python用softmax算法实现多摇臂老虎机强化学习的问题
我在网上找到了用贪心算法实现多摇臂老虎机强化学习的代码，把贪心算法换成了softmax算法，但是关于老虎机各个摇臂被选中的概率有一些疑问，代码如下：import numpy as np
import random
#老虎机的定义
class Bandit:
def __init__(self, arms_prob):
self.arms_prob = arms_prob
self.size = len(self.arms_prob)
def play(self, i):
if not 0 <= i < self.size:
return -1
else:
if random.uniform(0, 1) < self.arms_prob:
return 1
else:
return 0
class Model:
def __init__(self, bandit, temperature=0.2, training_epochs=10000):
self.bandit = bandit #摇臂老虎机
self.temperature = temperature #温度参数
self.training_epochs = training_epochs #尝试次数
self.values = np.zeros(bandit.size) #摇臂的平均奖赏
self.times = np.zeros(bandit.size) #尝试次数
self.p = #摇臂的概率分布
self.size = bandit.size #摇臂数
self.result = 0 #回报率最高的摇臂编号
#计算摇臂选择的概率分布
def probability(self,index):
self.a = 0
for i in range(bandit.size):
self.a += np.e ** (self.values / self.temperature)
return (np.e ** (self.values / self.temperature)) / self.a
def train(self):
i = 0
r = 0 #最终的奖赏
while i <= self.training_epochs:
index = random.choices(,self.p)#根据权重选取摇臂
reward = self.bandit.play(index)
assert reward >= 0 and index >= 0
self.times += 1
self.values = 1.0 / (self.times+1) *(self.values * self.times + reward)
for j in range(bandit.size):
self.p = self.probability(j)
# self.p = self.probability(index)
i += 1
print("Round %d, choose %d, reward %d " % (i, index, reward))
print(repr(self.values))
r += reward #累加奖赏
# print(repr(self.p))
sum_ = 0
for z in range (self.size):
sum_ += self.p
print(sum_)
self.result = self.values.argmax(axis=0)
print('累积奖赏：',r,' 回报率：', r / self.training_epochs)
bandit = Bandit()
model = Model(bandit, temperature=0.1, training_epochs=30000)
# bandit = Bandit()
# model = Model(bandit, temperature=0.01, training_epochs=3000)
model.train()
print("最佳选择是 %d." % model.result)
在47-49的三行里，我写了两种算摇臂被选中的概率的方法，前一种是每选择一次摇臂，就根据公式将所有摇臂的概率更新一次，我觉得这种似乎比较合理，后一种被我注释掉的，是只更新当次被选中的摇臂的概率。两种实验下来结果都差不多，我想问一下到底应该怎么写。
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