Multiplying elements in a sparse array with rows in matrix

Question

If you have a sparse matrix X:

>> X = csr_matrix([[0,2,0,2],[0,2,0,1]])
>> print type(X)    
>> print X.todense()    
<class 'scipy.sparse.csr.csr_matrix'>
[[0 2 0 2]
 [0 2 0 1]]

And a matrix Y:

>> print type(Y)
>> print text_scores
<class 'numpy.matrixlib.defmatrix.matrix'>
[[8]
 [5]]

...How can you multiply each element of X by the rows of Y. For example:

[[0*8 2*8 0*8 2*8]
 [0*5 2*5 0*5 1*5]]

or:

[[0 16 0 16]
 [0 10 0 5]]

I've tired this but obviously it doesn't work as the dimensions dont match: Z = X.data * Y

seberg · Accepted Answer · 2012-09-02 17:48:01Z

9

Unfortunatly the .multiply method of the CSR matrix seems to densify the matrix if the other one is dense. So this would be one way avoiding that:

# Assuming that Y is 1D, might need to do Y = Y.A.ravel() or such...

# just to make the point that this works only with CSR:
if not isinstance(X, scipy.sparse.csr_matrix):
    raise ValueError('Matrix must be CSR.')

Z = X.copy()
# simply repeat each value in Y by the number of nnz elements in each row: 
Z.data *= Y.repeat(np.diff(Z.indptr))

This does create some temporaries, but at least its fully vectorized, and it does not densify the sparse matrix.

For a COO matrix the equivalent is:

Z.data *= Y[Z.row] # you can use np.take which is faster then indexing.

For a CSC matrix the equivalent would be:

Z.data *= Y[Z.indices]

edited Sep 2, 2012 at 17:48

answered Sep 2, 2012 at 17:22

seberg

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3 Comments

seberg Over a year ago

No, for COO, you would need to do Z.data *= Y[Z.row] I think, or np.take instead of indexing if you care about speed.

Zach Over a year ago

That works. It does that without densifying the matrix right?

seberg Over a year ago

I think I will add it to the answer. Like the other one it only makes Y as large as the nonzero elements of Z, it does not densify Z.

ngazagna · Accepted Answer · 2020-08-23 15:41:22Z

1

Something I use to perform row-wise (resp. column-wise) multiplication is to use matrix multiplication with a diagonal matrix on the left (resp. on the right):

import numpy as np
import scipy.sparse as sp

X = sp.csr_matrix([[0,2,0,2],
                   [0,2,0,1]])
Y = np.array([8, 5])

D = sp.diags(Y) # produces a diagonal matrix which entries are the values of Y
Z = D.dot(X) # performs D @ X, multiplication on the left for row-wise action

Sparsity is preserved (in CSR format):

print(type(Z))
>>> <class 'scipy.sparse.csr.csr_matrix'>

And the output is also correct:

print(Z.toarray()) # Z is still sparse and gives the right output
>>> print(Z.toarray()) # Z is still sparse and gives the right output
[[ 0. 16.  0. 16.]
 [ 0. 10.  0.  5.]]

edited Aug 23, 2020 at 15:41

answered Aug 22, 2020 at 17:01

ngazagna

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Comments

Andrea · Accepted Answer · 2019-01-05 17:37:23Z

I had same problem. Personally I didn't find the documentation of scipy.sparse very helpful, neither found function that handles it directly. So I tried to write it myself and this solved for me:

Z = X.copy()
for row_y_idx in range(Y.shape[0]):
    Z.data[Z.indptr[row_y_idx]:Z.indptr[row_y_idx+1]] *= Y[row_y_idx, 0]

The idea is: for each element of Y in position row_y_idx-th, perform a scalar multiplication with the row_y_idx-th row of X. More info about accessing elements in CSR matrices here (where data is A, IA is indptr).

Given X and Y as you defined:

import numpy as np
import scipy.sparse as sps

X = sps.csr_matrix([[0,2,0,2],[0,2,0,1]])
Y = np.matrix([[8], [5]])

Z = X.copy()
for row_y_idx in range(Y.shape[0]):
    Z.data[Z.indptr[row_y_idx]:Z.indptr[row_y_idx+1]] *= Y[row_y_idx, 0]

print(type(Z))
print(Z.todense())

The output is the same as yours:

<class 'scipy.sparse.csr.csr_matrix'>
 [[ 0 16  0 16]
  [ 0 10  0  5]]

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