如何连接coo_matrix与列numpy阵列

Question

我有coo_matrixa形状(40106, 2048)和列numpy阵列b形状(40106,)。

我想要做的是简单地连接矩阵和数组（即结果数据结构将具有形状(40106, 2049)）。 我试着使用hstack如下图所示

concat = hstack([a, b]) 

，但我得到了以下错误：

File "/Users/usr/anaconda/lib/python3.5/site-packages/scipy/sparse/construct.py", line 464, in hstack 
    return bmat([blocks], format=format, dtype=dtype) 
File "/Users/usr/anaconda/lib/python3.5/site-packages/scipy/sparse/construct.py", line 581, in bmat 
    'row dimensions' % i) 
ValueError: blocks[0,:] has incompatible row dimensions 

我不太明白为什么尺寸不匹配，因为这两个a和b有相同的行数。

Answer 1

我认为这是sparse.hstack。转换为矩阵时，您的b将为(1,40106)。尝试将它转换为正确的稀疏矩阵，然后传递给hstack。 hstack传递作业bmat，从而结束了加入所有的输入矩阵的coo属性，从而使一个新的矩阵

In [66]: from scipy import sparse 
In [67]: A = sparse.coo_matrix(np.eye(3)) 
In [68]: b = np.ones(3) 
In [69]: sparse.hstack((A,b)) 
.... 
ValueError: blocks[0,:] has incompatible row dimensions 
In [70]: B=sparse.coo_matrix(b) 
In [71]: B 
Out[71]: 
<1x3 sparse matrix of type '<class 'numpy.float64'>' 
    with 3 stored elements in COOrdinate format> 
In [72]: sparse.hstack((A,B.T)) 
Out[72]: 
<3x4 sparse matrix of type '<class 'numpy.float64'>' 
    with 6 stored elements in COOrdinate format> 
In [73]: _.A 
Out[73]: 
array([[ 1., 0., 0., 1.], 
     [ 0., 1., 0., 1.], 
     [ 0., 0., 1., 1.]]) 

这也适用（如Divakar的答案）：

In [74]: sparse.hstack((A,b[:,None])) 
Out[74]: 
<3x4 sparse matrix of type '<class 'numpy.float64'>' 
    with 6 stored elements in COOrdinate format> 

我hastack作用：

return bmat([blocks], format=format, dtype=dtype) 

所以直接调用BMAT也适用

In [93]: sparse.bmat([[A, B.T]]) 
Out[93]: 
<3x4 sparse matrix of type '<class 'numpy.float64'>' 
    with 6 stored elements in COOrdinate format> 

sparse.bmat([A, B.T])产生你的blocks must be 2d错误。

Answer 2

转换的第二阵列，这是1D到2D和使用，然后hstack -

hstack([A,B[:,None]]) 

采样运行 -

In [86]: from scipy.sparse import coo_matrix, hstack 

# Sample inputs as a coo_matrix and an array 
In [87]: A = coo_matrix([[1, 2, 0], [3, 0, 4]]) 
    ...: B = np.array([5, 6]) 
    ...: 

# Use proposed solution 
In [88]: out = hstack([A,B[:,None]]) 

# Print the dense version to visually verify 
In [89]: out.toarray() 
Out[89]: 
array([[1, 2, 0, 5], 
     [3, 0, 4, 6]])