概述
一、UCB和汤普森算法比较¶
二、导入标准库
In [1]:
# Importing the libraries 导入库 import numpy as np import matplotlib.pyplot as plt import pandas as pd # 使图像能够调整 %matplotlib notebook #中文字体显示 plt.rc('font', family='SimHei', size=8)
三、导入数据
In [2]:
dataset = pd.read_csv('Ads_CTR_Optimisation.csv') # 数据表示虚拟环境,模拟我将投放哪些广告 dataset
Out[2]:
Ad 1 | Ad 2 | Ad 3 | Ad 4 | Ad 5 | Ad 6 | Ad 7 | Ad 8 | Ad 9 | Ad 10 | |
---|---|---|---|---|---|---|---|---|---|---|
0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
2 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
3 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
4 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
5 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
6 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
7 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
8 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
9 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
11 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
12 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
13 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
14 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
15 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
16 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
17 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
18 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
19 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
20 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
21 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
22 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
23 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 |
24 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
25 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
26 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
27 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
28 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
29 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 |
... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... |
9970 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
9971 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
9972 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
9973 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
9974 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 |
9975 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |
9976 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
9977 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
9978 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
9979 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
9980 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
9981 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
9982 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
9983 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 |
9984 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
9985 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
9986 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
9987 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
9988 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
9989 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
9990 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
9991 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
9992 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
9993 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
9994 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
9995 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
9996 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
9997 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
9998 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
9999 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
10000 rows × 10 columns
四、汤普森算法在多臂老虎机中的实现过程
五、实现
In [8]:
import random N = 10000 # 1000个用户 d = 10 # 10个广告 ads_selected = [] # 广告选择 numbers_of_rewards_1 = [0] * d # 广告i奖励1的总和 numbers_of_rewards_0 = [0] * d # 广告i奖励0的总和 total_reward = 0 # 总奖励 for n in range(0, N): # 每个用户循环 ad = 0 max_random = 0 for i in range(0, d): # 第i个广告 random_beta = random.betavariate(numbers_of_rewards_1[i] + 1, numbers_of_rewards_0[i] + 1) # beta分布 if random_beta > max_random: # 寻找最大随机量,最大随机量对应的广告为我们选择投放的广告 max_random = random_beta ad = i ads_selected.append(ad) reward = dataset.values[n, ad] # 获得奖励 if reward == 1: numbers_of_rewards_1[ad] = numbers_of_rewards_1[ad] + 1 else: numbers_of_rewards_0[ad] = numbers_of_rewards_0[ad] + 1 total_reward = total_reward + reward print(total_reward)
2625
六、画图展示
In [5]:
# 画图 plt.hist(ads_selected) plt.title(u'广告选择直方图') plt.xlabel(u'广告') plt.ylabel(u'每个广告的点击数') plt.show()
看来还是汤普森更加NB。随机生成的总奖励1282(见上一个专题),置信区间上界算法总奖励2358(见上一个专题),汤普森的总奖励2625
七、项目地址
https://coding.net/u/RuoYun/p/Python-of-machine-learning/git/tree/master/00%E6%9C%BA%E5%99%A8%E5%AD%A6%E4%B9%A0/5.%E8%BF%9B%E9%98%B6%E7%AE%97%E6%B3%95/1.%E5%BC%BA%E5%8C%96%E5%AD%A6%E4%B9%A0/2.%E6%B1%A4%E6%99%AE%E6%A3%AE%E7%AE%97%E6%B3%95?public=true
最后
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