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。
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]
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)
