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当我发现以下奇怪时，我正在将以下代码块与 networkx 一起使用。在第一种情况下，我在一个稀疏矩阵上使用了 ufunc multiply(*)，它意外地正确地给了我一个度数序列。但是，当对普通矩阵执行相同操作时，它给了我一个 10 x 10 矩阵，并且正如预期的那样 np.dot(...) 给了我正确的结果。

import numpy as np
import networks as nx

ba = nx.barabasi_albert_graph(n=10, m=2)

A = nx.adjacency_matrix(ba)
# <10x10 sparse matrix of type '<class 'numpy.int64'>'
# with 32 stored elements in Compressed Sparse Row format>

A * np.ones(10)

# output: array([ 5.,  3.,  4.,  5.,  4.,  3.,  2.,  2.,  2.,  2.])

nx.degree(ba)

# output {0: 5, 1: 3, 2: 4, 3: 5, 4: 4, 5: 3, 6: 2, 7: 2, 8: 2, 9: 2}

B = np.ones(100).reshape(10, 10)

B * np.ones(10)

array([[ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.],
   [ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.],
   [ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.],
   [ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.],
   [ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.],
   [ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.],
   [ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.],
   [ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.],
   [ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.],
   [ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.]])

np.dot(B, np.ones(10))
# array([ 10.,  10.,  10.,  10.,  10.,  10.,  10.,  10.,  10.,  10.])

我期待我应该这样做np.dot(A, np.ones(10))，但它返回一个由 10、10 x 10 矩阵组成的数组

array([ <10x10 sparse matrix of type '<class 'numpy.float64'>'
with 32 stored elements in Compressed Sparse Row format>,
   <10x10 sparse matrix of type '<class 'numpy.float64'>'
with 32 stored elements in Compressed Sparse Row format>,
   <10x10 sparse matrix of type '<class 'numpy.float64'>'
with 32 stored elements in Compressed Sparse Row format>,
   <10x10 sparse matrix of type '<class 'numpy.float64'>'
with 32 stored elements in Compressed Sparse Row format>,
   <10x10 sparse matrix of type '<class 'numpy.float64'>'
with 32 stored elements in Compressed Sparse Row format>,
   <10x10 sparse matrix of type '<class 'numpy.float64'>'
with 32 stored elements in Compressed Sparse Row format>,
   <10x10 sparse matrix of type '<class 'numpy.float64'>'
with 32 stored elements in Compressed Sparse Row format>,
   <10x10 sparse matrix of type '<class 'numpy.float64'>'
with 32 stored elements in Compressed Sparse Row format>,
   <10x10 sparse matrix of type '<class 'numpy.float64'>'
with 32 stored elements in Compressed Sparse Row format>,
   <10x10 sparse matrix of type '<class 'numpy.float64'>'
with 32 stored elements in Compressed Sparse Row format>], dtype=object)

这里有什么细微差别？

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1 回答 1

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对于常规的 numpy 数组，*乘法是逐个元素（使用broadcasting）。np.dot是矩阵乘积，乘积之和。对于np.matrix子类*是矩阵乘积，dot. sparse.matrix不是子类，但它是以此为模型的。*是矩阵乘积。

In [694]: A = sparse.random(10,10,.2, format='csr')
In [695]: A
Out[695]: 
<10x10 sparse matrix of type '<class 'numpy.float64'>'
    with 20 stored elements in Compressed Sparse Row format>
In [696]: A *np.ones(10)
Out[696]: 
array([ 0.6349177 ,  0.        ,  1.25781168,  1.12021258,  2.43477065,
        1.10407149,  1.95096264,  0.6253589 ,  0.44242708,  0.50353061])

稀疏矩阵具有dot方法，其行为相同：

In [698]: A.dot(np.ones(10))
Out[698]: 
array([ 0.6349177 ,  0.        ,  1.25781168,  1.12021258,  2.43477065,
        1.10407149,  1.95096264,  0.6253589 ,  0.44242708,  0.50353061])

密集版：

In [699]: np.dot(A.A,np.ones(10))
Out[699]: 
array([ 0.6349177 ,  0.        ,  1.25781168,  1.12021258,  2.43477065,
        1.10407149,  1.95096264,  0.6253589 ,  0.44242708,  0.50353061])

我认为np.dot应该正确处理稀疏矩阵，这与他们自己的方法不同。但np.dot(A,np.ones(10))这样做不对，产生了 2 个稀疏矩阵的对象数组。我可以深入研究为什么，但现在，避免它。

通常，使用稀疏矩阵的稀疏函数和方法。不要假设numpy函数会正确使用它们。

np.dot当两个数组都很稀疏时工作正常，

In [702]: np.dot(A,A)
Out[702]: 
<10x10 sparse matrix of type '<class 'numpy.float64'>'
    with 32 stored elements in Compressed Sparse Row format>
In [703]: np.dot(A,A.T)
Out[703]: 
<10x10 sparse matrix of type '<class 'numpy.float64'>'
    with 31 stored elements in Compressed Sparse Row format>

In [705]: np.dot(A, sparse.csr_matrix(np.ones(10)).T)
Out[705]: 
<10x1 sparse matrix of type '<class 'numpy.float64'>'
    with 9 stored elements in Compressed Sparse Row format>
In [706]: _.A
Out[706]: 
array([[ 0.6349177 ],
       [ 0.        ],
       [ 1.25781168],
       [ 1.12021258],
       [ 2.43477065],
       [ 1.10407149],
       [ 1.95096264],
       [ 0.6253589 ],
       [ 0.44242708],
       [ 0.50353061]])

值得sum用这种矩阵产品执行稀疏：

In [708]: A.sum(axis=1)
Out[708]: 
matrix([[ 0.6349177 ],
        [ 0.        ],
        [ 1.25781168],
        [ 1.12021258],
        [ 2.43477065],
        [ 1.10407149],
        [ 1.95096264],
        [ 0.6253589 ],
        [ 0.44242708],
        [ 0.50353061]])
于 2017-07-03T00:38:07.863 回答