如何连接Python中的两个矩阵？

Question

我有两个csr_matrix,uniFeature和biFeature。

我想要一个新的矩阵Feature = [uniFeature, biFeature]。但是，如果我以这种方式直接连接它们，那么表示矩阵Feature是一个列表。我怎样才能实现矩阵连接，并仍然得到相同类型的矩阵，即csr_matrix？

而且如果我这样做了连接后它不工作：Feature = csr_matrix(Feature) 它给人的错误：

Traceback (most recent call last): 
    File "yelpfilter.py", line 91, in <module> 
    Feature = csr_matrix(Feature) 
    File "c:\python27\lib\site-packages\scipy\sparse\compressed.py", line 66, in __init__ 
    self._set_self(self.__class__(coo_matrix(arg1, dtype=dtype))) 
    File "c:\python27\lib\site-packages\scipy\sparse\coo.py", line 185, in __init__ 
    self.row, self.col = M.nonzero() 
TypeError: __nonzero__ should return bool or int, returned numpy.bool_ 

Answer 1

的scipy.sparse模块包括的功能hstack和vstack。

例如：

In [44]: import scipy.sparse as sp 

In [45]: c1 = sp.csr_matrix([[0,0,1,0], 
    ...:      [2,0,0,0], 
    ...:      [0,0,0,0]]) 

In [46]: c2 = sp.csr_matrix([[0,3,4,0], 
    ...:      [0,0,0,5], 
    ...:      [6,7,0,8]]) 

In [47]: h = sp.hstack((c1, c2), format='csr') 

In [48]: h 
Out[48]: 
<3x8 sparse matrix of type '<type 'numpy.int64'>' 
    with 8 stored elements in Compressed Sparse Row format> 

In [49]: h.A 
Out[49]: 
array([[0, 0, 1, 0, 0, 3, 4, 0], 
     [2, 0, 0, 0, 0, 0, 0, 5], 
     [0, 0, 0, 0, 6, 7, 0, 8]]) 

In [50]: v = sp.vstack((c1, c2), format='csr') 

In [51]: v 
Out[51]: 
<6x4 sparse matrix of type '<type 'numpy.int64'>' 
    with 8 stored elements in Compressed Sparse Row format> 

In [52]: v.A 
Out[52]: 
array([[0, 0, 1, 0], 
     [2, 0, 0, 0], 
     [0, 0, 0, 0], 
     [0, 3, 4, 0], 
     [0, 0, 0, 5], 
     [6, 7, 0, 8]])