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。
粗糙的，但可能足够:
import random
weighted_choice = lambda s : random.choice(sum(([v]*wt for v,wt in s),[]))
这有用吗?
# define choices and relative weights
choices = [("WHITE",90), ("RED",8), ("GREEN",2)]
# initialize tally dict
tally = dict.fromkeys(choices, 0)
# tally up 1000 weighted choices
for i in xrange(1000):
tally[weighted_choice(choices)] += 1
print tally.items()
打印:
[('WHITE', 904), ('GREEN', 22), ('RED', 74)]
假设所有权重都是整数。它们的和不一定是100，我这么做只是为了让测试结果更容易理解。(如果权重是浮点数，则将它们都乘以10，直到所有权重>= 1。)
weights = [.6, .2, .001, .199]
while any(w < 1.0 for w in weights):
weights = [w*10 for w in weights]
weights = map(int, weights)
如果你有一个加权字典而不是一个列表，你可以这样写
items = { "a": 10, "b": 5, "c": 1 }
random.choice([k for k in items for dummy in range(items[k])])
注意(k, k范围的虚拟物品(物品[k])]产生这个列表(' a ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' c ', ' b ', ' b ', ' b ', ' b ', ' b ']
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
如果不介意使用numpy，可以使用numpy.random.choice。
例如:
import numpy
items = [["item1", 0.2], ["item2", 0.3], ["item3", 0.45], ["item4", 0.05]
elems = [i[0] for i in items]
probs = [i[1] for i in items]
trials = 1000
results = [0] * len(items)
for i in range(trials):
res = numpy.random.choice(items, p=probs) #This is where the item is selected!
results[items.index(res)] += 1
results = [r / float(trials) for r in results]
print "item\texpected\tactual"
for i in range(len(probs)):
print "%s\t%0.4f\t%0.4f" % (items[i], probs[i], results[i])
如果你知道你需要提前做多少选择，你可以不像这样循环:
numpy.random.choice(items, trials, p=probs)
这取决于你想对分布进行多少次抽样。
假设要对分布进行K次抽样。当n是分布中的项数时，每次使用np.random.choice()的时间复杂度为O(K(n + log(n)))。
在我的例子中，我需要对相同的分布进行多次采样，阶数为10^3其中n阶数为10^6。我使用了下面的代码，它预先计算了累积分布，并在O(log(n))中对其进行采样。总体时间复杂度为O(n+K*log(n))。
import numpy as np
n,k = 10**6,10**3
# Create dummy distribution
a = np.array([i+1 for i in range(n)])
p = np.array([1.0/n]*n)
cfd = p.cumsum()
for _ in range(k):
x = np.random.uniform()
idx = cfd.searchsorted(x, side='right')
sampled_element = a[idx]
