问题
I want to divide a sparse matrix's rows by scalars given in an array.
For example : I have a csr_matrix C :
C = [[2,4,6], [5,10,15]]
D = [2,5]
I want the result of C after division to be :
result = [[1, 2, 3], [1, 2, 3]]
I have tried this using the method that we use for numpy arrays :
result = C / D[:,None]
But this seems really slow. How to do this efficiently in sparse matrices ?
回答1:
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
回答2:
If you first cast D to type numpy.matrix (which I'm assuming you can do unless D is too big to fit into memory), then you can just run
C.multiply(1.0 / D.T)
to get what you want.
回答3:
one line code: result = [[C[i][j]/D[i] for j in range(len(C[0]))] for i in range(len(D))]
C = [[2,4,6], [5,10,15]] #len(C[0]) = 3
D = [2,5] # len(D) = 2
result = [[C[i][j]/D[i] for j in range(len(C[0]))] for i in range(len(D))]
print result
来源：https://stackoverflow.com/questions/49254111/row-division-in-scipy-sparse-matrix