I have just started thinking about probabilities. A Problem that come up was how to calculate all the potential arrangements for a given sequence. By arrangement I mean unique permutation.

I initially used this method:

from itertools import permutations

sequence = '11223344'
# 2520

But for long sequences this can take a long time! (or run out of memory)

I came up with this function arrangements

from math import factorial
from functools import reduce
from operator import mul

def arrangements(sequence):
    return factorial(len(sequence))/reduce(mul,
           [factorial(sequence.count(i)) for i in set(sequence)])

# arrangements(sequence)
# 2520.0

My thinking is this:

For a given length sequence with all unique items there are factorial(len(sequence)) permutations.

For every repeated item in the sequence there will be factorial(#repeats) that will result in the same permutation.

My function calculates all permutations / all repeated permutations.

I'm sure I have reinvented an already existing standard python function somewhere. I'd like to know if my thinking is sound and the implementation makes sense.

Wouldn't itertools.arrangements be cool?



  • I'd expect arrangements to return the unique permutations of sequence, not just how many there are.
  • If it returns a number, it should be an integer.
  • You could use collections.Counter instead of counting the integers again and again.
  • You're right, it would be nice to have itertools.unique_permutations. In the meantime, I often come back to this SO answer.

Possible refactoring

from math import factorial
from functools import reduce
from collections import Counter
from operator import mul

def count_unique_permutations(sequence):
    count_permutations = factorial(len(sequence))
    repetitions = (factorial(v) for v in Counter(sequence).values())
    return count_permutations // reduce(mul, repetitions)
  • \$\begingroup\$ Hadn't considered using Counter, it looks better :) \$\endgroup\$ – James Schinner Mar 20 '18 at 21:25

The keyword to search for is "multinomial coefficient". That returns this answer: https://stackoverflow.com/questions/46374185/does-python-have-a-function-which-computes-multinomial-coefficients

  • \$\begingroup\$ Good to know. The linked code isn't very readable IMHO but it has the merit of not needing 4 imports for 4 lines of code. \$\endgroup\$ – Eric Duminil Mar 20 '18 at 15:16
  • \$\begingroup\$ Thanks for the Terminology. @EricDuminil at least all imports are from the standard library. \$\endgroup\$ – James Schinner Mar 20 '18 at 21:28

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