theano点积是怎么广播的

Question

任何人都可以举例说明我是如何传播 theano 点积的。它似乎与 numpy

不同

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

下面是我用 theano 做的实验。

#   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 像 numpy 一样进行广播。为了演示，这段代码直接比较了 Theano 和 numpy：

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))

输出为

(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 和 numpy 在您描述的每种情况下都会产生相同形状的结果。

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