将一列零添加到csr_matrix

Question

我有一个MxN稀疏csr_matrix,我想在矩阵的右边添加一些只有零的列.原则上,阵列indptr,indices并data保持相同的,所以我只是想改变矩阵的尺寸.但是,这似乎没有实现.

>>> A = csr_matrix(np.identity(5), dtype = int)
>>> A.toarray()
array([[1, 0, 0, 0, 0],
       [0, 1, 0, 0, 0],
       [0, 0, 1, 0, 0],
       [0, 0, 0, 1, 0],
       [0, 0, 0, 0, 1]])
>>> A.shape
(5, 5)
>>> A.shape = ((5,7))
NotImplementedError: Reshaping not implemented for csr_matrix.

水平堆叠零矩阵似乎也不起作用.

>>> B = csr_matrix(np.zeros([5,2]), dtype = int)
>>> B.toarray()
array([[0, 0],
       [0, 0],
       [0, 0],
       [0, 0],
       [0, 0]])
>>> np.hstack((A,B))
array([ <5x5 sparse matrix of type '<type 'numpy.int32'>'
    with 5 stored elements in Compressed Sparse Row format>,
       <5x2 sparse matrix of type '<type 'numpy.int32'>'
    with 0 stored elements in Compressed Sparse Row format>], dtype=object)

这是我最终想要实现的目标.有没有快速重塑我的方法csr_matrix而不复制其中的所有内容？

>>> C = csr_matrix(np.hstack((A.toarray(), B.toarray())))
>>> C.toarray()
array([[1, 0, 0, 0, 0, 0, 0],
       [0, 1, 0, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 1, 0, 0, 0],
       [0, 0, 0, 0, 1, 0, 0]])

Answer 1

您可以使用scipy.sparse.vstack或scipy.sparse.hstack更快地执行此操作:

from scipy.sparse import csr_matrix, vstack, hstack

B = csr_matrix((5, 2), dtype=int)
C = csr_matrix((5, 2), dtype=int)
D = csr_matrix((10, 10), dtype=int)

B2 = vstack((B, C))
#<10x2 sparse matrix of type '<type 'numpy.int32'>'
#        with 0 stored elements in COOrdinate format>

hstack((B2, D))
#<10x12 sparse matrix of type '<type 'numpy.int32'>'
#        with 0 stored elements in COOrdinate format>

注意,输出是一个coo_matrix,它可以被有效地转化成CSR或CSC格式.

Answer 2

你想要做的并不是真正的numpy或scipy理解为重塑.但是,对于您的特定情况下,你可以创建一个新的CSR矩阵重用data,indices并indptr从原来的一个,无需将其复制:

import scipy.sparse as sps

a = sps.rand(10000, 10000, density=0.01, format='csr')

In [19]: %timeit sps.csr_matrix((a.data, a.indices, a.indptr),
...                             shape=(10000, 10020), copy=True)
100 loops, best of 3: 6.26 ms per loop

In [20]: %timeit sps.csr_matrix((a.data, a.indices, a.indptr),
...                             shape=(10000, 10020), copy=False)
10000 loops, best of 3: 47.3 us per loop

In [21]: %timeit sps.csr_matrix((a.data, a.indices, a.indptr),
...                             shape=(10000, 10020))
10000 loops, best of 3: 48.2 us per loop

因此,如果您不再需要原始矩阵a,则默认为copy=False:

a = sps.csr_matrix((a.data, a.indices, a.indptr), shape=(10000, 10020))