Hi,

Yes, it is an actual problem that we never managed to fix in a
satisfactory way. The current behaviour is inconsistent.

Doing the substitution one at a time is a workaround, I think Blocks
does that for dropout, but it can be cumbersome to have everything
cloned over and over again.

Another option, still experimental, may be the `map_variables` function
in scan_modules/scan_utils.

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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