Question

I need a number of unique random permutations of a list without replacement, efficiently. My current approach:

total_permutations = math.factorial(len(population))
permutation_indices = random.sample(xrange(total_permutations), k)
k_permutations = [get_nth_permutation(population, x) for x in permutation_indices]

where get_nth_permutation does exactly what it sounds like, efficiently (meaning O(N)). However, this only works for len(population) <= 20, simply because 21! is so mindblowingly long that xrange(math.factorial(21)) won’t work:

OverflowError: Python int too large to convert to C long

Is there a better algorithm to sample k unique permutations without replacement in O(N)?

Answer 1

Instead of using xrange simply keep generating random numbers until you have as many as you need. Using a set makes sure they’re all unique.

permutation_indices = set()
while len(permutation_indices) < k:
    permutation_indices.add(random.randrange(total_permutations))