我需要写一个加权版的random。选择(列表中的每个元素有不同的被选择的概率)。这是我想到的:
def weightedChoice(choices):
"""Like random.choice, but each element can have a different chance of
being selected.
choices can be any iterable containing iterables with two items each.
Technically, they can have more than two items, the rest will just be
ignored. The first item is the thing being chosen, the second item is
its weight. The weights can be any numeric values, what matters is the
relative differences between them.
"""
space = {}
current = 0
for choice, weight in choices:
if weight > 0:
space[current] = choice
current += weight
rand = random.uniform(0, current)
for key in sorted(space.keys() + [current]):
if rand < key:
return choice
choice = space[key]
return None
这个函数对我来说太复杂了，而且很丑。我希望这里的每个人都能提供一些改进的建议或其他方法。对我来说，效率没有代码的整洁和可读性重要。
我可能已经来不及提供任何有用的东西了，但这里有一个简单，简短，非常有效的片段:
def choose_index(probabilies):
cmf = probabilies[0]
choice = random.random()
for k in xrange(len(probabilies)):
if choice <= cmf:
return k
else:
cmf += probabilies[k+1]
不需要排序你的概率或用你的cmf创建一个向量，它一旦找到它的选择就会终止。内存:O(1)，时间:O(N)，平均运行时间~ N/2。
如果你有权重，只需添加一行:
def choose_index(weights):
probabilities = weights / sum(weights)
cmf = probabilies[0]
choice = random.random()
for k in xrange(len(probabilies)):
if choice <= cmf:
return k
else:
cmf += probabilies[k+1]
将权重排列成a
累积分布。
使用random.random()来选择一个随机的
浮点0.0 <= x < total。
搜索
用等分法进行分布。二等分的
如http://docs.python.org/dev/library/bisect.html#other-examples中的示例所示。
from random import random
from bisect import bisect
def weighted_choice(choices):
values, weights = zip(*choices)
total = 0
cum_weights = []
for w in weights:
total += w
cum_weights.append(total)
x = random() * total
i = bisect(cum_weights, x)
return values[i]
>>> weighted_choice([("WHITE",90), ("RED",8), ("GREEN",2)])
'WHITE'
如果需要做出多个选择，可以将其分成两个函数，一个用于构建累积权重，另一个用于对随机点进行等分。
我不喜欢它们的语法。我只想具体说明这些项目是什么以及每项的权重是多少。我意识到我可以用随机。选项，但我很快就写了下面的类。
import random, string
from numpy import cumsum
class randomChoiceWithProportions:
'''
Accepts a dictionary of choices as keys and weights as values. Example if you want a unfair dice:
choiceWeightDic = {"1":0.16666666666666666, "2": 0.16666666666666666, "3": 0.16666666666666666
, "4": 0.16666666666666666, "5": .06666666666666666, "6": 0.26666666666666666}
dice = randomChoiceWithProportions(choiceWeightDic)
samples = []
for i in range(100000):
samples.append(dice.sample())
# Should be close to .26666
samples.count("6")/len(samples)
# Should be close to .16666
samples.count("1")/len(samples)
'''
def __init__(self, choiceWeightDic):
self.choiceWeightDic = choiceWeightDic
weightSum = sum(self.choiceWeightDic.values())
assert weightSum == 1, 'Weights sum to ' + str(weightSum) + ', not 1.'
self.valWeightDict = self._compute_valWeights()
def _compute_valWeights(self):
valWeights = list(cumsum(list(self.choiceWeightDic.values())))
valWeightDict = dict(zip(list(self.choiceWeightDic.keys()), valWeights))
return valWeightDict
def sample(self):
num = random.uniform(0,1)
for key, val in self.valWeightDict.items():
if val >= num:
return key
下面是Python 3.6标准库中包含的版本:
import itertools as _itertools
import bisect as _bisect
class Random36(random.Random):
"Show the code included in the Python 3.6 version of the Random class"
def choices(self, population, weights=None, *, cum_weights=None, k=1):
"""Return a k sized list of population elements chosen with replacement.
If the relative weights or cumulative weights are not specified,
the selections are made with equal probability.
"""
random = self.random
if cum_weights is None:
if weights is None:
_int = int
total = len(population)
return [population[_int(random() * total)] for i in range(k)]
cum_weights = list(_itertools.accumulate(weights))
elif weights is not None:
raise TypeError('Cannot specify both weights and cumulative weights')
if len(cum_weights) != len(population):
raise ValueError('The number of weights does not match the population')
bisect = _bisect.bisect
total = cum_weights[-1]
return [population[bisect(cum_weights, random() * total)] for i in range(k)]
来源:https://hg.python.org/cpython/file/tip/Lib/random.py l340
假设你有
items = [11, 23, 43, 91]
probability = [0.2, 0.3, 0.4, 0.1]
你有一个函数，它生成一个介于[0,1)之间的随机数(我们可以在这里使用random.random())。
现在求概率的前缀和
prefix_probability=[0.2,0.5,0.9,1]
现在，我们只需取一个0-1之间的随机数，然后使用二分搜索来查找该数字在prefix_probability中的位置。这个索引就是你的答案
代码是这样的
return items[bisect.bisect(prefix_probability,random.random())]
