Approach #1
Here's a sparse matrix solution using manual replication with indexing
-
from scipy.sparse import csr_matrix
r,c = C.nonzero()
rD_sp = csr_matrix(((1.0/D)[r], (r,c)), shape=(C.shape))
out = C.multiply(rD_sp)
The output is a sparse matrix as well as opposed to the output from C / D[:,None]
that creates a full matrix. As such, the proposed approach saves on memory.
Possible performance boost with replication using np.repeat
instead of indexing -
val = np.repeat(1.0/D, C.getnnz(axis=1))
rD_sp = csr_matrix((val, (r,c)), shape=(C.shape))
Approach #2
Another approach could involve data
method of the sparse matrix that gives us a flattened view into the sparse matrix for in-place
results and also avoid the use of nonzero
, like so -
val = np.repeat(D, C.getnnz(axis=1))
C.data /= val
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