scipy.sparse 从 csc_matrix 中找到前 3 个值

Question

I have a very large sparse matrix, I want to retrieve the row and column value of the top 10 value. I have created a small sample matrix down below to simulate this case. Any idea how to get the top 3 in the following example?

import numpy as np
from scipy.sparse import csc_matrix

a = np.matrix([[7,2,0],[0,0,6],[1,0,4]])
m = csc_matrix(a)

  (0, 0)    7
  (2, 0)    1
  (0, 1)    2
  (1, 2)    6
  (2, 2)    4

Expected

  (0, 0)    7
  (1, 2)    6
  (2, 2)    4

Answer 1

Does this help?

If you just want the values:

n = 3
np.partition(np.asarray(a), a.size - n, axis=None)[-n:]

Output

array([4, 6, 7])

If you need the position

n = 3
[np.where(a == x) for x in np.partition(np.asarray(a), 
                                        a.size - n, 
                                        axis=None)[-n:]]

Output

[(array([2], dtype=int64), array([2], dtype=int64)),
 (array([1], dtype=int64), array([2], dtype=int64)),
 (array([0], dtype=int64), array([0], dtype=int64))]

Answer 2

In [32]: a = np.array([[7,2,0],[0,0,6],[1,0,4]])
In [33]: M = sparse.coo_matrix(a)
In [34]: M
Out[34]: 
<3x3 sparse matrix of type '<class 'numpy.int64'>'
    with 5 stored elements in COOrdinate format>
In [35]: print(M)
  (0, 0)    7
  (0, 1)    2
  (1, 2)    6
  (2, 0)    1
  (2, 2)    4
In [36]: M.data
Out[36]: array([7, 2, 6, 1, 4])
In [37]: idx = np.argsort(M.data)
In [38]: idx
Out[38]: array([3, 1, 4, 2, 0])
In [39]: idx = idx[-3:]
In [40]: M.data[idx]
Out[40]: array([4, 6, 7])
In [41]: M1 = sparse.coo_matrix((M.data[idx], (M.row[idx], M.col[idx])), M.shape
    ...: )
In [42]: M1
Out[42]: 
<3x3 sparse matrix of type '<class 'numpy.int64'>'
    with 3 stored elements in COOrdinate format>
In [43]: M1.A
Out[43]: 
array([[7, 0, 0],
       [0, 0, 6],
       [0, 0, 4]])
In [44]: print(M1)
  (2, 2)    4
  (1, 2)    6
  (0, 0)    7

I'm using the coo format because it's easier to get the row/col values given the data idx . For csr/csc indices match with data , but the indptr values will be harder to recreate.