.        ,  0.        ,  0.        ],
        [ 0.        ,  1.        ,  0.        ,  0.01443313,  0.        ],
        [ 0.        ,  0.        ,  1.        ,  0.        ,  2.1444144 ],
        [ 0.        ,  0.        ,  1.65583764,  1.        ,  1.01552603],
        [ 0.        ,  0.        ,  0.        ,  0.        ,  1.        ]])

或者除以行-（乘以一个列向量）

In [391]: M1/M1.diagonal()[:,None]

哎呀，这些太密集了；让对角线稀疏

In [408]: md = sparse.csr_matrix(1/M1.diagonal())  # do the inverse here
In [409]: md
Out[409]: 
<1x5 sparse matrix of type '<class 'numpy.float64'>'
    with 5 stored elements in Compressed Sparse Row format>
In [410]: M.multiply(md)
Out[410]: 
<5x5 sparse matrix of type '<class 'numpy.float64'>'
    with 5 stored elements in Compressed Sparse Row format>
In [411]: M.multiply(md).A
Out[411]: 
array([[ 0.        ,  1.94983185,  0.        ,  0.        ,  0.        ],
       [ 0.        ,  0.        ,  0.        ,  0.01443313,  0.        ],
       [ 0.        ,  0.        ,  0.        ,  0.        ,  2.1444144 ],
       [ 0.        ,  0.        ,  1.65583764,  0.        ,  1.01552603],
       [ 0.        ,  0.        ,  0.        ,  0.        ,  0.        ]])

md.multiply(M) 为列版本。

- 类似，只是它使用行的总和而不是对角线。处理更多潜在的 'divide-by-zero' 问题。

用对角线归一化 sparse.csc_matrix

Normalizing sparse.csc_matrix by its diagonals

numpy

scipy

sparse-matrix

python-3.x