从python中的稀疏矩阵列出非零元素

Question

How to list, in a simple and one line code (and fast!), all non zero elements of a csr_matrix ?

I'm using this code:

edges_list = list([tuple(row) for row in np.transpose(A.nonzero())])
weight_list = [A[e] for e in edges_list]

but it is taking quite a long time to execute.

Answer 1

For a CSR matrix in canonical form, access the data array directly:

A.data

but be aware that matrices not in canonical form may include explicit zeros or duplicate entries in their representation, which will need special handling. For example,

# Merge duplicates and remove explicit zeros. Both operations modify A.
# We sum duplicates first because they might sum to zero - for example,
# if a 5 and a -5 are in the same spot, we have to sum them to 0 and then remove the 0.
A.sum_duplicates()
A.eliminate_zeros()

# Now use A.data
do_whatever_with(A.data)

Answer 2

You can use A.nonzero() to index into A directly:

In [19]: A = np.random.randint(0, 3, (3, 3))

In [20]: A
Out[20]: 
array([[2, 1, 1],
       [1, 2, 2],
       [0, 1, 0]])

In [21]: A[A.nonzero()]
Out[21]: array([2, 1, 1, 1, 2, 2, 1])

The result is the same as with your approach:

In [22]: edges_list = list([tuple(row) for row in np.transpose(A.nonzero())])

In [23]: [A[e] for e in edges_list]
Out[23]: [2, 1, 1, 1, 2, 2, 1]

And obviously quite a bit faster (and more so if the matrix gets bigger):

In [25]: %timeit [A[e] for e in list([tuple(row) for row in np.transpose(A.nonzero())])]
10000 loops, best of 3: 48 µs per loop

In [26]: %timeit A[A.nonzero()]
100000 loops, best of 3: 10.7 µs per loop

Also works with scipy csr_matrix , although there are better methods for those, as shown in other answers:

In [30]: M = scipy.sparse.csr_matrix(A)

In [31]: M[M.nonzero()]
Out[31]: matrix([[2, 1, 1, 1, 2, 2, 1]], dtype=int32)

Answer 3

Just use A.data

In [16]: from scipy.sparse import csr_matrix

In [17]: A = csr_matrix([[1,0,0],[0,2,0]])

In [18]: A.data
Out[18]: array([1, 2])

If the sparse matrix has been modified or to be safe, you should use: A.eliminate_zeros()

In [19]: A[0,0] = 0

In [20]: A.data
Out[20]: array([0, 2])

In [21]: A.eliminate_zeros()

In [22]: A.data
Out[22]: array([2])

Answer 4

You could use scipy.sparse.find like this:

>>> from scipy.sparse import csr_matrix, find
>>> A = csr_matrix([[7.0, 8.0, 0],[0, 0, 9.0]])
>>> find(A)
(array([0, 0, 1], dtype=int32), array([0, 1, 2], 
dtype=int32), array([ 7.,  8.,  9.]))

https://docs.scipy.org/doc/scipy/reference/generated/scipy.sparse.find.html