[python-uk] memoize & ordering of kwargs.items()

[Jonathan]

On 11/11/2011 11:14, Jonathan wrote:
> On 11/11/2011 09:35, Jonathan wrote:
>> On 11/11/2011 09:24, Duncan Booth wrote:
>>> pick keys that that hash to the same value modulo the size of the 
>>> dictionary. Since the dictionary copy copies the keys in the order 
>>> they are stored you will get the same hash conflict in the copy as 
>>> the original.
>>
>> Brilliant, thank-you. So I was just being 'lucky' in my choices of 
>> dict keys. I believe this is the answer I am looking for. René & 
>> Ross's point about being mindful of different implementations is 
>> still very pertinent in my mind, but I'm happy for now.
>>
>> Thanks everyone!
>>
>>     Jonathan
>>
>
> For completeness, my final test is therefore:
>
>     def test_not_dependant_on_order_of_kwargs(self):
>         calls = []
>
>         @memoize
>         def counter(**kwargs):
>             calls.append(1)
>
>         for first in ascii_letters:
>             for second in ascii_letters:
>                 forward = {first:0, second:0}
>                 reverse = {second:0, first:0}
>
>                 del calls[:]
>                 counter(**forward)
>                 counter(**reverse)
>                 self.assertEqual(len(calls), 1,
>                     '%d for %s,%s' % (len(calls), first, second))
>

oops, but that doesn't pass with the 'sorted'. This does:

         for index, first in enumerate(ascii_letters):
             for second in ascii_letters[index:]:

(i.e. I was testing each combination forwards and backwards, but then 
later in the iteration also testing them again, in the opposite order, 
which resulted in 0 extra calls to counter.)

-- 
Jonathan Hartley    tartley at tartley.com    http://tartley.com
Made of meat.       +44 7737 062 225       twitter/skype: tartley