On 7 March 2010 12:58, 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
> --
> ________________________________
>
> la vita e estrany
>
> spir.wikidot.com
>
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You are calling child.__iter__(), but it's return value is being thrown away.
What you want to be doing is something like
def __iter__(self):
if self.holdsEntry:
yield self.entry
for child in self.children:
print "<"
for val in child: #implicit call to child.__iter__()
yield val
print ">"
Then, when the child.__iter__() is called, the returned iterator is
iterated, and the values are passed up the call stack. There's
probably a more terse way of doing this using itertools, but I think
this is probably more readable.
Hope this clears things up (a little, anyway...)
--
Rich "Roadie Rich" Lovely
Just because you CAN do something, doesn't necessarily mean you SHOULD.
In fact, more often than not, you probably SHOULDN'T. Especially if I
suggested it.
10 re-discover BASIC
20 ???
30 PRINT "Profit"
40 GOTO 10
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