Right, that's a bug (https://github.com/Theano/Theano/issues/5790).
However, for some reason `seq` seems to be defined as int64, which I
doubt is what you want since you are taking a gradient wrt it, which
would be 0.
Moreover, the expression for the gradient does not look right. It
currently returns 0 for the boundaries, which is not correct.
It should be 0 for param2 (changing it by epsilon would not change
the output most of the time), and the sum of the output gradients for
param1, at least when step > 0, which is your case.
On Mon, Mar 27, 2017, Julia K wrote:
> Hi,
>
> I'm getting an error when trying to get the gradient of a function.
>
> Basically, I want to use two parameters to define the lower and upper bound
> of an iterator.
> For easier understanding I simplified my code:
>
> from theano import theano, tensor as tt
>
> def get_sum(param1, param2):
> seq = tt.arange(param1, param2)
>
> components, _ = theano.scan(lambda index: index**2,
> outputs_info=None,
> sequences=[seq])
> return tt.sum(components)
>
>
> param1 = tt.dscalar('param1')
> param1.tag.test_value = 0.0
>
> param2 = tt.dscalar('param2')
> param2.tag.test_value = 10.0
>
> objective = get_sum(param1, param2)
>
> obj_function = theano.function(inputs=[param1, param2],outputs=objective)
>
> obj_gradient = theano.gradient.jacobian(objective,[param1])
> gradient_function = theano.function(inputs=[param1, param2],
> outputs=obj_gradient)
>
> ---------------------------------------------------------------------------
>
> ...
>
> TypeError: ARange{dtype='int64'}.grad illegally returned an integer-valued
> variable. (Input index 0, dtype int64)
>
>
> The function that is called in theano.scan (which is index**2 in the
> example) could be any function that uses the index as one of its parameters.
>
> What do I need to change about the code to make it work?
>
> The same error occurs when using theano version 0.8.2 or
> 0.9.0.dev-b3d19ce8e8d7cdf87c3f84599be81647803aa1f6
>
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Pascal
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