Related to, but not quite the same as Steven D'Aprano's point, I would find it very strange for itertools.permutations() to return a list that was narrowed to equal-but-not-identical items.This is why I've raised the example of 'items=[Fraction(3,1), Decimal(3.0), 3.0]' several times. I've created the Fraction, Decimal, and float for distinct reasons to get different behaviors and available methods. When I want to look for the permutations of those I don't want "any old random choice of equal values" since presumably I've given them a type for a reason.On the other hand, I can see a little bit of sense that 'itertools.permutations([3,3,3,3,3,3,3])' doesn't *really* need to tell me a list of 7!==5040 things that are exactly the same as each other. On the other hand, I don't know how to generalize that, since my feeling is far less clear for 'itertools.permutations([1,2,3,4,5,6,6])' ... there's redundancy, but there's also important information in the probability and count of specific sequences.My feeling, however, is that if one were to trim down the results from a permutations-related function, it is more interesting to me to only eliminate IDENTICAL items, not to eliminate merely EQUAL ones.On Fri, Oct 11, 2013 at 7:37 PM, Neil Girdhar <mistersheik@gmail.com> wrote:
I think it's pretty indisputable that permutations are formally defined this way (and I challenge you to find a source that doesn't agree with that). I'm sure you know that your idea of using permutations to evaluate a multinomial distribution is not efficient. A nicer way to evaluate probabilities is to pass your set through a collections.Counter, and then use the resulting dictionary with scipy.stats.multinomial (if it exists yet).I believe most people will be surprised that len(permutations(iterable)) does count unique permutations.Best,Neil
On Fri, Oct 11, 2013 at 10:06 PM, Steven D'Aprano <steve@pearwood.info> wrote:
On Fri, Oct 11, 2013 at 11:38:33AM -0700, Neil Girdhar wrote:I dispute this entire premise. Take a simple (and stereotypical)
> "It is universally agreed that a list of n distinct symbols has n!
> permutations. However, when the symbols are not distinct, the most common
> convention, in mathematics and elsewhere, seems to be to count only
> distinct permutations." —
example, picking balls from an urn.
Say that you have three Red and Two black balls, and randomly select
without replacement. If you count only unique permutations, you get only
four possibilities:
py> set(''.join(t) for t in itertools.permutations('RRRBB', 2))
{'BR', 'RB', 'RR', 'BB'}
which implies that drawing RR is no more likely than drawing BB, which
is incorrect. The right way to model this experiment is not to count
distinct permutations, but actual permutations:
py> list(''.join(t) for t in itertools.permutations('RRRBB', 2))
['RR', 'RR', 'RB', 'RB', 'RR', 'RR', 'RB', 'RB', 'RR', 'RR', 'RB', 'RB',
'BR', 'BR', 'BR', 'BB', 'BR', 'BR', 'BR', 'BB']
which makes it clear that there are two ways of drawing BB compared to
six ways of drawing RR. If that's not obvious enough, consider the case
where you have two thousand red balls and two black balls -- do you
really conclude that there are the same number of ways to pick RR as BB?
So I disagree that counting only distinct permutations is the most
useful or common convention. If you're permuting a collection of
non-distinct values, you should expect non-distinct permutations.
I'm trying to think of a realistic, physical situation where you would
only want distinct permutations, and I can't.
Absolutely not. Even if you were right that it should return unique
> Should we consider fixing itertools.permutations and to output only unique
> permutations (if possible, although I realize that would break code).
permutations, and I strongly disagree that you were, the fact that it
would break code is a deal-breaker.
--
Steven
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