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我的问题在某个时间间隔 (0,1) 上具有单一状态和无限量的动作。经过一段时间的谷歌搜索，我发现了一些关于一种称为缩放算法的算法的论文，它可以解决连续动作空间的问题。但是我的实现不善于利用。因此，我正在考虑添加一种 epsilon-greedy 行为。

结合不同的方法是否合理？

你知道我的问题的其他方法吗？

代码示例：

import portion as P
def choose_action(self, i_ph):
    # Activation rule
    not_covered = P.closed(lower=0, upper=1)
    for arm in self.active_arms:
        confidence_radius = calc_confidence_radius(i_ph, arm)
        confidence_interval = P.closed(arm.norm_value - confidence_radius, arm.norm_value + confidence_radius)
        not_covered = not_covered - confidence_interval
    
    if not_covered != P.empty():
        rans = []
        height = 0
        heights = []
        for i in not_covered:
            rans.append(np.random.uniform(i.lower, i.upper))
            height += i.upper - i.lower
            heights.append(i.upper - i.lower)
        ran_n = np.random.uniform(0, height)
        j = 0
        ran = 0
        for i in range(len(heights)):
            if j < ran_n < j + heights[i]:
                ran = rans[i]
            j += heights[i]
        self.active_arms.append(Arm(len(self.active_arms), ran * (self.sigma_square - lower) + lower, ran))

    # Selection rule
    max_index = float('-inf')
    max_index_arm = None
    for arm in self.active_arms:
        confidence_radius = calc_confidence_radius(i_ph, arm)

        # indexfunction from zooming algorithm
        index = arm.avg_learning_reward + 2 * confidence_radius

        if index > max_index:
            max_index = index
            max_index_arm = arm
    action = max_index_arm.value
    self.current_arm = max_index_arm
    return action

def learn(self, action, reward):
    arm = self.current_arm
    arm.avg_reward = (arm.pulled * arm.avg_reward + reward) / (arm.pulled + 1)
    if reward > self.max_profit:
        self.max_profit = reward
    elif reward < self.min_profit:
        self.min_profit = reward

    # normalize reward to [0, 1]
    high = 100
    low = -75
    if reward >= high:
        reward = 1
        self.high_count += 1
    elif reward <= low:
        reward = 0
        self.low_count += 1
    else:
        reward = (reward - low)/(high - low)

    arm.avg_learning_reward = (arm.pulled * arm.avg_learning_reward + reward) / (arm.pulled + 1)
    arm.pulled += 1

# zooming algorithm confidence radius
def calc_confidence_radius(i_ph, arm: Arm):
    return math.sqrt((8 * i_ph)/(1 + arm.pulled))
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1 回答 1

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您可能会发现这里有用的完整算法描述。他们将探针均匀地网格化，从而告知这种选择（例如，在著名的高能臂上正常居中）也是可能的（但这可能会使我不确定的一些界限无效）。

于 2020-10-30T18:26:25.987 回答