强化学习6-MC与TD的比较-实战

# encoding:utf-8
import numpy as np
import matplotlib.pylab as plt

'''
随机行走问题
0 - 1 - 2 - 3 - 4 - 5 - 6
e           s           e
0终点r为0. 6终点r为1
中间每个选择r为0

策略 [-1, 1] 每种选择0.5， -1向左，1向右
这个策略下，理论上数字越大回报越高
'''

stats = range(7)
start = 3
end = [0, 6]
actions = [-1, 1]

r = 1   # 衰减因子
alpha = 0.5 # 学习率
echos = [5, 10, 50, 100, 500, 1000, 10000]

def choose_act(stat):
    # 策略
    if np.random.rand() > 0.5:
        return 1
    else:
        return -1

v = np.zeros([len(stats)])

for i in echos:
    for j in range(i):
        act = choose_act(start)
        stat_ = start + act

        if stat_ in end:
            if stat_ == 6:
                v[start] += alpha * (1 + v[stat_] - v[start])
            else:
                v[start] += alpha * (v[stat_] - v[start])
            start = np.random.randint(1,6)
        else:
            v[start] += alpha * (v[stat_] - v[start])
            start = np.random.randint(1,6)

    plt.plot(v[1:-1])
    plt.text(stats[-4], v[-3], j+1)

plt.xlabel('state')
plt.ylabel('v')
plt.text(1, 0.8, 'alpha = %s'%alpha)
plt.show()

可以看到 随着学习率的增大，效果越来越好，当学习率为0.5时，已经明显过拟合了

这个是单步的，书上是单回合的，所以不同，后续有空会更新代码

# encoding:utf-8
from __future__ import division
import numpy as np
import matplotlib.pylab as plt

stats = range(7)
end = [0, 6]
actions = [-1, 1]
r = 1   # 衰减因子

def choose_act(stat):
    # 策略
    if np.random.rand() > 0.5:
        return 1
    else:
        return -1

v_t = [0, 1/6, 1/3, 1/2, 2/3, 5/6, 0]
alpha_td = [0.1, 0.15, 0.2] # 学习率
alpha_mc = [0.01, 0.02, 0.04]
for c in range(3):
    # TD
    alpha = alpha_td[c]
    # v = np.random.rand(len(stats))
    # v = np.zeros(len(stats))
    v = [0.2] * len(stats)
    errors = []
    start = 3

    for j in range(100):
        act = choose_act(start)
        stat_ = start + act

        if stat_ in end:
            if stat_ == 6:
                v[start] += alpha * (1 + v[stat_] - v[start])
            else:
                v[start] += alpha * (v[stat_] - v[start])
            start = np.random.randint(1,6)
        else:
            v[start] += alpha * (v[stat_] - v[start])
            start = stat_   # np.random.randint(1,6)

        error = np.sqrt(sum([pow(value - v_t[index], 2) for index, value in enumerate(v)]))
        errors.append(error)

    plt.plot(range(100), errors)
    index = np.random.randint(40,100)
    plt.text(index-3, errors[index], 'alpha_td = %s'%alpha)

    # MC
    alpha = alpha_mc[c]
    # v_mc = np.random.rand(len(stats))
    # v_mc = np.zeros(len(stats))
    v_mc = [0.2] * len(stats)
    count_mc = np.zeros(len(stats))
    errors = []
    for j in range(100):
        process = []
        start = 3   # np.random.randint(1, 6)
        while True:
            if start in end:
                process.append([start])
                break
            act = choose_act(start)
            if start == 5 and act == 1:
                r = 1
            else:
                r = 0
            process.append([start, act, r])
            start = start + act

        T = len(process[:-1])
        s_all = [i[0] for i in process[:-1]]
        s_dealed = []
        for k in range(T):
            sar = process[k]
            s = sar[0]
            if s in s_dealed:continue

            # first visit
            t = s_all.index(s)     # 该s 首次出现的位置
            num = s_all.count(s)   # 该s 总共出现的次数
            r_all = sum([i[2] for i in process[t:-1]]) / num
            v_mc[s] += alpha * (r_all - v_mc[s])
            # v_mc[s] = (v_mc[s] * count_mc[s] + r_all) / (count_mc[s] + 1)
            # count_mc[s] += 1

            s_dealed.append(s)
        error = np.sqrt(sum([pow(value - v_t[index], 2) for index, value in enumerate(v_mc)]))
        errors.append(error)
    plt.plot(range(100), errors, '.')
    index = np.random.randint(40,100)
    plt.text(index-3, errors[index], 'alpha_mc = %s'%alpha)

plt.xlabel('echo')
plt.ylabel('mse')
plt.show()

随机行走有个特殊性：两个终点，有一个终点奖励为0，也就是说在前几个回合中，单步更新的TD如果一开始向左走，需要好多步才能到达右边终点，而MC由于是整个回合，要么左，要么右，先到右边终点的概率要大得多，所以，前几步MC收敛明显比TD快

但是从总体来看，TD收敛比MC要快，而且收敛值要小，故TD效率更高

上述代码的问题

1.TD 是单步计算MSE，而MC是单回合计算MSE，比较的前提不同

2.在计算MSE时，只是计算了一次评估的误差，并不是平均误差

更新代码