[Python-ideas] Extremely weird itertools.permutations

[Nick Coghlan]

On 12 Oct 2013 17:18, "Neil Girdhar" <mistersheik at gmail.com> wrote:
>
> I'm sorry, but I can't find a reference supporting the statement that the
current permutations function is consistent with the mathematical
definition.  Perhaps you would like to find a reference? A quick search
yielded the book "the Combinatorics of Permutations":
http://books.google.ca/books?id=Op-nF-mBR7YC&lpg=PP1   Please look in the
chapter "Permutation of multisets".

Itertools effectively produces the permutation of (index, value) pairs.
Hence Steven's point about the permutations of a list not being
mathematically defined, so you have to decide what set to map it to in
order to decide what counts as a unique value. The mapping itertools uses
considers position in the iterable relevant so exchanging two values that
are themselves equivalent is still considered a distinct permutation since
their original position is taken into account. Like a lot of mathematics,
it's a matter of paying close attention to which entities are actually
being manipulated and how the equivalence classes are being defined :)

Hence the current proposal amounts to adding another variant that provides
the permutations of an unordered multiset instead of those of a set of
(index, value) 2-tuples (with the indices stripped from the results).

One interesting point is that combining collections.Counter.elements() with
itertools.permutations() currently does the wrong thing, since
itertools.permutations() *always* considers iterable order significant,
while for collections.Counter.elements() it's explicitly arbitrary.

Cheers,
Nick.

>
> Best,
>
> Neil
>
>
> On Sat, Oct 12, 2013 at 2:34 AM, Steven D'Aprano <steve at pearwood.info>
wrote:
>>
>> On Fri, Oct 11, 2013 at 10:55:06PM -0400, Neil Girdhar wrote:
>> > I honestly think that Python should stick to the mathematical
definition of
>> > permutations rather than some kind of consensus of the tiny minority of
>> > people here.
>>
>> So do I. And that is exactly what itertools.permutations already does.
>>
>>
>>
>> --
>> Steven
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