Convert integer to a random but deterministically repeatable choice

Using hash and modulo

import hashlib

def id_to_choice(id_num, num_choices):
    id_bytes = id_num.to_bytes((id_num.bit_length() + 7) // 8, 'big')
    id_hash = hashlib.sha512(id_bytes)
    id_hash_int = int.from_bytes(id_hash.digest(), 'big')  # Uses explicit byteorder for system-agnostic reproducibility
    choice = id_hash_int % num_choices  # Use with small num_choices only
    return choice

>>> id_to_choice(123, 3)
0
>>> id_to_choice(456, 3)
1

Notes:

• The built-in hash method must not be used because it can preserve the input's distribution, e.g. with hash(123). Alternatively, it can return values that differ when Python is restarted, e.g. with hash('123').

• For converting an int to bytes, bytes(id_num) works but is grossly inefficient as it returns an array of null bytes, and so it must not be used. Using int.to_bytes is better. Using str(id_num).encode() works but wastes a few bytes.

• Admittedly, using modulo doesn't offer exactly uniform probability,[1][2] but this shouldn't bias much for this application because id_hash_int is expected to be very large and num_choices is assumed to be small.

Using random

The random module can be used with id_num as its seed, while addressing concerns surrounding both thread safety and continuity. Using randrange in this manner is comparable to and simpler than hashing the seed and taking modulo.

With this approach, not only is cross-language reproducibility a concern, but reproducibility across multiple future versions of Python could also be a concern. It is therefore not recommended.

import random

def id_to_choice(id_num, num_choices):
    localrandom = random.Random(id_num)
    choice = localrandom.randrange(num_choices)
    return choice

>>> id_to_choice(123, 3)
0
>>> id_to_choice(456, 3)
2