Element wise(column wise) multiply a single column sparse tensor with a rectangular tensor(same numb-CodePudding

I have the following code:

a = tf.constant([[10],
                 [0],
                 [4]])
s = tf.sparse.from_dense(a)
b = tf.constant([[3, 4],
                 [1, 2],
                 [2, 5]])
# print(tf.multiply(s, b))  # this didn't work either though it works if s is dense
print(s.__mul__(b))

I want to multiply a with each column of b getting:

[[30, 40],
 [ 0,  0],
 [ 8, 20]]

But the above code gives me an error that the sparse tensor cant be broadcasted to make the shapes match.

tensorflow.python.framework.errors_impl.InvalidArgumentError: SparseDenseBinaryOpShared broadcasts dense to sparse only; got incompatible shapes: [3,1] vs. [3,2] [Op:SparseDenseCwiseMul]

How do I achieve the above efficiently without materializing the sparse tensor a into a dense one?

CodePudding user response：

Not sure how efficient this is and if it is a viable option for you, but you could try using tf.TensorArray with tf.sparse.sparse_dense_matmul:

import tensorflow as tf

a = tf.constant([[10],
                 [0],
                 [4]])
s = tf.sparse.from_dense(a)
b = tf.constant([[3, 4],
                 [1, 2],
                 [2, 5]])

ta = tf.TensorArray(dtype=tf.int32, size=0, dynamic_size=True)
for i in range(len(b)):
  if i == len(b):
    ta = ta.write(ta.size(), tf.sparse.sparse_dense_matmul(s, b[i:])[i])
  else:
    ta = ta.write(ta.size(), tf.sparse.sparse_dense_matmul(s, b[i:i   1])[i])
print(ta.stack())

tf.Tensor(
[[30 40]
 [ 0  0]
 [ 8 20]], shape=(3, 2), dtype=int32)