[PyTorch][chapter 60][强化学习-2-有模型学习2]-编程知识

前言：

前面我们讲了一下策略评估的原理,以及例子.

强化学习核心是找到最优的策略，这里

重点讲解两个知识点：

策略改进

策略迭代与值迭代

最后以下面环境E 为例，给出Python 代码

。

1：策略改进

2：策略迭代与值迭代

3：策略迭代代码实现 Python 代码

一策略改进

理想的策略应该能够最大化累积奖赏：

$\pi^{*}= arg max_{\pi} \sum_{x \in X} V^{\pi}(x)$

最优策略对应的值函数 $V^{*}$ 称为最优值函数

$\forall x\in X: V^{*}(x)= V^{\pi^*}(x)$

状态值函数（Bellman 等式）:

动作求和

$V_{T}^{\pi}=\sum_{a \in A}\pi(x,a)\sum_{x^{'}\in X}P_{x\rightarrow x^{'}}^a(\frac{1}{T}R_{x \rightarrow x^{'}}^a+\frac{T-1}{T}V_{T-1}^{\pi}(x^{'}))$ ......16.9

$V_{\gamma}^{\pi}=\sum_{a \in A}\pi(x,a)\sum_{x^{'}\in X}P_{x\rightarrow x^{'}}^a(R_{x \rightarrow x^{'}}^a+\gamma V_{\gamma}^{\pi}(x^{'}))$ ......16.9

状态-动作值函数

状态值函数（Bellman 等式）: 动作求和

$Q_{T}^{\pi}(x,a)=\sum_{x^{'}\in X}P_{x\rightarrow x^{'}}^a(\frac{1}{T}R_{x \rightarrow x^{'}}^a+\frac{T-1}{T}V_{T-1}^{\pi}(x^{'}))$ ...16.10

$V_{\gamma}^{\pi}(x,a)=\sum_{x^{'}\in X}P_{x\rightarrow x^{'}}^a(R_{x \rightarrow x^{'}}^a+\gamma V_{\gamma}^{\pi}(x^{'}))$ ...16.10

由于最优值的累计奖赏已经最大，可以对前面的Bellman 等式做改动，

即使对动作求和改为取最优

最优

$V_{T}^{\pi}=max_{a \in A}\sum_{x^{'}\in X}P_{x\rightarrow x^{'}}^a(\frac{1}{T}R_{x \rightarrow x^{'}}^a+\frac{T-1}{T}V_{T-1}^{*}(x^{'}))$ ....16.13

$V_{\gamma}^{\pi}=\sum_{x^{'}\in X}P_{x\rightarrow x^{'}}^a(R_{x \rightarrow x^{'}}^a+\gamma V_{\gamma}^{*}(x^{'}))$ ...16.13

则 $V^{*}(x)=max_{a\in A}Q^{\pi^{*}}(x,a)$ ....16.14 带入16.10

$Q_{T}^{*}(x,a)=\sum_{x^{'}\in X}P_{x\rightarrow x^{'}}^a(\frac{1}{T}R_{x \rightarrow x^{'}}^a+\frac{T-1}{T}max_{a^{'} \in A}Q_{T-1}^{*}(x^{'},a^{'}))$ ...16.10

$V_{\gamma}^{\pi}(x,a)=\sum_{x^{'}\in X}P_{x\rightarrow x^{'}}^a(R_{x \rightarrow x^{'}}^a+\gamma max_{a^{'} \in A} Q_{\gamma}^{*}(x^{'},a^{'}))$ ...16.10

最优Bellman 等式揭示了非最优策略的改进方式：

将策略选择的动作改变为当前的最优动作。这样改进能使策略更好

策略为 $\pi^{'}$ ,改变动作的条件为： $Q^{\pi}(x,\pi^{'}(x)) \geq V^{\pi}(x)$

带入16.10,可以得到递推不等式

$V^{\pi}(x)\leq Q^{\pi}(x,\pi^{'}(x))$

$=\sum_{x^{'} \in X}P_{x\rightarrow x^{'}}^{\pi^{'}(x)}(R_{x\rightarrow x^{'}}^{\pi^{'}(x)}+\gamma V^{\pi}(x^{'}))$

$=\sum_{x^{'} \in X}P_{x\rightarrow x^{'}}^{\pi^{'}(x)}(R_{x\rightarrow x^{'}}^{\pi^{'}(x)}+\gamma Q^{\pi}(x^{'},\pi^{'}(x^{'})))$

$=V^{\pi^{*}}(x)$ 16.16

二策略迭代与值迭代

可以看出：策略迭代法在每次改进策略后都要对策略进行重新评估，因此比较耗时。

由公式16.16 $V^{\pi}(x)\leq Q^{\pi}(x,\pi^{'}(x))\leq V^{\pi^{*}}(x)$ 策略改进与值函数的改进是一致的

由公式16.13可得

$V_{T}(x)=max_{a \in A}\sum_{x^{'}\in X}P_{x\rightarrow x^{'}}^a(\frac{1}{T}R_{x \rightarrow x^{'}}^a+\frac{T-1}{T}V_{T-1}^{*}(x^{'}))$

$V_{\gamma}^{\pi}=max_{a\in A}\sum_{x^{'}\in X}P_{x\rightarrow x^{'}}^a(R_{x \rightarrow x^{'}}^a+\gamma V_{\gamma}^{*}(x^{'}))$

于是可得值迭代（value iteration）算法.

三策略迭代代码实现

# -*- coding: utf-8 -*-
"""
Created on Wed Nov  1 19:34:00 2023@author: cxf
"""# -*- coding: utf-8 -*-
"""
Created on Mon Oct 30 15:38:17 2023@author: chengxf2
"""
import numpy as np
from enum import Enum
import copyclass State(Enum):#状态空间X    shortWater =1 #缺水health = 2   #健康overflow = 3 #凋亡apoptosis = 4 #溢水class Action(Enum):#动作空间Awater = 1 #浇水noWater = 2 #不浇水class Env():def __init__(self):#状态空间self.X = [State.shortWater, State.health,State.overflow, State.apoptosis]   #动作空间self.A = [Action.water,Action.noWater]   #从状态x出发,执行动作a,转移到新的状态x'，得到的奖赏 r为已知道self.Q ={}self.Q[State.shortWater] =          [[Action.water,0.5,   State.shortWater,-1],[Action.water,0.5,   State.health,1],[Action.noWater,0.4, State.shortWater,-1],[Action.noWater,0.6, State.overflow,-100]]self.Q[State.health] =                [[Action.water,0.6,  State.health,1],[Action.water,0.4,   State.overflow,-1],[Action.noWater,0.6, State.shortWater,-1],[Action.noWater,0.4, State.health,1]]self.Q[State.overflow] =                [[Action.water,0.6,   State.overflow,-1],[Action.water,0.4,   State.apoptosis,-100],[Action.noWater,0.6, State.health,1],[Action.noWater,0.4, State.overflow,-1]]self.Q[State.apoptosis] =[[Action.water,1, State.apoptosis,-100],[Action.noWater,1, State.apoptosis,-100]]self.curV ={} #前面的累积奖赏,t时刻的累积奖赏self.V ={} #累积奖赏,t-1时刻的累积奖赏for x in self.X:    self.V[x] =0self.curV[x]=0def GetX(self):#获取状态空间return self.Xdef GetAction(self):#获取动作空间return self.Adef GetQTabel(self):#获取状态转移概率return self.Qclass LearningAgent():def initStrategy(self):   #初始化策略stragegy ={}stragegy[State.shortWater] = Action.waterstragegy[State.health] =    Action.waterstragegy[State.overflow] = Action.waterstragegy[State.apoptosis] = Action.waterself.stragegy = stragegydef __init__(self):env = Env()self.X = env.GetX()self.A = env.GetAction()self.QTabel = env.GetQTabel()self.curV ={} #前面的累积奖赏self.V ={} #累积奖赏for x in self.X:    self.V[x] =0self.curV[x]=0def  evaluation(self,T):#策略评估for t in range(1,T):#当前策略下面的累积奖赏for  state in self.X: #状态空间reward = 0.0action = self.stragegy[state]QTabel= self.QTabel[state]for Q in QTabel:if action == Q[0]:#在状态x 下面执行了动作a,转移到了新的状态，得到的rnewstate = Q[2] p_a_ss =   Q[1]r_a_ss =   Q[-1]#print("\n p_a_ss",p_a_ss, "\t r_a_ss ",r_a_ss)reward += p_a_ss*((1.0/t)*r_a_ss + (1.0-1/t)*self.V[newstate])self.curV[state] = reward               if (T+1)== t:breakelse:self.V = self.curVdef  improve(self,T):#策略改进stragegy ={}for  state in self.X:QTabel= self.QTabel[state]max_reward = -float('inf') #计算每种Q(state, action)for action in self.A:reward = 0.0for Q in QTabel:if action == Q[0]:#在状态x 下面执行了动作a,转移到了新的状态，得到的rnewstate = Q[2] p_a_ss =   Q[1]r_a_ss =   Q[-1]#print("\n p_a_ss",p_a_ss, "\t r_a_ss ",r_a_ss)reward += p_a_ss*((1.0/T)*r_a_ss + (1.0-1/T)*self.V[newstate])if reward> max_reward:max_reward = rewardstragegy[state] = action#print("\n state ",state, "\t action ",action, "\t reward %4.2f"%reward)return stragegydef compare(self,dict1, dict2):#策略比较for key in dict1:if dict1[key] != dict2.get(key):return Falsereturn Truedef learn(self,T):#随机初始化策略self.initStrategy()n = 0while True:self.evaluation(T-1) #策略评估n = n+1print("\n 迭代次数 %d"%n ,State.shortWater.name, "\t 奖赏: %4.2f "%self.V[State.shortWater],State.health.name, "\t 奖赏: %4.2f "%self.V[State.health],State.overflow.name, "\t 奖赏: %4.2f "%self.V[State.overflow],State.apoptosis.name, "\t 奖赏: %4.2f "%self.V[State.apoptosis],)strategyN =self.improve(T) #策略改进#print("\n ---cur---\n",self.stragegy,"\n ---new-- \n ",strategyN )if self.compare(self.stragegy,strategyN):print("\n ----- 最终策略 -----\n ")for state in self.X:print("\n state ",state, "\t action: ",self.stragegy[state])breakelse:for state in self.X:self.stragegy[state] = strategyN[state]if __name__ == "__main__":T =10agent = LearningAgent()agent.learn(T)

参考：

机器学习.周志华《16 强化学习》_51CTO博客_机器学习周志华

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