sorry, forgot to forward this to the list.
On Sun, Mar 7, 2010 at 1:58 PM, spir <denis.s...@gmail.com> wrote:
> Hello,
>
> Is it possible at all to have a recursive generator? I think at a iterator
> for a recursive data structure (here, a trie). The following code does not
> work: it only yields a single value. Like if child.__iter__() were never
> called.
>
> def __iter__(self):
> ''' Iteration on (key,value) pairs. '''
> print '*',
> if self.holdsEntry:
> yield (self.key,self.value)
> for child in self.children:
> print "<",
> child.__iter__()
> print ">",
> raise StopIteration
>
> With the debug prints in code above, "for e in t: print e" outputs:
>
> * ('', 0)
> < > < > < > < > < > < > < > < >
>
> print len(t.children) # --> 9
>
> Why is no child.__iter__() executed at all? I imagine this can be caused by
> the special feature of a generator recording current execution point. (But
> then, is it at all possible to have a call in a generator? Or does the issue
> only appear whan the callee is a generator itself?) Else, the must be an
> obvious error in my code.
>
>
> Denis
remember that child.__iter__ returns a generator object. Merely
calling the function won't do anything. To actually get elements from
a generator object, you need to call next() on the returned iterator,
or iterate over it in another way (e.g. a for loop). I would write
this method more or less like so:
from itertools import chain
def __iter__(self):
this_entry = [(self.key, self.value)] is self.holds_entry else []
return chain(this_entry, *[iter(c) for c in self.children])
(Warning! untested! I'm pretty sure chain will work like this, but you
might have to do a little tweaking)
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