45
votes
import random
pos = ["A", "B", "C"]
x = random.choice["A", "B", "C"]

This code gives me either "A", "B" or "C" with equal probability. Is there a nice way to express it when you want "A" with 30%, "B" with 40% and "C" with 30% probability?

6
Searching found several similar/identical questions here and here - snapshoe
@ma3: One of the weak points of this site is that as old questions tend to be ignored, there's no mechanism or motivation to improve on them. I think my answer is significantly better than at least the higher answers on those questions--havn't read the lower ones--but nobody would ever see it if I posted it on those. - Glenn Maynard
Starting with Python 3.6 you can use random.choices to submit a list of weights/probabilities. - Nick

6 Answers

52
votes

Weights define a probability distribution function (pdf). Random numbers from any such pdf can be generated by applying its associated inverse cumulative distribution function to uniform random numbers between 0 and 1.

See also this SO explanation, or, as explained by Wikipedia:

If Y has a U[0,1] distribution then F⁻¹(Y) is distributed as F. This is used in random number generation using the inverse transform sampling-method.

import random
import bisect
import collections

def cdf(weights):
    total = sum(weights)
    result = []
    cumsum = 0
    for w in weights:
        cumsum += w
        result.append(cumsum / total)
    return result

def choice(population, weights):
    assert len(population) == len(weights)
    cdf_vals = cdf(weights)
    x = random.random()
    idx = bisect.bisect(cdf_vals, x)
    return population[idx]

weights=[0.3, 0.4, 0.3]
population = 'ABC'
counts = collections.defaultdict(int)
for i in range(10000):
    counts[choice(population, weights)] += 1
print(counts)

# % test.py
# defaultdict(<type 'int'>, {'A': 3066, 'C': 2964, 'B': 3970})

The choice function above uses bisect.bisect, so selection of a weighted random variable is done in O(log n) where n is the length of weights.


Note that as of version 1.7.0, NumPy has a Cythonized np.random.choice function. For example, this generates 1000 samples from the population [0,1,2,3] with weights [0.1, 0.2, 0.3, 0.4]:

import numpy as np
np.random.choice(4, 1000, p=[0.1, 0.2, 0.3, 0.4])

np.random.choice also has a replace parameter for sampling with or without replacement.


A theoretically better algorithm is the Alias Method. It builds a table which requires O(n) time, but after that, samples can be drawn in O(1) time. So, if you need to draw many samples, in theory the Alias Method may be faster. There is a Python implementation of the Walker Alias Method here, and a numpy version here.

30
votes

Not... so much...

pos = ['A'] * 3 + ['B'] * 4 + ['C'] * 3
print random.choice(pos)

or

pos = {'A': 3, 'B': 4, 'C': 3}
print random.choice([x for x in pos for y in range(pos[x])])
9
votes

Here's a class to expose a bunch of items with relative probabilities, without actually expanding the list:

import bisect
class WeightedTuple(object):
    """
    >>> p = WeightedTuple({'A': 2, 'B': 1, 'C': 3})
    >>> len(p)
    6
    >>> p[0], p[1], p[2], p[3], p[4], p[5]
    ('A', 'A', 'B', 'C', 'C', 'C')
    >>> p[-1], p[-2], p[-3], p[-4], p[-5], p[-6]
    ('C', 'C', 'C', 'B', 'A', 'A')
    >>> p[6]
    Traceback (most recent call last):
    ...
    IndexError
    >>> p[-7]
    Traceback (most recent call last):
    ...
    IndexError
    """
    def __init__(self, items):
        self.indexes = []
        self.items = []
        next_index = 0
        for key in sorted(items.keys()):
            val = items[key]
            self.indexes.append(next_index)
            self.items.append(key)
            next_index += val

        self.len = next_index

    def __getitem__(self, n):
        if n < 0:
            n = self.len + n
        if n < 0 or n >= self.len:
            raise IndexError

        idx = bisect.bisect_right(self.indexes, n)
        return self.items[idx-1]

    def __len__(self):
        return self.len

Now, just say:

data = WeightedTuple({'A': 30, 'B': 40, 'C': 30})
random.choice(data)
7
votes

You can also make use this form, which does not create a list arbitrarily big (and can work with either integral or decimal probabilities):

pos = [("A", 30), ("B", 40), ("C", 30)]


from random import uniform
def w_choice(seq):
    total_prob = sum(item[1] for item in seq)
    chosen = random.uniform(0, total_prob)
    cumulative = 0
    for item, probality in seq:
        cumulative += probality
        if cumulative > chosen:
            return item
5
votes

There are some good solutions offered here, but I would suggest that you look at Eli Bendersky's thorough discussion of this issue, which compares various algorithms to achieve this (with implementations in Python) before choosing one.

1
votes

Try this:

import random
from decimal import Decimal

pos = {'A': Decimal("0.3"), 'B': Decimal("0.4"), 'C': Decimal("0.3")}
choice = random.random()
F_x = 0
for k, p in pos.iteritems():
    F_x += p
    if choice <= F_x:
        x = k
        break