网上找到的一个实现:
地址:
https://gist.github.com/HenryJia/23db12d61546054aa43f8dc587d9dc2c
稍微修改后的代码:
import numpy as np import gym def sigmoid(x): return 1.0 / (1.0 + np.exp(-x)) env = gym.make('CartPole-v1') desired_state = np.array([0, 0, 0, 0]) desired_mask = np.array([0, 0, 1, 0]) P, I, D = 0.1, 0.01, 0.5 ### N_episodes = 10 N_steps = 50000 for i_episode in range(N_episodes): state, _ = env.reset() integral = 0 derivative = 0 prev_error = 0 for t in range(N_steps): # print(f"step: {t}") env.render() error = state - desired_state integral += error derivative = error - prev_error prev_error = error pid = np.dot(P * error + I * integral + D * derivative, desired_mask) action = sigmoid(pid) action = np.round(action).astype(np.int32) # print(P * error + I * integral + D * derivative, pid, action) # print(state, action, ) state, reward, done, info, _ = env.step(action) if done or t==N_steps-1: print("Episode finished after {} timesteps".format(t+1)) break env.close()
运行效果:

这个表现是极为神奇的,如果不考虑泛化性的话,不考虑使用AI算法和机器学习算法的话,那么不使用强化学习和遗传算法以外的算法,那么使用自动化的算法或许也是不错的选择,并且从这个表现来看这个效果远比使用AI类的算法表现好。
上面的这个代码只考虑小车平衡杆的角度与0的偏差,就可以获得如此高的表现。
根据原地址的讨论:
https://gist.github.com/HenryJia/23db12d61546054aa43f8dc587d9dc2c

我们可以知道,如果通过调整PID算法的系数,那么可以获得更为优秀的性能表现,为此我们修改代码如下:
点击查看代码
import numpy as np import gym def sigmoid(x): return 1.0 / (1.0 + np.exp(-x)) env = gym.make('CartPole-v1') desired_state = np.array([0, 0, 0, 0]) # desired_mask = np.array([0, 0, 1, 0]) desired_mask = np.array([1, 1, 1, 1]) # P, I, D = 0.1, 0.01, 0.5 ### P, I, D = [1/150, 1/950, 0.1, 0.01], [0.0005, 0.001, 0.01, 0.0001], [0.2, 0.0001, 0.5, 0.005] N_episodes = 10 N_steps = 1000000 for i_episode in range(N_episodes): state, _ = env.reset() integral = 0 derivative = 0 prev_error = 0 for t in range(N_steps): # print(f"step: {t}") env.render() error = state - desired_state integral += error derivative = error - prev_error prev_error = error pid = np.dot(P * error + I * integral + D * derivative, desired_mask) action = sigmoid(pid) action = np.round(action).astype(np.int32) # print(P * error + I * integral + D * derivative, pid, action) # print(state, action, ) state, reward, done, info, _ = env.step(action) if done or t==N_steps-1: print("Episode finished after {} timesteps".format(t+1)) break env.close()
性能表现:

根据这个PID的系数来运行gym下的cartpole游戏,可以认为这个游泳永远不会终止,因为这里我们已经将运行长度设置为100万步。
PS:
需要注意的是PID算法的这个P,I,D系数才是影响算法的关键,而如何获得这个系数也是一个极为难的问题,很多时候是需要使用试错的方法来进行的,可以说有的P,I,D系数可以运行几十步,有的可以运行几百步或几千步,而下面的系数却可以运行上百万步,甚至是永远一直运行,可以说这种PID系数的求解才是真正的关键。
P, I, D = [1/150, 1/950, 0.1, 0.01], [0.0005, 0.001, 0.01, 0.0001], [0.2, 0.0001, 0.5, 0.005]
附:
另一个实现:
https://ethanr2000.medium.com/using-pid-to-cheat-an-openai-challenge-f17745226449
代码实现:
import gym from matplotlib import pyplot as plt env = gym.make("CartPole-v1") observation, _ = env.reset() Kp = 135 Ki = 96.5 Kd = 47.5 force = 0 integral = 0 for step in range(10000000): print("step: ", step) env.render() observation, reward, done, info, _ = env.step(force) velocity = observation[1] angle = observation[2] angular_velocity = observation[3] integral = integral + angle F = Kp*(angle) + Kd*(angular_velocity) + Ki*(integral) force = 1 if F > 0 else 0 if done: observation, _ = env.reset() integral = 0 env.close()
运行结果:
