How is theano dot product broadcasted

Question

Could anyone example how i theano dot product broadcast. It seems it is different from numpy

import numpy
import theano
import theano.tensor as T

theano.config.compute_test_value = 'off'

W1val = numpy.random.rand(2, 5, 10, 4).astype(theano.config.floatX)

W1 = theano.shared(W1val, 'W1')

x  = T.tensor3('x')

func_of_W1 = W1

h1 = T.dot(x, func_of_W1)

f = theano.function([x], h1)

print f(numpy.random.rand(3, 5, 10)).shape

Here are the experiments I tried with theano.

#   T.dot(x shape ,  W1 shape) = result shape

# (3, 5, 10) * (2, 5, 10, 4) = (3, 5, 2, 5, 4)

# (3, 10) * (2, 5, 10, 4) = (3, 2, 5, 4)

# (3, 10) * (10 ,4) = (3, 4)

# (3, 10) * (2, 10 ,4) = (3, 2, 4)

# (5,10) * (2, 10 ,10) = (5, 2, 10)

Answer 1

Theano does broadcasting just like numpy. To demonstrate, this code compares Theano and numpy directly:

import numpy

import theano
import theano.tensor as T

TENSOR_TYPES = dict([(0, T.scalar), (1, T.vector), (2, T.matrix), (3, T.tensor3), (4, T.tensor4)])

rand = numpy.random.rand


def theano_dot(x, y):
    sym_x = TENSOR_TYPES[x.ndim]('x')
    sym_y = TENSOR_TYPES[y.ndim]('y')
    return theano.function([sym_x, sym_y], theano.dot(sym_x, sym_y))(x, y)


def compare_dot(x, y):
    print theano_dot(x, y).shape, numpy.dot(x, y).shape


print compare_dot(rand(3, 5, 10), rand(2, 5, 10, 4))
print compare_dot(rand(3, 10), rand(2, 5, 10, 4))
print compare_dot(rand(3, 10), rand(10, 4))
print compare_dot(rand(3, 10), rand(2, 10, 4))
print compare_dot(rand(5, 10), rand(2, 10, 10))

The output is

(3L, 5L, 2L, 5L, 4L) (3L, 5L, 2L, 5L, 4L)
(3L, 2L, 5L, 4L) (3L, 2L, 5L, 4L)
(3L, 4L) (3L, 4L)
(3L, 2L, 4L) (3L, 2L, 4L)
(5L, 2L, 10L) (5L, 2L, 10L)

Theano and numpy produce results with the same shape in every case you describe.