Python Weighted Random [duplicate]

Question

I need to return different values based on a weighted round-robin such that 1 in 20 gets A, 1 in 20 gets B, and the rest go to C.

So:

A => 5%
B => 5%
C => 90%

Here's a basic version that appears to work:

import random

x = random.randint(1, 100)

if x <= 5:
    return 'A'
elif x > 5 and x <= 10:
    return 'B'
else:
    return 'C'

Is this algorithm correct? If so, can it be improved?

Answer 1

Your algorithm is correct, how about something more elegant:

import random
my_list = ['A'] * 5 + ['B'] * 5 + ['C'] * 90
random.choice(my_list)

Answer 2

that's fine. more generally, you can define something like:

from collections import Counter
from random import randint

def weighted_random(pairs):
    total = sum(pair[0] for pair in pairs)
    r = randint(1, total)
    for (weight, value) in pairs:
        r -= weight
        if r <= 0: return value

results = Counter(weighted_random([(1,'a'),(1,'b'),(18,'c')])
                  for _ in range(20000))
print(results)

which gives

Counter({'c': 17954, 'b': 1039, 'a': 1007})

which is as close to 18:1:1 as you can expect.