MDP、MRP的python实现

Question

import numpy as np

np.random.seed(0)
# 定义状态转移概率矩阵P
P = [
    [0.9, 0.1, 0.0, 0.0, 0.0, 0.0],
    [0.5, 0.0, 0.5, 0.0, 0.0, 0.0],
    [0.0, 0.0, 0.0, 0.6, 0.0, 0.4],
    [0.0, 0.0, 0.0, 0.0, 0.3, 0.7],
    [0.0, 0.2, 0.3, 0.5, 0.0, 0.0],
    [0.0, 0.0, 0.0, 0.0, 0.0, 1.0],
]
P = np.array(P)

rewards = [-1, -2, -2, 10, 1, 0]  # 定义奖励函数
gamma = 0.5  # 定义折扣因子


# 给定一条序列,计算从某个索引（起始状态）开始到序列最后（终止状态）得到的回报
def compute_return(start_index, chain, gamma):
    G = 0
    for i in reversed(range(start_index, len(chain))):
        # TODO ~1: 实现回报函数
        # G = 
    return G


# 一个状态序列,s1-s2-s3-s6
chain = [1, 2, 3, 6]
start_index = 0
G = compute_return(start_index, chain, gamma)
print("根据本序列计算得到回报为：%s。" % G)


def compute(P, rewards, gamma, states_num):
    value=0
    ''' 利用贝尔曼方程的矩阵形式计算解析解,states_num是MRP的状态数 '''
    # TODO ~X: 在完成矩阵形式的解析解推导后，用代码实现解析解
    # 这个TODO直接给出答案了，没写进实验文档
    rewards = np.array(rewards).reshape((-1, 1))  # 将rewards写成列向量形式
    value = np.dot(np.linalg.inv(np.eye(states_num, states_num) - gamma * P),
                   rewards)
    return value


V = compute(P, rewards, gamma, 6)
print("MRP中每个状态价值分别为\n", V)

S = ["s1", "s2", "s3", "s4", "s5"]  # 状态集合
A = ["保持s1", "前往s1", "前往s2", "前往s3", "前往s4", "前往s5", "概率前往"]  # 动作集合
# 状态转移函数
P = {
    "s1-保持s1-s1": 1.0,
    "s1-前往s2-s2": 1.0,
    "s2-前往s1-s1": 1.0,
    "s2-前往s3-s3": 1.0,
    "s3-前往s4-s4": 1.0,
    "s3-前往s5-s5": 1.0,
    "s4-前往s5-s5": 1.0,
    "s4-概率前往-s2": 0.2,
    "s4-概率前往-s3": 0.4,
    "s4-概率前往-s4": 0.4,
}
# 奖励函数
R = {
    "s1-保持s1": -1,
    "s1-前往s2": 0,
    "s2-前往s1": -1,
    "s2-前往s3": -2,
    "s3-前往s4": -2,
    "s3-前往s5": 0,
    "s4-前往s5": 10,
    "s4-概率前往": 1,
}
gamma = 0.5  # 折扣因子
MDP = (S, A, P, R, gamma)

# 策略1,随机策略
Pi_1 = {
    "s1-保持s1": 0.5,
    "s1-前往s2": 0.5,
    "s2-前往s1": 0.5,
    "s2-前往s3": 0.5,
    "s3-前往s4": 0.5,
    "s3-前往s5": 0.5,
    "s4-前往s5": 0.5,
    "s4-概率前往": 0.5,
}
# 策略2
Pi_2 = {
    "s1-保持s1": 0.6,
    "s1-前往s2": 0.4,
    "s2-前往s1": 0.3,
    "s2-前往s3": 0.7,
    "s3-前往s4": 0.5,
    "s3-前往s5": 0.5,
    "s4-前往s5": 0.1,
    "s4-概率前往": 0.9,
}


# 把输入的两个字符串通过“-”连接,便于使用上述定义的P、R变量
def join(str1, str2):
    return str1 + '-' + str2

gamma = 0.5
# 转化后的MRP的状态转移矩阵
# TODO ~2: 动手修改状态转移函数等参数，体会决策过程变化
P_from_mdp_to_mrp = [
    [0.5, 0.5, 0.0, 0.0, 0.0],
    [0.5, 0.0, 0.5, 0.0, 0.0],
    [0.0, 0.0, 0.0, 0.5, 0.5],
    [0.0, 0.1, 0.2, 0.2, 0.5],
    [0.0, 0.0, 0.0, 0.0, 1.0],
]

P_from_mdp_to_mrp = np.array(P_from_mdp_to_mrp)
R_from_mdp_to_mrp = [-0.5, -1.5, -1.0, 5.5, 0]

V = compute(P_from_mdp_to_mrp, R_from_mdp_to_mrp, gamma, 5)
print("MDP中每个状态价值分别为\n", V)


def sample(MDP, Pi, timestep_max, number):
    ''' 采样函数,策略Pi,限制最长时间步timestep_max,总共采样序列数number '''
    S, A, P, R, gamma = MDP
    episodes = []
    for _ in range(number):
        episode = []
        timestep = 0
        s = S[np.random.randint(4)]  # 随机选择一个除s5以外的状态s作为起点
        # 当前状态为终止状态或者时间步太长时,一次采样结束
        while s != "s5" and timestep <= timestep_max:
            timestep += 1
            rand, temp = np.random.rand(), 0
            # 在状态s下根据策略选择动作
            for a_opt in A:
                temp += Pi.get(join(s, a_opt), 0)
                if temp > rand:
                    a = a_opt
                    r = R.get(join(s, a), 0)
                    break
            rand, temp = np.random.rand(), 0
            # 根据状态转移概率得到下一个状态s_next
            for s_opt in S:
                temp += P.get(join(join(s, a), s_opt), 0)
                if temp > rand:
                    s_next = s_opt
                    break
            episode.append((s, a, r, s_next))  # 把（s,a,r,s_next）元组放入序列中
            s = s_next  # s_next变成当前状态,开始接下来的循环
        episodes.append(episode)
    return episodes


# 采样5次,每个序列最长不超过20步
episodes = sample(MDP, Pi_1, 20, 5)
print('第一条序列\n', episodes[0])
print('第二条序列\n', episodes[1])
print('第五条序列\n', episodes[4])


# 对所有采样序列计算所有状态的价值
def MC(episodes, V, N, gamma):
    for episode in episodes:
        G = 0
        for i in range(len(episode) - 1, -1, -1):  #一个序列从后往前计算
            (s, a, r, s_next) = episode[i]
            # TODO ~3: 代码填空
            # G = 
            # N[s] = 
            # V[s] = 


timestep_max = 20
# 采样1000次,可以自行修改
episodes = sample(MDP, Pi_1, timestep_max, 1000)
gamma = 0.5
V = {"s1": 0, "s2": 0, "s3": 0, "s4": 0, "s5": 0}
N = {"s1": 0, "s2": 0, "s3": 0, "s4": 0, "s5": 0}
MC(episodes, V, N, gamma)
print("使用蒙特卡洛方法计算MDP的状态价值为\n", V)

Answer 1

https://blog.csdn.net/lesileqin/article/details/121534911