将稀疏数组中的元素与矩阵中的行相乘

问题描述

如果矩阵X稀疏:

>> 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]]

还有一个矩阵Y:

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

...如何将X的每个元素乘以Y的行.例如:

...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]]

或:

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

我对此感到厌倦，但是显然由于尺寸不匹配而无法正常工作: Z = X.data * Y

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

推荐答案

不幸的是，如果另一个CSR矩阵是密集的，则CSR矩阵的.multiply方法似乎使该矩阵致密.因此，这是避免这种情况的一种方法:

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))

这确实会创建一些临时对象，但至少将其完全矢量化，并且不会使稀疏矩阵致密化.

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

对于COO矩阵，等效值为:

For a COO matrix the equivalent is:

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

对于CSC矩阵，等效值为:

For a CSC matrix the equivalent would be:

Z.data *= Y[Z.indices]

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