Reputation: 7841
I have a big sparse matrix. I want to take log4
for all element in that sparse matrix.
I try to use numpy.log()
but it doesn't work with matrices.
I can also take logarithm row by row. Then I crush old row with a new one.
# Assume A is a sparse matrix (Linked List Format) with float values as data
# It is only for one row
import numpy as np
c = np.log(A.getrow(0)) / numpy.log(4)
A[0, :] = c
This was not as quick as I'd expected. Is there a faster way to do this?
Upvotes: 7
Views: 5497
Reputation: 7841
I think I solve it with very easy way. It is very strange that no one could answer immediately.
# Let A be a COO_matrix
import numpy as np
from scipy.sparse import coo_matrix
new_data = np.log(A.data+3)/np.log(4) #3 is not so important. It can be 1 too.
A = coo_matrix((new_data, (A.row, A.col)), shape=A.shape)
Upvotes: 0
Reputation: 28846
You can modify the data
attribute directly:
>>> a = np.array([[5,0,0,0,0,0,0],[0,0,0,0,2,0,0]])
>>> coo = coo_matrix(a)
>>> coo.data
array([5, 2])
>>> coo.data = np.log(coo.data)
>>> coo.data
array([ 1.60943791, 0.69314718])
>>> coo.todense()
matrix([[ 1.60943791, 0. , 0. , 0. , 0. ,
0. , 0. ],
[ 0. , 0. , 0. , 0. , 0.69314718,
0. , 0. ]])
Note that this doesn't work properly if the sparse format has repeated elements (which is valid in the COO format); it'll take the logs individually, and log(a) + log(b) != log(a + b)
. You probably want to convert to CSR or CSC first (which is fast) to avoid this problem.
You'll also have to add checks if the sparse matrix is in a different format, of course. And if you don't want to modify the matrix in-place, just construct a new sparse matrix as you did in your answer, but without adding 3
because that's completely unnecessary here.
Upvotes: 12