scipy.sparse.coo_matrix how to fast find all zeros column, fill with 1 and normalize

Question

For a matrix, i want to find columns with all zeros and fill with 1s, and then normalize the matrix by column. I know how to do that with np.arrays

[[0 0 0 0 0]
         


        

Answer 1

This will be a lot easier with the lil format, and working with rows rather than columns:

In [1]: from scipy import sparse
In [2]: A=np.array([[0,0,0,0,0],[0,0,1,0,0],[1,0,0,1,0],[0,0,0,0,1],[1,0,0,0,0]])
In [3]: A
Out[3]: 
array([[0, 0, 0, 0, 0],
       [0, 0, 1, 0, 0],
       [1, 0, 0, 1, 0],
       [0, 0, 0, 0, 1],
       [1, 0, 0, 0, 0]])
In [4]: At=A.T                # switch to work with rows

In [5]: M=sparse.lil_matrix(At)

Now it is obvious which row is all zeros

In [6]: M.data
Out[6]: array([[1, 1], [], [1], [1], [1]], dtype=object)
In [7]: M.rows
Out[7]: array([[2, 4], [], [1], [2], [3]], dtype=object)

And lil format allows us to fill that row:

In [8]: M.data[1]=[1,1,1,1,1]
In [9]: M.rows[1]=[0,1,2,3,4]
In [10]: M.A
Out[10]: 
array([[0, 0, 1, 0, 1],
       [1, 1, 1, 1, 1],
       [0, 1, 0, 0, 0],
       [0, 0, 1, 0, 0],
       [0, 0, 0, 1, 0]], dtype=int32)

I could have also used M[1,:]=np.ones(5,int)

The coo format is great for creating the array from the data/row/col arrays, but doesn't implement indexing or math. It has to be transformed to csr for that. And csc for column oriented stuff.

The row that I filled isn't so obvious in the csr format:

In [14]: Mc=M.tocsr()
In [15]: Mc.data
Out[15]: array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1], dtype=int32)
In [16]: Mc.indices
Out[16]: array([2, 4, 0, 1, 2, 3, 4, 1, 2, 3], dtype=int32)
In [17]: Mc.indptr
Out[17]: array([ 0,  2,  7,  8,  9, 10], dtype=int32)

On the other hand normalizing is probably easier in this format.

In [18]: Mc.sum(axis=1)
Out[18]: 
matrix([[2],
        [5],
        [1],
        [1],
        [1]], dtype=int32)
In [19]: Mc/Mc.sum(axis=1)
Out[19]: 
matrix([[ 0. ,  0. ,  0.5,  0. ,  0.5],
        [ 0.2,  0.2,  0.2,  0.2,  0.2],
        [ 0. ,  1. ,  0. ,  0. ,  0. ],
        [ 0. ,  0. ,  1. ,  0. ,  0. ],
        [ 0. ,  0. ,  0. ,  1. ,  0. ]])

Notice that it's converted the sparse matrix to a dense one. The sum is dense, and math involving sparse and dense usually produces dense.

I have to use a more round about calculation to preserve the sparse status:

In [27]: Mc.multiply(sparse.csr_matrix(1/Mc.sum(axis=1)))
Out[27]: 
<5x5 sparse matrix of type '<class 'numpy.float64'>'
    with 10 stored elements in Compressed Sparse Row format>

Here's a way of doing this with the csc format (on A)

In [40]: Ms=sparse.csc_matrix(A)
In [41]: Ms.sum(axis=0)
Out[41]: matrix([[2, 0, 1, 1, 1]], dtype=int32)

Use sum to find the all-zeros column. Obviously this could be wrong if the columns have negative values and happen to sum to 0. If that's a concern I can see making a copy of the matrix with all data values replaced by 1.

In [43]: Ms[:,1]=np.ones(5,int)[:,None]
/usr/lib/python3/dist-packages/scipy/sparse/compressed.py:730: SparseEfficiencyWarning: Changing the sparsity structure of a csc_matrix is expensive. lil_matrix is more efficient.
  SparseEfficiencyWarning)
In [44]: Ms.A
Out[44]: 
array([[0, 1, 0, 0, 0],
       [0, 1, 1, 0, 0],
       [1, 1, 0, 1, 0],
       [0, 1, 0, 0, 1],
       [1, 1, 0, 0, 0]])

The warning matters more if you do this sort of change repeatedly. Notice I have to adjust the dimension of the LHS array. Depending on the number of all-zero columns this action can change the sparsity of the matrix substantially.

==================

I could search the col of coo format for missing values with:

In [69]: Mo=sparse.coo_matrix(A)
In [70]: Mo.col
Out[70]: array([2, 0, 3, 4, 0], dtype=int32)

In [71]: Mo.col==np.arange(Mo.shape[1])[:,None]
Out[71]: 
array([[False,  True, False, False,  True],
       [False, False, False, False, False],
       [ True, False, False, False, False],
       [False, False,  True, False, False],
       [False, False, False,  True, False]], dtype=bool)

In [72]: idx = np.nonzero(~(Mo.col==np.arange(Mo.shape[1])[:,None]).any(axis=1))[0]
In [73]: idx
Out[73]: array([1], dtype=int32)

I could then add a column of 1s at this idx with:

In [75]: N=Mo.shape[0]
In [76]: data = np.concatenate([Mo.data, np.ones(N,int)])
In [77]: row = np.concatenate([Mo.row, np.arange(N)])
In [78]: col = np.concatenate([Mo.col, np.ones(N,int)*idx])
In [79]: Mo1 = sparse.coo_matrix((data,(row, col)), shape=Mo.shape)
In [80]: Mo1.A
Out[80]: 
array([[0, 1, 0, 0, 0],
       [0, 1, 1, 0, 0],
       [1, 1, 0, 1, 0],
       [0, 1, 0, 0, 1],
       [1, 1, 0, 0, 0]])

As written it works for just one column, but it could be generalized to several. I also created a new matrix rather than update Mo. But this in-place seems to work as well:

Mo.data,Mo.col,Mo.row = data,col,row

The normalization still requires csr conversion, though I think sparse can hide that for you.

In [87]: Mo1/Mo1.sum(axis=0)
Out[87]: 
matrix([[ 0. ,  0.2,  0. ,  0. ,  0. ],
        [ 0. ,  0.2,  1. ,  0. ,  0. ],
        [ 0.5,  0.2,  0. ,  1. ,  0. ],
        [ 0. ,  0.2,  0. ,  0. ,  1. ],
        [ 0.5,  0.2,  0. ,  0. ,  0. ]])

Even when I take the extra work of maintaining the sparse nature, I still get a csr matrix:

In [89]: Mo1.multiply(sparse.coo_matrix(1/Mo1.sum(axis=0)))
Out[89]: 
<5x5 sparse matrix of type '<class 'numpy.float64'>'
    with 10 stored elements in Compressed Sparse Row format>

See

Find all-zero columns in pandas sparse matrix

for more methods of finding the 0 columns. It turns out Mo.col==np.arange(Mo.shape[1])[:,None] is too slow with large Mo. A test using np.in1d is much better.

1 - np.in1d(np.arange(Mo.shape[1]),Mo.col)