Equivalent for np.add.at in tensorflow

Question

How do I convert a np.add.at statement into tensorflow?

np.add.at(dW, self.x.ravel(), dout.reshape(-1, self.D))

Edit

s

Answer 1

For np.add.at, you probably want to look at tf.SparseTensor, which represents a tensor by a list of values and a list of indices (which is more suitable for sparse data, hence the name).

So for your example:

np.add.at(dW, self.x.ravel(), dout.reshape(-1, self.D))

that would be (assuming dW, x and dout are tensors):

tf.sparse_add(dW, tf.SparseTensor(x, tf.reshape(dout, [-1])))

This is assuming x is of shape [n, nDims] (i.e. x is a 'list' of n indices, each of dimension nDims), and dout has shape [n].

Answer 2

Here's an example of what np.add.at does:

In [324]: a=np.ones((10,))
In [325]: x=np.array([1,2,3,1,4,5])
In [326]: b=np.array([1,1,1,1,1,1])
In [327]: np.add.at(a,x,b)
In [328]: a
Out[328]: array([ 1.,  3.,  2.,  2.,  2.,  2.,  1.,  1.,  1.,  1.])

If instead I use +=

In [331]: a1=np.ones((10,))
In [332]: a1[x]+=b
In [333]: a1
Out[333]: array([ 1.,  2.,  2.,  2.,  2.,  2.,  1.,  1.,  1.,  1.])

note that a1[1] is 2, not 3.

If instead I use an iterative solution

In [334]: a2=np.ones((10,))
In [335]: for i,j in zip(x,b):
     ...:     a2[i]+=j
     ...:     
In [336]: a2
Out[336]: array([ 1.,  3.,  2.,  2.,  2.,  2.,  1.,  1.,  1.,  1.])

it matches.

If x does not have duplicates then += works just fine. But with the duplicates, the add.at is required to match the iterative solution.