rewards = np.zeros() 什么意思

这行代码的意思是创建一个由0组成的数组，其中数组的大小没有指定。在Python中，通常会使用NumPy库来创建和处理数组。np.zeros()是NumPy中的一个函数，用于创建一个由0组成的数组。示例代码如下：

import numpy as np

# 创建一个大小为5的由0组成的数组
rewards = np.zeros(5)
print(rewards)

输出结果为：

array([0., 0., 0., 0., 0.])

这意味着我们创建了一个有5个元素的数组，每个元素都是0。如果您想要创建不同大小的数组，只需在np.zeros()中传入相应的大小即可。例如：np.zeros(10)将创建一个由10个0组成的数组。

相关问题

implement the UCB algorithm and plot the expected regret as a function of 𝑇 using 1000 sample path simulations with python, total_try = 2000, data = np.linspace(0, 19, 20), reward = data * [10-0.5*data + 𝜖], 𝜖~𝑁(0, 0.04)

Sure, here's a Python implementation of the UCB algorithm with the given parameters:

import numpy as np
import matplotlib.pyplot as plt

# Define the UCB algorithm
def ucb(total_try, data, c):
    n_arms = len(data)
    Q = np.zeros(n_arms)
    N = np.zeros(n_arms)
    regret = []
    for t in range(total_try):
        a = np.argmax(Q + c * np.sqrt(np.log(t+1) / (N+1e-6)))
        r = reward(data[a])
        regret.append(max(data) - data[a])
        N[a] += 1
        Q[a] += (r - Q[a]) / N[a]
    return regret

# Define the reward function
def reward(x):
    return x * (10 - 0.5 * x + np.random.normal(0, 0.04))

# Set the parameters
total_try = 2000
data = np.linspace(0, 19, 20)
c = 2

# Run the simulation 1000 times and plot the results
regret_all = np.zeros((1000, total_try))
for i in range(1000):
    regret_all[i] = ucb(total_try, data, c)
regret_mean = np.mean(regret_all, axis=0)
plt.plot(regret_mean)
plt.xlabel('T')
plt.ylabel('Expected Regret')
plt.show()

This code will simulate 1000 sample paths of the UCB algorithm with the given parameters, and plot the expected regret as a function of T (the number of tries). The plot should show that the regret decreases as T increases, which is expected since the algorithm learns more about the rewards as it tries more arms. The regret should converge to zero as T goes to infinity since the rewards are less stochastic than in the previous example.

implement the UCB algorithm and plot the expected regret as a function of 𝑇 using 1000 sample path simulations with python, total_try = 2000, data = np.linspace(0, 19, 20), reward = data * [10-0.5*data + 𝜖], 𝜖~𝑁(0, 0.22)

Sure, I can help you with that. Here's a Python implementation of the UCB algorithm with the given parameters:

import numpy as np
import matplotlib.pyplot as plt

# Define the UCB algorithm
def ucb(total_try, data, c):
    n_arms = len(data)
    Q = np.zeros(n_arms)
    N = np.zeros(n_arms)
    regret = []
    for t in range(total_try):
        a = np.argmax(Q + c * np.sqrt(np.log(t+1) / (N+1e-6)))
        r = reward(data[a])
        regret.append(max(data) - data[a])
        N[a] += 1
        Q[a] += (r - Q[a]) / N[a]
    return regret

# Define the reward function
def reward(x):
    return x * (10 - 0.5 * x + np.random.normal(0, 0.22))

# Set the parameters
total_try = 2000
data = np.linspace(0, 19, 20)
c = 2

# Run the simulation 1000 times and plot the results
regret_all = np.zeros((1000, total_try))
for i in range(1000):
    regret_all[i] = ucb(total_try, data, c)
regret_mean = np.mean(regret_all, axis=0)
plt.plot(regret_mean)
plt.xlabel('T')
plt.ylabel('Expected Regret')
plt.show()

This code will simulate 1000 sample paths of the UCB algorithm with the given parameters, and plot the expected regret as a function of T (the number of tries). The plot should show that the regret decreases as T increases, which is expected since the algorithm learns more about the rewards as it tries more arms. However, the regret may not converge to zero since the rewards are stochastic.