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。
如果你没有提前定义你想要选择多少项(所以，你没有做k=10这样的事情)，你只有概率，你可以做下面的事情。注意，你的概率加起来不需要等于1，它们可以相互独立:
soup_items = ['pepper', 'onion', 'tomato', 'celery']
items_probability = [0.2, 0.3, 0.9, 0.1]
selected_items = [item for item,p in zip(soup_items,items_probability) if random.random()<p]
print(selected_items)
>>>['pepper','tomato']
我看了指向的其他线程，并在我的编码风格中提出了这种变化，这返回了用于计数的索引，但返回字符串很简单(注释返回替代):
import random
import bisect
try:
range = xrange
except:
pass
def weighted_choice(choices):
total, cumulative = 0, []
for c,w in choices:
total += w
cumulative.append((total, c))
r = random.uniform(0, total)
# return index
return bisect.bisect(cumulative, (r,))
# return item string
#return choices[bisect.bisect(cumulative, (r,))][0]
# define choices and relative weights
choices = [("WHITE",90), ("RED",8), ("GREEN",2)]
tally = [0 for item in choices]
n = 100000
# tally up n weighted choices
for i in range(n):
tally[weighted_choice(choices)] += 1
print([t/sum(tally)*100 for t in tally])
下面是使用numpy的另一个版本的weighted_choice。传入weights向量，它将返回一个由0组成的数组，其中包含一个1，表示所选择的bin。该代码默认只进行一次绘制，但您可以传入绘制的数量，并且将返回每个绘制的bin的计数。
如果权重向量的和不等于1，它将被规范化，使之等于1。
import numpy as np
def weighted_choice(weights, n=1):
if np.sum(weights)!=1:
weights = weights/np.sum(weights)
draws = np.random.random_sample(size=n)
weights = np.cumsum(weights)
weights = np.insert(weights,0,0.0)
counts = np.histogram(draws, bins=weights)
return(counts[0])
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())]
