Recursive generator

Ben C spamspam at spam.eggs
Wed Feb 13 20:15:52 CET 2008


On 2008-02-12, Paul Hankin <paul.hankin at gmail.com> wrote:
> On Feb 12, 10:17 pm, Ben C <spams... at spam.eggs> wrote:
>> On 2008-02-12, Paul Rubin <> wrote:
>>
>> > Paul Hankin <paul.han... at gmail.com> writes:
>> >> def genDescendants(self):
>> >>     return chain([self], *[child.genDescendants()
>> >>         for child in self.children])
>>
>> > That is scary.  It generates an in-memory list the size of the
>> > whole subtree, at every level.  Total memory consumption is maybe
>> > even quadratic, depending on the tree shape, but even if it's
>> > only linear, it's way ugly.
>>
>> This would probably be better (in terms of memory if not beauty):
>>
>>     def genDescendants(self):
>>         return chain([self], *(child.genDescendants()
>>             for child in self.children))
>
> No, it's the same I think.

Yes I think you're right. The problem is really the *. The generator
comprehension (child.genDescendants() ...) will have to be run to
completion to build the arguments to pass to chain. So no laziness
there.

> forgot that a generator using yield isn't the same as the same
> function that returns an equivalent generator, because one is lazy and
> the other isn't. To get the lazy behaviour back, you'd have to
> write....
>
> def genDescendants(self):
>     for g in chain([self], *(child.genDescendants()
>             for child in self.children)):
>         yield g

Is that lazy? It still seems like (child.genDescendants() ...) would
have to be run all the way to the end before chain could be called.

Unless * is lazy. I don't think it is, but I don't know for sure.



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