[Python-ideas] Fwd: Extremely weird itertools.permutations

[David Mertz]

I realize after reading
http://stackoverflow.com/questions/6284396/permutations-with-unique-valuesthat
my version was ALMOST right:

def nonredundant_permutations(seq, r=None):
    last = ()
    for perm in permutations(sorted(seq), r):
        if perm > last:
            yield perm
            last = perm

I can't look only for inequality, but must use the actual comparison.

>>> ["".join(x) for x in nonredundant_permutations('aaabb',3)]
['aaa', 'aab', 'aba', 'abb', 'baa', 'bab', 'bba']
>>> list(nonredundant_permutations([F(3,1), D(3.0), 3.0]))
[(Fraction(3, 1), Decimal('3'), 3.0)]

Of course, this approach DOES rely on the order in which
itertools.permutations() returns values.  However, it's a bit more compact
than MRAB's version.




On Fri, Oct 11, 2013 at 3:23 PM, Neil Girdhar <mistersheik at gmail.com> wrote:

> Beautiful!!
>
>
> On Fri, Oct 11, 2013 at 6:19 PM, MRAB <python at mrabarnett.plus.com> wrote:
>
>> On 11/10/2013 23:03, David Mertz wrote:
>>
>>> Bummer.  You are right, Neil.  I saw MRAB's suggestion about sorting,
>>> and falsely thought that would be general; but obviously it's not.
>>>
>>> So I guess the question is whether there is ANY way to do this without
>>> having to accumulate a 'seen' set (which can grow to size N!).  The
>>> answer isn't jumping out at me, but that doesn't mean there's not a way.
>>>
>>> I don't want itertools.permutations() to do "equality filtering", but
>>> assuming some other function in itertools were to do that, how could it
>>> do so algorithmically? Or whatever, same question if it is
>>> itertools.permutations(seq, distinct=True) as the API.
>>>
>>>  Here's an implementation:
>>
>> def unique_permutations(iterable, count=None, key=None):
>>     def perm(items, count):
>>         if count:
>>             prev_item = object()
>>
>>             for i, item in enumerate(items):
>>                 if item != prev_item:
>>                     for p in perm(items[ : i] + items[i + 1 : ], count -
>> 1):
>>                         yield [item] + p
>>
>>                 prev_item = item
>>
>>         else:
>>             yield []
>>
>>     if key is None:
>>         key = lambda item: item
>>
>>     items = sorted(iterable, key=key)
>>
>>     if count is None:
>>         count = len(items)
>>
>>     yield from perm(items, count)
>>
>>
>> And some results:
>>
>> >>> print(list("".join(x) for x in unique_permutations('aaabb', 3)))
>> ['aaa', 'aab', 'aba', 'abb', 'baa', 'bab', 'bba']
>> >>> print(list(unique_**permutations([0, 'a', 0], key=str)))
>> [[0, 0, 'a'], [0, 'a', 0], ['a', 0, 0]]
>>
>>
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