强化学习之Policy Gradient及代码是实现

导读

强化学习的目标是学习到一个策略 π θ ( s ) \pi_{\theta}(\mathrm{s}) πθ(s)来最大化期望回报,一种直接的方法就是在策略空间直接搜出最佳的策略,称为搜索策略。策略搜索的本质是一个优化问题,可以分为基于梯度的优化和无梯度优化。策略搜索与基于价值函数的方法相比,策略搜索不需要值函数,可以直接优化策略。参数化的策略能够处理连续状态和动作,可以直接学出随机性策略。
策略梯度(Policy Gradient)是一种基于梯度的强化学习方法.假设 π θ ( a ∣ s ) \pi_{\theta}(a|\mathrm{s}) πθ(as)是一个关于 的连续可微函数(神经网络),我们可以用梯度上升的方法来优化参数 使得目标函数()最大。

模型构建

我们构建是的关于 的连续可微函数为神经网络。
神经网络的输入:以向量或矩阵表示的机器Observation(state)
神经网络的输出:每个动作对应输出层的一个神经元。
强化学习之Policy Gradient及代码是实现_第1张图片

使用策略 π θ ( s ) \pi_{\theta}(\mathrm{s}) πθ(s)我们完成一次游戏,轨迹如下所示 τ = s 0 , a 0 , s 1 , r 1 , a 1 , ⋯   , s T − 1 , a T − 1 , s T , r T \tau=s_{0}, a_{0}, s_{1}, r_{1}, a_{1}, \cdots, s_{T-1}, a_{T-1}, s_{T}, r_{T} τ=s0,a0,s1,r1,a1,,sT1,aT1,sT,rT这个轨迹的全部回报为 R θ = ∑ t = 1 T r t R_{\theta}=\sum_{t=1}^{T} r_{t} Rθ=t=1Trt。因为输出为每个行动的概率,所以同一个神经网络,每次 R θ R_{\theta} Rθ都是不同的。因此我们定义 R ˉ θ \bar{R}_{\theta} Rˉθ R θ R_{\theta} Rθ的期望, R ˉ θ \bar{R}_{\theta} Rˉθ的大小可以评价策略 π θ ( s ) \pi_{\theta}(\mathrm{s}) πθ(s)的好坏,我们希望这个期望值越大越好。

一个Episode的轨迹 τ = s 0 , a 0 , s 1 , r 1 , a 1 , ⋯   , s T − 1 , a T − 1 , s T , r T \tau=s_{0}, a_{0}, s_{1}, r_{1}, a_{1}, \cdots, s_{T-1}, a_{T-1}, s_{T}, r_{T} τ=s0,a0,s1,r1,a1,,sT1,aT1,sT,rT则每个 τ \tau τ出现的概率 P ( τ ∣ θ ) = p ( s 1 ) p ( a 1 ∣ s 1 , θ ) p ( r 1 , s 2 ∣ s 1 , a 1 ) p ( a 2 ∣ s 2 , θ ) p ( r 2 , s 3 ∣ s 2 , a 2 ) ⋯ = p ( s 1 ) ∏ t = 1 T p ( a t ∣ s t , θ ) p ( r t , s t + 1 ∣ s t , a t ) \begin{array}{l} P(\tau \mid \theta)= p\left(s_{1}\right) p\left(a_{1} \mid s_{1}, \theta\right) p\left(r_{1}, s_{2} \mid s_{1}, a_{1}\right) p\left(a_{2} \mid s_{2}, \theta\right) p\left(r_{2}, s_{3} \mid s_{2}, a_{2}\right) \cdots \end{array}\\=p\left(s_{1}\right) \prod_{t=1}^{T} p\left(a_{t} \mid s_{t}, \theta\right) p\left(r_{t}, s_{t+1} \mid s_{t}, a_{t}\right) P(τθ)=p(s1)p(a1s1,θ)p(r1,s2s1,a1)p(a2s2,θ)p(r2,s3s2,a2)=p(s1)t=1Tp(atst,θ)p(rt,st+1st,at)

其中强化学习之Policy Gradient及代码是实现_第2张图片
因为每进行一次Episode,每一个轨迹 τ \tau τ都有可能被采样。我们将在神经网络参数 θ θ θ固定条件下,选择轨迹 τ \tau τ的概率为(|)。所以有 R ˉ θ ⋅ = ∑ τ R ( τ ) P ( τ ∣ θ ) ≈ 1 N ∑ n = 1 N R ( τ n ) \bar{R}_{\theta}^{\cdot}=\sum_{\tau} R(\tau) P(\tau \mid \theta) \approx \frac{1}{N} \sum_{n=1}^{N} R\left(\tau^{n}\right) Rˉθ=τR(τ)P(τθ)N1n=1NR(τn)穷举所有的轨迹不可能的,这里的近似采用了蒙特卡洛采样的方法,我们使用策略 π θ ( s ) \pi_{\theta}(\mathrm{s}) πθ(s),完成 N N N次Episode,会得到 N N N个轨迹,即 { τ 1 , τ 2 , ⋯   , τ N } \left\{\tau^{1}, \tau^{2}, \cdots, \tau^{N}\right\} {τ1,τ2,,τN},将将这N次得到回报取平均,当N趋向于无穷时候,两者近似。


我们的目标是最大化期望回报 θ ∗ = arg ⁡ max ⁡ θ R ˉ θ \theta^{*}=\arg \max _{\theta} \bar{R}_{\theta} θ=argθmaxRˉθ我们可以采用梯度上升的方法进行求解,强化学习之Policy Gradient及代码是实现_第3张图片(1) 已知条件 R ˉ θ ⋅ = ∑ τ R ( τ ) P ( τ ∣ θ ) \bar{R}_{\theta}^{\cdot}=\sum_{\tau} R(\tau) P(\tau \mid \theta) Rˉθ=τR(τ)P(τθ),我们求 ∇ R ˉ θ \nabla \bar{R}_{\theta} Rˉθ?
∇ R ˉ θ = ∑ τ R ( τ ) ∇ P ( τ ∣ θ ) = ∑ τ R ( τ ) P ( τ ∣ θ ) ∇ P ( τ ∣ θ ) P ( τ ∣ θ ) = ∑ τ R ( τ ) P ( τ ∣ θ ) ∇ log ⁡ P ( τ ∣ θ ) ≈ 1 N ∑ n = 1 N R ( τ n ) ∇ log ⁡ P ( τ n ∣ θ ) \nabla \bar{R}_{\theta}=\sum_{\tau} R(\tau) \nabla P(\tau \mid \theta)=\sum_{\tau} R(\tau) P(\tau \mid \theta) \frac{\nabla P(\tau \mid \theta)}{P(\tau \mid \theta)} \\\\ \begin{array}{l} =\sum_{\tau} R(\tau) P(\tau \mid \theta) \nabla \log P(\tau \mid \theta) \\\\ \approx \frac{1}{N} \sum_{n=1}^{N} R\left(\tau^{n}\right) \nabla \log P\left(\tau^{n} \mid \theta\right) \end{array} Rˉθ=τR(τ)P(τθ)=τR(τ)P(τθ)P(τθ)P(τθ)=τR(τ)P(τθ)logP(τθ)N1n=1NR(τn)logP(τnθ)注意 dlog ⁡ ( f ( x ) ) d x = 1 f ( x ) d f ( x ) d x \frac{\operatorname{dlog}(f(x))}{d x}=\frac{1}{f(x)} \frac{d f(x)}{d x} dxdlog(f(x))=f(x)1dxdf(x)
(2)接下来我们对 ∇ log ⁡ P ( τ ∣ θ ) \nabla \log P\left(\tau \mid \theta\right) logP(τθ)进行推导式求解?

有上文 P ( τ ∣ θ ) = p ( s 1 ) ∏ t = 1 T p ( a t ∣ s t , θ ) p ( r t , s t + 1 ∣ s t , a t ) P\left(\tau \mid \theta\right) = p\left(s_{1}\right) \prod_{t=1}^{T} p\left(a_{t} \mid s_{t}, \theta\right) p\left(r_{t}, s_{t+1} \mid s_{t}, a_{t}\right) P(τθ)=p(s1)t=1Tp(atst,θ)p(rt,st+1st,at),所以有
log ⁡ P ( τ ∣ θ ) = log ⁡ p ( s 1 ) + ∑ t = 1 T log ⁡ p ( a t ∣ s t , θ ) + log ⁡ p ( r t , s t + 1 ∣ s t , a t ) ∇ log ⁡ P ( τ ∣ θ ) = ∑ t = 1 T ∇ log ⁡ p ( a t ∣ s t , θ ) \begin{array}{l} \log P(\tau \mid \theta) =\log p\left(s_{1}\right)+\sum_{t=1}^{T} \log p\left(a_{t} \mid s_{t}, \theta\right)+\log p\left(r_{t}, s_{t+1} \mid s_{t}, a_{t}\right) \\\\ \nabla \log P(\tau \mid \theta)=\sum_{t=1}^{T} \nabla \log p\left(a_{t} \mid s_{t}, \theta\right) \end{array} logP(τθ)=logp(s1)+t=1Tlogp(atst,θ)+logp(rt,st+1st,at)logP(τθ)=t=1Tlogp(atst,θ)
(3)我们对参数 θ θ θ进行更新,由 θ new ← θ old + η ∇ R ˉ θ old \theta^{\text {new}} \leftarrow \theta^{\text {old}}+\eta \nabla \bar{R}_{\theta^{\text {old}}} θnewθold+ηRˉθold
强化学习之Policy Gradient及代码是实现_第4张图片
更新过程如下图所示:
强化学习之Policy Gradient及代码是实现_第5张图片
总结下来就是:增加带来正激励的概率;减少带来负激励的概率,并且我们考虑的是整个轨迹的回报。

Reinforce算法

因为 R θ = ∑ t = 1 T r t R_{\theta}=\sum_{t=1}^{T} r_{t} Rθ=t=1Trt,所以 ∇ R ˉ θ = 1 N ∑ n = 1 N ∑ t = 1 T n ∇ log ⁡ p ( a t n ∣ s t n , θ ) ( ∑ t = 1 T r t ) \nabla \bar{R}_{\theta}=\frac{1}{N} \sum_{n=1}^{N} \sum_{t=1}^{T_{n}} \nabla \log p\left(a_{t}^{n} \mid s_{t}^{n}, \theta\right)(\sum_{t=1}^{T} r_{t}) Rˉθ=N1n=1Nt=1Tnlogp(atnstn,θ)(t=1Trt)结合随机梯度上升算法,我们可以每次采集一条轨迹,计算每个时刻的梯度并更新参数,这称为REINFORCE算法[Williams, 1992],此时

强化学习之Policy Gradient及代码是实现_第6张图片

代码实现

import argparse
import gym
import numpy as np
from itertools import count

import torch
import torch.nn as nn
import torch.nn.functional as F
import torch.optim as optim
from torch.distributions import Categorical


parser = argparse.ArgumentParser(description='PyTorch REINFORCE example')
parser.add_argument('--gamma', type=float, default=0.99, metavar='G',
                    help='discount factor (default: 0.99)')
parser.add_argument('--seed', type=int, default=543, metavar='N',
                    help='random seed (default: 543)')
parser.add_argument('--render', action='store_true',
                    help='render the environment')
parser.add_argument('--log-interval', type=int, default=10, metavar='N',
                    help='interval between training status logs (default: 10)')
args = parser.parse_args()


env = gym.make('CartPole-v1')
env.seed(args.seed)
torch.manual_seed(args.seed)


class Policy(nn.Module):
    #[1,4]-->[1,128]-->[1,2]-->softmax
    def __init__(self):
        super(Policy, self).__init__()
        self.affine1 = nn.Linear(4, 128)
        self.dropout = nn.Dropout(p=0.6)
        self.affine2 = nn.Linear(128, 2)

        self.saved_log_probs = []
        self.rewards = []

    def forward(self, x):
        x = self.affine1(x)
        x = self.dropout(x)
        x = F.relu(x)
        action_scores = self.affine2(x)
        return F.softmax(action_scores, dim=1)


policy = Policy()
optimizer = optim.Adam(policy.parameters(), lr=1e-2)
eps = np.finfo(np.float32).eps.item()


def select_action(state):
    state = torch.from_numpy(state).float().unsqueeze(0)  #将observation转换成tensor
    probs = policy(state)
    m = Categorical(probs)
    action = m.sample()  #按照概率进行采样
    policy.saved_log_probs.append(m.log_prob(action))#m.log_prob(value)函数则是公式中的log部分
    return action.item()


def finish_episode():
    R = 0
    policy_loss = []
    returns = []
    #计算的每个动作时刻收到的回报
    for r in policy.rewards[::-1]:
        R = r + args.gamma * R
        returns.insert(0, R)
    returns = torch.tensor(returns)
    returns = (returns - returns.mean()) / (returns.std() + eps) #标准化
    for log_prob, R in zip(policy.saved_log_probs, returns):
        policy_loss.append(-log_prob * R)
    optimizer.zero_grad()
    policy_loss = torch.cat(policy_loss).sum()
    policy_loss.backward()
    optimizer.step()
    del policy.rewards[:]
    del policy.saved_log_probs[:]


def main():
    running_reward = 10
    for i_episode in count(1):
        state, ep_reward = env.reset(), 0
        for t in range(1, 10000):  # Don't infinite loop while learning
            action = select_action(state)
            state, reward, done, _ = env.step(action)
            if args.render:
                env.render()
            policy.rewards.append(reward) #记录每一步的奖励
            ep_reward += reward  #回报相加
            if done:  #如果为True,这个eposid结束
                break
        #平均reward奖励
        running_reward = 0.05 * ep_reward + (1 - 0.05) * running_reward
        finish_episode()
        if i_episode % args.log_interval == 0:
            print('Episode {}\tLast reward: {:.2f}\tAverage reward: {:.2f}'.format(
                  i_episode, ep_reward, running_reward))
        if running_reward > env.spec.reward_threshold:
            print("Solved! Running reward is now {} and "
                  "the last episode runs to {} time steps!".format(running_reward, t))
            break


if __name__ == '__main__':
    main()
    
$ python temp.py

Episode 10      Last reward: 26.00      Average reward: 16.00
Episode 20      Last reward: 16.00      Average reward: 14.85
Episode 30      Last reward: 49.00      Average reward: 20.77
Episode 40      Last reward: 45.00      Average reward: 27.37
Episode 50      Last reward: 44.00      Average reward: 30.80
Episode 60      Last reward: 111.00     Average reward: 42.69
Episode 70      Last reward: 141.00     Average reward: 70.65
Episode 80      Last reward: 138.00     Average reward: 100.27
Episode 90      Last reward: 30.00      Average reward: 86.27
Episode 100     Last reward: 114.00     Average reward: 108.18
Episode 110     Last reward: 175.00     Average reward: 156.48
Episode 120     Last reward: 141.00     Average reward: 143.86
Episode 130     Last reward: 101.00     Average reward: 132.91
Episode 140     Last reward: 19.00      Average reward: 89.99
Episode 150     Last reward: 30.00      Average reward: 63.52
Episode 160     Last reward: 27.00      Average reward: 49.30
Episode 170     Last reward: 79.00      Average reward: 47.18
Episode 180     Last reward: 77.00      Average reward: 51.38
Episode 190     Last reward: 80.00      Average reward: 63.45
Episode 200     Last reward: 55.00      Average reward: 62.21
Episode 210     Last reward: 38.00      Average reward: 57.16
Episode 220     Last reward: 50.00      Average reward: 54.56
Episode 230     Last reward: 31.00      Average reward: 53.13
Episode 240     Last reward: 115.00     Average reward: 67.47
Episode 250     Last reward: 204.00     Average reward: 137.96
Episode 260     Last reward: 195.00     Average reward: 191.86
Episode 270     Last reward: 214.00     Average reward: 218.05
Episode 280     Last reward: 500.00     Average reward: 294.96
Episode 290     Last reward: 358.00     Average reward: 350.75
Episode 300     Last reward: 327.00     Average reward: 319.03
Episode 310     Last reward: 93.00      Average reward: 260.24
Episode 320     Last reward: 500.00     Average reward: 258.35
Episode 330     Last reward: 83.00      Average reward: 290.17
Episode 340     Last reward: 201.00     Average reward: 259.37
Episode 350     Last reward: 482.00     Average reward: 270.48
Episode 360     Last reward: 500.00     Average reward: 337.46
Episode 370     Last reward: 91.00      Average reward: 375.76
Episode 380     Last reward: 500.00     Average reward: 402.21
Episode 390     Last reward: 260.00     Average reward: 377.44
Episode 400     Last reward: 242.00     Average reward: 327.93
Episode 410     Last reward: 500.00     Average reward: 318.99
Episode 420     Last reward: 390.00     Average reward: 339.36
Episode 430     Last reward: 476.00     Average reward: 364.77
Episode 440     Last reward: 500.00     Average reward: 368.53
Episode 450     Last reward: 82.00      Average reward: 370.17
Episode 460     Last reward: 500.00     Average reward: 404.95
Episode 470     Last reward: 500.00     Average reward: 443.09
Episode 480     Last reward: 500.00     Average reward: 465.92
Solved! Running reward is now 476.20365101578085 and the last episode runs to 500 time steps!

参考

李宏毅强化学习
邱锡鹏《神经网络与深度学习》

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