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Deterministic-ally get activation_index, fixed identation, added support for python3 #4
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deterministically get activation index
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self.compute_rank is now a function that returns a function (hook). It looks like the pytorch hook will call compute_rank, it will return hook as a function object (but won't run it), and self.filter_ranks won't be computed anywhere.
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self.compute_rank now returns a function (hook). So when self.compute_rank(activation_index) is called, hook (a partial function with the local variable activation_index) is passed in as the call back function for register_hook.
So when the gradients are updated, hook is called, but doesn't need to calculate the activation_index because it's given when you called (self.compute_ranks(INDEX))
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Thanks.
But if so, then wasn't the intention to do:
x.register_hook(self.compute_rank(activation_index))
self.activations.append(x)
Othwerwise x isn't appended to self.activations and can't be used from within hook, and pytorch isn't registering the gradient callback to the partial function from self.compute_rank.