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选择概率不同的列表元素的方法[重复]

python

Christian · 技术社区 · 14 年前

这个问题已经有了答案:

A weighted version of random.choice 20个答案

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

这段代码给了我相等概率的“a”、“b”或“c”。当你想要30%的“A”,40%的“B”,30%的概率的“C”时,有没有更好的表达方式?

6 回复 | 直到 6 年前

jpyams David Thompson 6 年前

applying its associated inverse cumulative distribution function

SO explanation Wikipedia

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})

choice bisect.bisect O(log n) n weights

np.random.choice function [0,1,2,3] [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 replace

Alias Method O(n) O(1) here numpy version here

Ignacio Vazquez-Abrams 14 年前

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

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

Glenn Maynard 14 年前

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

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

jsbueno 14 年前

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

Jeet 14 年前

Eli Bendersky's thorough discussion

Jeff Bradberry 14 年前

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