I see, thanks Pascal! Shame map_variables doesn't do the trick in this
case. I think I'll go with the manual approach you recommended as it seems
the most efficient and relatively straight forward in my case.
On Friday, October 14, 2016 at 5:13:53 PM UTC-7, Pascal Lamblin wrote:
>
> On Sat, Oct 15, 2016, Pascal Lamblin wrote:
> > Another option, still experimental, may be the `map_variables` function
> > in scan_modules/scan_utils.
>
> There seem to be some challenges regarding scalar constants with that
> function, but I was able to do the following:
>
> >>> theano.tensor.basic.constant.enable = False
> >>> v = theano.tensor.lscalar('v')
> >>> exp1 = 2 * v
> >>> exp1.name = 'exp1'
> >>> exp2 = 4 * exp1
> >>> exp2.name = 'exp2'
> >>> exp3 = 6 * exp2
> >>> exp3.name = 'exp3'
> >>> exp4 = 8 * exp3
> >>> exp4.name = 'exp4'
> >>> replace_dict = {'exp1': (3*exp1), 'exp2': (5*exp2), 'exp3': (7*exp3)}
> >>> def replace(var):
> ... return replace_dict.get(var.name, var)
> >>> exp5, = theano.scan_module.scan_utils.map_variables(replace, [exp4])
> >>> theano.printing.debugprint(exp5)
> Elemwise{mul,no_inplace} [id A] 'exp4'
> |TensorConstant{8} [id B]
> |Elemwise{mul,no_inplace} [id C] ''
> |TensorConstant{7} [id D]
> |Elemwise{mul,no_inplace} [id C] ''
>
> The issue is that it introduced a cycle in the graph: it replaced exp3
> by 7*exp3, where exp3 is the new one...
>
> I guess that illustrates the challenge of getting replacements right.
>
> >
> > Finally, it is actually possible to replace Apply nodes inputs manually.
> > In your case, you could do something like:
> >
> > >>> exp2.owner.inputs[1] = 3*exp1
> > >>> exp3.owner.inputs[1] = 5*exp2
> > >>> exp4.owner.inputs[1] = 7*exp3
> > >>> print(theano.pp(exp4))
> > (TensorConstant{8} * (TensorConstant{7} * (TensorConstant{6} *
> (TensorConstant{5} * (TensorConstant{4} * (TensorConstant{3} *
> (TensorConstant{2} * <TensorType(int64, scalar)>)))))))
> > >>> exp4.eval({v: 1})
> > array(40320)
> >
> > But it can get hard to get right if the same expression is re-used
> > several times.
> >
> > On Fri, Oct 14, 2016, John Coolidge wrote:
> > > Hello,
> > >
> > > I'm trying to use theano.clone to implement dropout in my MLP network.
> > > Because I want to apply dropout at multiple layers, I pass the clone
> call
> > > multiple key value pairs to its replacement parameter:
> > > replace={layer1:mask*layer1, layer2:mask*layer2, etc} however the
> graph
> > > that's returned seems to have only actually made one of the
> replacements.
> > > I suspect this is because clone is doing the replacements
> sequentially and
> > > once it's done one replacement it generates a new graph for which the
> other
> > > key value pairs no longer correspond.
> > >
> > > Here is some example code that demonstrates the unexpected behavior:
> > >
> > > v = T.lscalar()
> > > exp1 = 2*v
> > > exp2 = 4*exp1
> > > exp3 = 6*exp2
> > > exp4 = 8*exp3
> > >
> > > print theano.pp(exp4)
> > > exp5 = theano.clone(exp4, replace={exp1:(3*exp1), exp2:(5*exp2),
> > > exp3:(7*exp3)})
> > > print theano.pp(exp5)
> > > t = theano.function(inputs=[v], outputs=exp5)
> > > print t(1)
> > >
> > >
> > > The output is:
> > > (TensorConstant{8} * (TensorConstant{6} * (TensorConstant{4} *
> > > (TensorConstant{2} * <TensorType(int64, scalar)>))))
> > > (TensorConstant{8} * (TensorConstant{7} * (TensorConstant{6} *
> > > (TensorConstant{4} * (TensorConstant{2} * <TensorType(int64,
> scalar)>)))))
> > > 2688
> > >
> > > Although the clone adds the 7 factor to the new graph, it does not add
> the
> > > 3 or 5 factors such that the output for an input value of 1 is
> 8*7*6*4*2*1
> > > instead of 8! as I would have expected.
> > >
> > > I'm guessing this is how the clone function is supposed to work, but
> does
> > > anyone see how to get the desired behavior I'm looking for? Perhaps I
> > > could apply the replacements one at a time and after each replacement
> > > update the remaining replacement key value pairs to point to
> corresponding
> > > points in the new graph, but I'm not sure how to find these
> corresponding
> > > points. Or perhaps there's a function like the clone but that
> actually
> > > makes the replacements in place so that the other replacement key
> value
> > > pairs would not be invalidated after the first replacement? Any ideas
> > > would be greatly appreciated!
> > >
> > > --
> > >
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> >
> >
> > --
> > Pascal
> >
> > --
> >
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> --
> Pascal
>
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