Create a shared row in theano

Question

I noticed that in theano, when one creates a shared variable based on 1D numpy array, this becomes a vector, but not a row:

import theano.tensor as T
import theano, numpy

shared_vector = theano.shared(numpy.zeros((10,)))
print(shared_vector.type)
# TensorType(float64, vector)
print(shared_vector.broadcastable)
# (False,)

The same goes for a 1xN matrix, it becomes a matrix but not a row:

shared_vector = theano.shared(numpy.zeros((1,10,)))
print(shared_vector.type)
# TensorType(float64, matrix)
print(shared_vector.broadcastable)
# (False, False)

This is troublesome when I want to add a M x N matrix to a 1 X N row-vector, because the shared vector is not broadcastable in the first dimension. First of all, this will not work:

row = T.row('row')
mat=T.matrix('matrix')
f=theano.function(
    [],
    mat + row,
    givens={
        mat: numpy.zeros((20,10), dtype=numpy.float32),
        row: numpy.zeros((10,), dtype=numpy.float32)
    },
    on_unused_input='ignore'
)

With the error:

TypeError: Cannot convert Type TensorType(float32, vector) (of Variable <TensorType(float32, vector)>) into Type TensorType(float32, row). You can try to manually convert <TensorType(float32, vector)> into a TensorType(float32, row).

Ok, that's clear, we can't assign vectors to rows. Unfortunately, this is also not fine:

row = T.matrix('row')
mat=T.matrix('matrix')
f=theano.function(
    [],
    mat + row,
    givens={
        mat: numpy.zeros((20,10), dtype=numpy.float32),
        row: numpy.zeros((1,10,), dtype=numpy.float32)
    },
    on_unused_input='ignore'
)
f()

With the error:

ValueError: Input dimension mis-match. (input[0].shape[0] = 20, input[1].shape[0] = 1)
Apply node that caused the error: Elemwise{add,no_inplace}(<TensorType(float32, matrix)>, <TensorType(float32, matrix)>)
Inputs types: [TensorType(float32, matrix), TensorType(float32, matrix)]
Inputs shapes: [(20, 10), (1, 10)]
Inputs strides: [(40, 4), (40, 4)]
Inputs values: ['not shown', 'not shown']

Backtrace when the node is created:
  File "<ipython-input-55-0f03bee478ec>", line 5, in <module>
    mat + row,

HINT: Use the Theano flag 'exception_verbosity=high' for a debugprint and storage map footprint of this apply node.

So we can't just use a 1 x N matrix as a row as well (because the first dimension of a 1 x N matrix is not broadcastable).

The question remains, what 'can' we do? How can I create a shared variable of type row, such that is i broadcastable using matrix-row addition?

Eugene

I seems that I have found out, seconds after posting. I think I can use .reshape on a vector, to create a 1 x N matrix, which has broadcastable (True, False) – Eugene

Answer 1

An alternative to using reshape(1, N) is to use dimshuffle('x', 0) as described in the documentation.

Here's a demo of the two approaches:

import numpy
import theano

x = theano.shared(numpy.arange(10))
print x
print x.dimshuffle('x', 0).type
print x.dimshuffle(0, 'x').type
print x.reshape((1, x.shape[0])).type
print x.reshape((x.shape[0], 1)).type

f = theano.function([], outputs=[x, x.dimshuffle('x', 0), x.reshape((1, x.shape[0]))])
theano.printing.debugprint(f)

This prints

<TensorType(int32, vector)>
TensorType(int32, row)
TensorType(int32, col)
TensorType(int32, row)
TensorType(int32, col)
DeepCopyOp [@A] ''   2
 |<TensorType(int32, vector)> [@B]
DeepCopyOp [@C] ''   4
 |InplaceDimShuffle{x,0} [@D] ''   1
   |<TensorType(int32, vector)> [@B]
DeepCopyOp [@E] ''   6
 |Reshape{2} [@F] ''   5
   |<TensorType(int32, vector)> [@B]
   |MakeVector{dtype='int64'} [@G] ''   3
     |TensorConstant{1} [@H]
     |Shape_i{0} [@I] ''   0
       |<TensorType(int32, vector)> [@B]

Demonstrating that the dimshuffle is probably preferable as it involves less work than the reshape.

Rufous

I always preferred reshape for readability reasons, but I will consider dimshuffle now. Thank you! – Eugene

Answer 2

I would use:

shared_row = theano.shared(numpy.zeros((1,10,)), broadcastable=(True, False))
print(shared_row.type)
# TensorType(float64, row)
print(shared_row.broadcastable)
(True, False)

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