how to show sparse matrix with increasing data

Question

how can we have the sparse matrix to be displayed in a way that the value part of a sparse matrix is increasing? I think by default, column part is increasing.

Answer 1

Here's a variation on the basic approach of sorting the data, and displaying the coordinates in the same order

First make a sample random matrix with integer values (for ease of display). I'm using coo format because that's the one that shows the simplest relation between data and coordinates.

In [9]: M = (sparse.random(10,10,.1, 'coo')*10).astype(int)
In [10]: M
Out[10]: 
<10x10 sparse matrix of type '<class 'numpy.int32'>'
    with 10 stored elements in COOrdinate format>
In [11]: print(M)
  (1, 3)    1
  (3, 4)    7
  (5, 4)    1
  (7, 4)    8
  (5, 6)    9
  (6, 6)    7
  (5, 8)    4
  (6, 8)    1
  (4, 9)    5
  (9, 9)    1

In ipython, the basic display is the repr version, while print shows the str version.

Now get the sort order of the data:

In [12]: idx = np.argsort(M.data)

and make a new matrix with row and column in the same order:

In [13]: M1 = sparse.coo_matrix((M.data[idx],(M.row[idx], M.col[idx])),shape=M.shape)    
In [14]: M1
Out[14]: 
<10x10 sparse matrix of type '<class 'numpy.int32'>'
    with 10 stored elements in COOrdinate format>

In [15]: print(M1)
  (1, 3)    1
  (5, 4)    1
  (6, 8)    1
  (9, 9)    1
  (5, 8)    4
  (4, 9)    5
  (3, 4)    7
  (6, 6)    7
  (7, 4)    8
  (5, 6)    9

It's the same matrix, but the data array is sorted

In [16]: M.A
Out[16]: 
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 7, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 5],
       [0, 0, 0, 0, 1, 0, 9, 0, 4, 0],
       [0, 0, 0, 0, 0, 0, 7, 0, 1, 0],
       [0, 0, 0, 0, 8, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 1]])

In [17]: M1.A
Out[17]: 
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 7, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 5],
       [0, 0, 0, 0, 1, 0, 9, 0, 4, 0],
       [0, 0, 0, 0, 0, 0, 7, 0, 1, 0],
       [0, 0, 0, 0, 8, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 1]])

In [18]: M1.data
Out[18]: array([1, 1, 1, 1, 4, 5, 7, 7, 8, 9])

Converting the coo to csr format requires sorting (lexical for row and col), so both coo matrices are converted in the same way:

In [19]: M.tocsr().data
Out[19]: array([1, 7, 5, 1, 9, 4, 7, 1, 8, 1], dtype=int32)
In [20]: M1.tocsr().data
Out[20]: array([1, 7, 5, 1, 9, 4, 7, 1, 8, 1], dtype=int32)

The cannonical csr order by rows and columns within those:

In [23]: print(M1.tocsr())
  (1, 3)    1
  (3, 4)    7
  (4, 9)    5
  (5, 4)    1
  (5, 6)    9
  (5, 8)    4
  (6, 6)    7
  (6, 8)    1
  (7, 4)    8
  (9, 9)    1