Hi,
Thank you for the suggestion, actually inlining makes more sense for what I
am trying to do.
However, a casting issue arises when trying to compute the derivative wrt
to the continuous input. If I understood correctly, DisconnectedInput
should be returned as the gradient for integral inputs (or inputs wrt which
I don't need the derivative) right?
Below is the slightly modified code which illustrate this new issue:
import numpy as np
import theano.tensor as T
import theano
def make_ops():
x_var = T.vector()
m_var = T.bvector()
r = m_var.sum().astype('floatX')
z = x_var * m_var / r
def grad_op1(inputs, output_gradients):
return [
output_gradients[0], # computation delegated to op2
theano.gradient.DisconnectedType()(),
]
op1 = theano.OpFromGraph(
inputs=[x_var, m_var],
outputs=[z, r],
grad_overrides=grad_op1,
inline=True)
z_var = T.vector()
r_var = T.scalar()
def grad_op2(inputs, output_gradients):
_, m_, r_ = inputs
return [
m_ * r_,
theano.gradient.DisconnectedType()(),
theano.gradient.DisconnectedType()()
]
op2 = theano.OpFromGraph(
inputs=[z_var, m_var, r_var],
outputs=[z_var],
grad_overrides=grad_op2,
inline=True)
return op1, op2
op1, op2 = make_ops()
x_var = T.vector()
m_var = T.bvector()
z_, r = op1(x_var, m_var)
z = op2(z_, m_var, r)
g = theano.grad(T.sum(z), wrt=x_var)
print(g.eval({x_var: np.array([1., .3, .0, .2], dtype=np.float32),
m_var: np.array([1, 0, 1, 1], dtype=np.int8)}))
Le mardi 11 juillet 2017 11:32:50 UTC+2, nicolas....@gmail.com a écrit :
>
> Hi,
>
> I am trying to split an computation over two ops in order to avoid
> spurious computations when computing the gradient.
> My current attempt uses a first op which returns the desired result for
> the forward part and extra intermediate results. The second op just
> forwards the desired result, but its grad is overriden to compute the
> gradient based on the intermediate results.
>
> In this configuration, Theano complains about unused inputs in the forward
> computation because the intermediate results are not used for the forward
> method of the second op.
>
> Is this an expected behaviour or a bug?
>
> ----
>
> import numpy as np
> import theano.tensor as T
> import theano
>
>
> def make_ops():
> x_var = T.vector()
> m_var = T.bvector()
>
> r = m_var.sum().astype('floatX')
> z = x_var * m_var / r
>
>
> def grad_op1(inputs, output_gradients):
> return [
> output_gradients[0], # computation delegated to op2
> theano.gradient.DisconnectedType()()
> ]
>
>
> op1 = theano.OpFromGraph(
> inputs=[x_var, m_var],
> outputs=[z, r],
> grad_overrides=grad_op1)
>
>
> z_var = T.vector()
> r_var = T.scalar()
>
> def grad_op2(inputs, output_gradients):
> _, m_, r_ = inputs
> return [
> m_ * r_,
> theano.gradient.DisconnectedType()(),
> theano.gradient.DisconnectedType()()
> ]
>
> op2 = theano.OpFromGraph(
> inputs=[z_var, m_var, r_var],
> outputs=[z_var],
> grad_overrides=grad_op2)
>
> return op1, op2
>
>
> op1, op2 = make_ops()
> x_var = T.vector()
> m_var = T.bvector()
> z_, r = op1(x_var, m_var)
> z = op2(z_, m_var, r)
>
> print(z_.eval({x_var: np.array([1., .3, .0, .2], dtype=np.float32),
> m_var: np.array([1, 0, 1, 1], dtype=np.int8)}))
>
> f = theano.function([x_var, m_var], [z], on_unused_input='ignore') #
> raises anyway
>
> print(f(np.array([1., .3, .0, .2], dtype=np.float32),
> np.array([1, 0, 1, 1], dtype=np.int8)))
>
> # g = theano.grad(T.sum(z), wrt=x_var)
> # print(g.eval({x_var: np.array([1., .3, .0, .2], dtype=np.float32),
> # m_var: np.array([1, 0, 1, 1], dtype=np.int8)}))
>
>
>
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