[英]Converting dense matrix code to sparse matrix code
我正在尝试将此代码转换为使用scipy稀疏矩阵,因为真实矩阵非常大,但是遇到了麻烦。 有人可以帮忙吗?
import numpy as np
G = np.array([[0., 50., 50., 0.],
[10., 0., 10., 0.],
[0., 0., 0., 10.],
[2., 0., 2., 0.]])
s = G.sum(axis=0)
m = np.minimum(G, 1).transpose()
sm = s * m
sm_rnorm = (sm / sm.sum(axis=0))
smm = sm * sm_rnorm
G += smm.transpose()
print(G)
我尝试了以下方法:
import numpy as np
from scipy.sparse import csc_matrix
G = np.array([[0.,50.,50.,0.],
[10.,0.,10.,0.],
[0.,0.,0.,10.],
[2.,0.,2.,0.]])
G = csc_matrix(G, dtype=np.float)
s = csc_matrix(G.sum(axis=0))
m = csc_matrix.minimum(G, 1).transpose()
sm = s * m
sm_rnorm = (sm / csc_matrix(sm.sum(axis=0)))
smm = sm * sm_rnorm
G += smm.transpose()
print(G)
...但出现ValueError: dimension mismatch
我运行了您的密集代码,
In [224]: G = np.array([[0., 50., 50., 0.],
...: [10., 0., 10., 0.],
...: [0., 0., 0., 10.],
...: [2., 0., 2., 0.]])
...: s = G.sum(axis=0)
...: m = np.minimum(G, 1).transpose()
...: sm = s * m
...: sm_rnorm = (sm / sm.sum(axis=0))
...: smm = sm * sm_rnorm
...:
In [225]: s
Out[225]: array([12., 50., 62., 10.])
In [226]: m
Out[226]:
array([[0., 1., 0., 1.],
[1., 0., 0., 0.],
[1., 1., 0., 1.],
[0., 0., 1., 0.]])
In [227]: sm
Out[227]:
array([[ 0., 50., 0., 10.],
[12., 0., 0., 0.],
[12., 50., 0., 10.],
[ 0., 0., 62., 0.]])
然后启动稀疏版本:
In [192]: from scipy import sparse
In [228]: Gm = sparse.csr_matrix(G)
In [229]: Gm
Out[229]:
<4x4 sparse matrix of type '<class 'numpy.float64'>'
with 7 stored elements in Compressed Sparse Row format>
In [230]: s_m = Gm.sum(axis=0)
In [231]: s_m
Out[231]: matrix([[12., 50., 62., 10.]])
In [233]: m_m = Gm.minimum(1).T
In [234]: m_m.A
Out[234]:
array([[0., 1., 0., 1.],
[1., 0., 0., 0.],
[1., 1., 0., 1.],
[0., 0., 1., 0.]])
哎呀:
In [236]: s_m * m_m
Out[236]: matrix([[112., 74., 10., 74.]])
*
如果np.matrix
和稀疏矩阵的矩阵相乘
In [237]: s.dot(m)
Out[237]: array([112., 74., 10., 74.])
稀疏矩阵元素的明智乘法:
In [242]: sm_m = m_m.multiply(s_m)
In [243]: sm_m.A
Out[243]:
array([[ 0., 50., 0., 10.],
[12., 0., 0., 0.],
[12., 50., 0., 10.],
[ 0., 0., 62., 0.]])
现在匹配sm_rnorm
:
In [244]: sm_m.sum(axis=0)
Out[244]: matrix([[ 24., 100., 62., 20.]])
In [250]: sm_m / sm_m.sum(axis=0)
Out[250]:
matrix([[0. , 0.5, 0. , 0.5],
[0.5, 0. , 0. , 0. ],
[0.5, 0.5, 0. , 0.5],
[0. , 0. , 1. , 0. ]])
sparse/dense
在元素上起作用,但是sparse/sparse
存在问题:
In [252]: sm_m / sparse.csr_matrix(sm_m.sum(axis=0))
----> 1 sm_m / sparse.csr_matrix(sm_m.sum(axis=0))
--> 576 return self._divide(other, true_divide=True)
568 if true_divide and np.can_cast(self.dtype, np.float_):
ValueError: inconsistent shapes
我认为这是一个矩阵除法问题,但我想进一步确定。
sm_m.multiply(1 / sm_m.sum(axis=0))
给出具有正确值的稀疏矩阵,但速度较慢(至少在此示例中如此)。
smm_m = sm_m.multiply( sm_m / sm_m.sum(axis=0))
与smm
匹配。 和Gm += smm_m
起作用。 稀疏+=
不会引起效率错误,因为它不会改变稀疏性。
因此,关键问题是保持矩阵乘法和元素乘法直(以及相应的除法)。
sklearn.utils.sparsefuncs
具有一些稀疏的实用程序功能
上面的sm_m
是coo
格式数组(不确定原因):
In [366]: sm_m
Out[366]:
<4x4 sparse matrix of type '<class 'numpy.float64'>'
with 7 stored elements in COOrdinate format>
In [367]: sm_m.A
Out[367]:
array([[ 0., 50., 0., 10.],
[12., 0., 0., 0.],
[12., 50., 0., 10.],
[ 0., 0., 62., 0.]])
将其转换为csr
:
In [368]: sm_m1 = sm_m.tocsr()
In [369]: sm_m1
Out[369]:
<4x4 sparse matrix of type '<class 'numpy.float64'>'
with 7 stored elements in Compressed Sparse Row format>
导出列缩放数组:
In [370]: x = sm_m1.sum(axis=0)
In [371]: x
Out[371]: matrix([[ 24., 100., 62., 20.]])
In [372]: x = 1/x.A1 # .A1 makes a 1d array from np.matrix
在适当位置应用缩放:
In [373]: sklearn.utils.sparsefuncs.inplace_csr_column_scale(sm_m1,x)
In [374]: sm_m1.A
Out[374]:
array([[0. , 0.5, 0. , 0.5],
[0.5, 0. , 0. , 0. ],
[0.5, 0.5, 0. , 0.5],
[0. , 0. , 1. , 0. ]])
适当的column_scale很简单:
def inplace_csr_column_scale(X, scale):
# ....
X.data *= scale.take(X.indices, mode='clip')
m_m.multiply(s_m)
步骤也可以通过以下方式完成:
In [380]: m1_m = m_m.tocsr()
In [381]: sklearn.utils.sparsefuncs.inplace_csr_column_scale(m1_m,s_m.A1)
In [382]: m1_m.A
Out[382]:
array([[ 0., 50., 0., 10.],
[12., 0., 0., 0.],
[12., 50., 0., 10.],
[ 0., 0., 62., 0.]])
我怀疑代码可以清理,删除转置等。
G
本质上是正方形吗? 我喜欢使用非正方形数组来更好地跟踪形状,转置和尺寸和。 我试图将G
扩展为(5,4),并在s*m
步骤遇到问题。
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