Convert sparse matrice in list of list in python

Question

I want to convert a sparse matrice in list of non zero index like this :

>>> row = array([0,2,2,0,1,2])
>>> col = array([0,0,1,2,2,2])
>>> data = array([1,1,1,1,1,1])
>>> mat = csc_matrix( (data,(row,col)), shape=(3,3) )
>>> mat.todense() 
matrix([[1, 0, 1],
    [0, 0, 1],
    [1, 1, 1]])
>>> convert(mat)
[[0, 2],[2],[0, 1, 2]]

It's unclear to me how the output of convert relates to mat. — Fred Foo
– Fred Foo, Commented Mar 23, 2012 at 16:06
I'll state it more strongly: I have no idea how the output of convert relates to mat, or perhaps more importantly, to the input row/col/data. — Karl Knechtel
– Karl Knechtel, Commented Mar 23, 2012 at 16:09
Oh, right, you changed the third value of the output as well. Sorry, hadn't seen that. — Fred Foo
– Fred Foo, Commented Mar 23, 2012 at 16:14
@zedouard: Are you looking for the non-zero entries, or rather for the entries occurring in the sparsity pattern? — Sven Marnach
– Sven Marnach, Commented Mar 23, 2012 at 16:15

DSM · Accepted Answer · 2012-03-23 16:12:13Z

1

Maybe something like

>>> from numpy import array
>>> from scipy.sparse import csc_matrix
>>> 
>>> row = array([0,2,2,0,1,2])
>>> col = array([0,0,1,2,2,2])
>>> data = array([1,1,1,1,1,1])
>>> mat = csc_matrix( (data,(row,col)), shape=(3,3) )
>>> [list(line.nonzero()[1]) for line in mat]
[[0, 2], [2], [0, 1, 2]]

would help? You should probably look at nonzero, anyway.

answered Mar 23, 2012 at 16:12

DSM

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4 Comments

zedouard Over a year ago

Yes, i'm looking for something like that, but i wonder if there isn't exist a function to do this

Sven Marnach Over a year ago

Maybe you can make do with the output of mat.nonzero()?

zedouard Over a year ago

[list(mat[i].nonzero()[1]) for i in range(mat.shape[0])] "for line in mat" doesn't work

DSM Over a year ago

@zedouard: it should, unless there's an edge case I haven't thought of. In what way did it not work?

Sven Marnach · Accepted Answer · 2012-03-23 16:14:03Z

1

Maybe you are looking for something like this:

>>> [mat.indices[mat.indptr[i]:mat.indptr[i+1]]
     for i in range(len(mat.indptr) - 1)]
[array([0, 2]), array([2]), array([0, 1, 2])]

Not sure what this is supposed to be useful for, though. Chances are there are better ways of achieving what you are trying to do.

answered Mar 23, 2012 at 16:14

Sven Marnach

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2 Comments

Fred Foo Over a year ago

This is very CSC/CSR-format specific. @DSM's nonzero-based answer will work for all scipy.sparse types.

Sven Marnach Over a year ago

@larsmans: The important difference is that this answer gives the sparsity pattern, while DSM's answer gives the non-zero elements. I don't think there's a type-agnostic way to get the sparsity pattern. Also note that DSM's answer doesn't work for coo_matrix and dia_matrix.

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