Normalizing Sparse.csc_matrix By Its Diagonals
I have a scipy.sparse.csc_matrix with dtype = np.int32. I want to efficiently divide each column (or row, whichever faster for csc_matrix) of the matrix by the diagonal element in
Solution 1:
Make a sparse matrix:
In [379]: M = sparse.random(5,5,.2, format='csr')
In [380]: M
Out[380]:
<5x5 sparse matrix of type '<class 'numpy.float64'>'with5 stored elements in Compressed Sparse Row format>In [381]: M.diagonal()
Out[381]: array([ 0., 0., 0., 0., 0.])
too many 0s in the diagonal - lets add a nonzero diagonal:
In [382]: D=sparse.dia_matrix((np.random.rand(5),0),shape=(5,5))
In [383]: D
Out[383]:
<5x5 sparse matrix of type'<class 'numpy.float64'>'
with 5 stored elements (1 diagonals) in DIAgonal format>
In [384]: M1 = M+D
In [385]: M1
Out[385]:
<5x5 sparse matrix of type'<class 'numpy.float64'>'
with 10 stored elements in Compressed Sparse Row format>
In [387]: M1.A
Out[387]:
array([[ 0.35786668, 0.81754484, 0. , 0. , 0. ],
[ 0. , 0.41928992, 0. , 0.01371273, 0. ],
[ 0. , 0. , 0.4685924 , 0. , 0.35724102],
[ 0. , 0. , 0.77591294, 0.95008721, 0.16917791],
[ 0. , 0. , 0. , 0. , 0.16659141]])
Now it's trivial to divide each column by its diagonal (this is a matrix 'product')
In [388]: M1/M1.diagonal()
Out[388]:
matrix([[ 1. , 1.94983185, 0. , 0. , 0. ],
[ 0. , 1. , 0. , 0.01443313, 0. ],
[ 0. , 0. , 1. , 0. , 2.1444144 ],
[ 0. , 0. , 1.65583764, 1. , 1.01552603],
[ 0. , 0. , 0. , 0. , 1. ]])
Or divide the rows - (multiply by a column vector)
In [391]: M1/M1.diagonal()[:,None]
oops, these are dense; let's make the diagonal sparse
In [408]: md = sparse.csr_matrix(1/M1.diagonal()) # do the inverse here
In [409]: md
Out[409]:
<1x5 sparse matrix of type'<class 'numpy.float64'>'
with 5 stored elements in Compressed Sparse Row format>
In [410]: M.multiply(md)
Out[410]:
<5x5 sparse matrix of type'<class 'numpy.float64'>'
with 5 stored elements in Compressed Sparse Row format>
In [411]: M.multiply(md).A
Out[411]:
array([[ 0. , 1.94983185, 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0.01443313, 0. ],
[ 0. , 0. , 0. , 0. , 2.1444144 ],
[ 0. , 0. , 1.65583764, 0. , 1.01552603],
[ 0. , 0. , 0. , 0. , 0. ]])
md.multiply(M)
for the column version.
Division of sparse matrix - similar except it is using the sum of the rows instead of the diagonal. Deals a bit more with the potential 'divide-by-zero' issue.
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