{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# 2.2 决策树"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"决策树是一种通过**树形结构**进行分类的方法，使用层层推理来实现最终的分类。\n",
"\n",
"决策树由下面几种元素构成：\n",
"+ 根节点：最顶层的分类条件。\n",
"+ 决策节点（中间节点）：中间分类条件。\n",
"+ 叶子节点：代表标签类别。\n",
"\n",
"<img src=\"http://imgbed.momodel.cn//20200110170450.png\" width=400>\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"上面的说法过于抽象，下面来看一个实际的例子。构建一棵结构简单的决策树，用于预测贷款用户是否具有偿还贷款的能力。\n",
"\n",
"贷款用户主要具备三个属性：**是否拥有房产**，**是否结婚**，**平均月收入**。\n",
"\n",
"每一个内部节点都表示一个属性条件判断，叶子节点表示贷款用户是否具有偿还能力。\n",
"\n",
"<img src=\"http://imgbed.momodel.cn//20200110171836.png\" width=400>\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"首先判断贷款用户是否拥有房产，如果用户拥有房产，则说明该用户具有偿还贷款的能力；否则需要判断该用户是否结婚，如果已经结婚则具有偿还贷款的能力；否则需要判断该用户的收入大小，如果该用户月收入小于 4K 元，则该用户不具有偿还贷款的能力，否则该用户是具有偿还能力的。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**想一想**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"决策树的流程是什么？\n",
"\n",
"在有一个贷款用户A，其情况是月收入 3K、已经结婚、没有房产，那么他是否具有偿还贷款的能力呢？ \n",
"\n",
"上图中我们为什么要用“是否拥有房产”作根节点呢？可不可以用“是否结婚”和“平均月收入”做根节点呢？"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 2.2.1 决策树分类概念"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<center><video src=\"http://files.momodel.cn/decision_tree_playground_demo.mp4\" controls=\"controls\" width=800px></center>"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**数据**: 游乐场经营者提供**天气情况**（如晴、雨、多云）、**温度高低**、**湿度大小**、**风力强弱**等气象特点以及游客当天是否前往游乐场。\n",
"\n",
"**目标**: 预测游客是否来游乐场游玩。\n",
"\n",
"\n",
"|序号|天气|温度（℃）|湿度|是否有风|是（1）否（0）前往游乐场|\n",
"|:--:|:--:|:--:|:--:|:--:|:--:|\n",
"|1|晴|29|85|否|0|\n",
"|2|晴|26|88|是|0|\n",
"|3|多云|28|78|否|1\n",
"|4|雨|21|96|否|1|\n",
"|5|雨|20|80|否|1|\n",
"|6|雨|18|70|是|0|\n",
"|7|多云|18|65|是|1|\n",
"|8|晴|22|90|否|0|\n",
"|9|晴|21|68|否|1|\n",
"|10|雨|24|80|否|1|\n",
"|11|晴|24|63|是|1|\n",
"|12|多云|22|90|是|1|\n",
"|13|多云|27|75|否|1|\n",
"|14|雨|21|80|是|0|\n",
"\n",
"根据上表，绘制如图所示的决策树：\n",
"\n",
"<img src=\"http://imgbed.momodel.cn//20200110172806.png\" width=500>"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"根节点是天气状况，具有 **雨**、 **多云** 和 **晴** 三种属性取值。\n",
"+ **多云**: 样本子集是 { 3, 7, 12, 13 } ，仅有“前往游乐场游玩”一个类别，即肯定去游乐场。 \n",
" \n",
" \n",
"+ **晴**: 样本子集是 { 1, 2, 8, 9, 11 }\n",
" + 湿度大于 75：样本子集为 { 1, 2, 8 }，不前往游乐场。\n",
" + 湿度不大于 75：样本子集 { 9, 11 }，前往游乐场。\n",
" \n",
" \n",
"+ **雨**：样本子集为 { 4, 5, 6, 10, 14 }\n",
" + 有风：样本子集 { 6, 14 },不去游乐场。\n",
" + 无风：样本子集 { 4, 5, 10 }，前往游乐场。\n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**想一想**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"通过上面的例子，你观察到构建决策树的过程是哪几步？"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"下面，我们创建数据并进行一些预处理"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"import numpy as np\n",
"import pandas as pd\n",
"import matplotlib.pyplot as plt\n",
"%matplotlib inline\n",
"\n",
"import math\n",
"from math import log\n",
"import warnings\n",
"warnings.filterwarnings(\"ignore\")\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# 原始数据\n",
"datasets = [\n",
" ['晴', 29, 85, '否', '0'],\n",
" ['晴', 26, 88, '是', '0'],\n",
" ['多云', 28, 78, '否', '1'],\n",
" ['雨', 21, 96, '否', '1'],\n",
" ['雨', 20, 80, '否', '1'],\n",
" ['雨', 18, 70, '是', '0'],\n",
" ['多云', 18, 65, '是', '1'],\n",
" ['晴', 22, 90, '否', '0'],\n",
" ['晴', 21, 68, '否', '1'],\n",
" ['雨', 24, 80, '否', '1'],\n",
" ['晴', 24, 63, '是', '1'],\n",
" ['多云', 22, 90, '是', '1'],\n",
" ['多云', 27, 75, '否', '1'],\n",
" ['雨', 21, 80, '是', '0']\n",
"]\n",
"# 数据的列名\n",
"labels = ['天气', '温度', '湿度', '是否有风', '是否前往游乐场']\n",
"# 将湿度大小分为大于 75 和小于等于 75 这两个属性值，\n",
"# 将温度大小分为大于 26 和小于等于 26 这两个属性值\n",
"for i in range(len(datasets)):\n",
" if datasets[i][2] > 75:\n",
" datasets[i][2] = '>75'\n",
" else:\n",
" datasets[i][2] = '<=75'\n",
" if datasets[i][1] > 26:\n",
" datasets[i][1] = '>26'\n",
" else:\n",
" datasets[i][1] = '<=26'\n",
"# 构建 dataframe 并查看数据\n",
"df = pd.DataFrame(datasets, columns=labels)\n",
"df"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 2.2.2 构建决策树 "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<center><video src=\"http://files.momodel.cn/decision_tree_entropy.mp4\" controls=\"controls\" width=800px></center>"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**想一想**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"信息熵和信息增益分别是什么？"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"假设有 $K$ 个信息，其组成了集合样本 $D$ ，记第 $k$ 个信息发生的概率为 $p_k(1≤k≤K)$。 \n",
"\n",
"这 $K$ 个信息的信息熵该如何计算？ \n",
"\n",
"所有 $p_k$ 累加起来的和是多少？"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**动手练**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"根据公式编写代码计算信息熵"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"def calc_entropy(total_num, count_dict):\n",
" \"\"\"\n",
" 计算信息熵\n",
" :param total_num: 总样本数\n",
" :param count_dict: 每类样本及其对应数目的字典\n",
" :return: 信息熵\n",
" \"\"\"\n",
" # todo 使用公式计算信息熵\n",
" ent = \n",
" # 返回信息熵，精确到小数点后 4 位\n",
" return round(ent, 4)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n",
"\n",
"现在用**信息熵**来构建决策树。数据中 14 个样本分为 “游客来游乐场 （9 个样本）” 和 “游客不来游乐场（ 5 个样本）” 两个类别，即 K = 2。\n",
"\n",
"记 “游客来游乐场” 和 “游客不来游乐场” 的概率分别为 $p_1$ 和 $p_2$ ，显然 $p_1=\\frac{9}{14}$，$p_1=\\frac{5}{14}$，则这 14 个样本所蕴含的信息熵：\n",
"\n",
"$$E(D)=-\\sum_{k=1}^{2}p_{k}log_{2}{p_k}=-(\\frac{9}{14}×log_{2}{\\frac{9}{14}}+\\frac{5}{14}×log_{2}{\\frac{5}{14}})=0.940$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"我们可以用下面这种方式对 DataFrame 的数据按条件进行筛选。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# 例如：按 是否前往游乐场 == 0 进行筛选\n",
"df[df['是否前往游乐场']=='0']\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"使用上面的方法，可以得到计算信息熵所需的总样本数，以及每类样本及其对应数目的字典，然后计算信息熵。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# 总样本数\n",
"total_num = df.shape[0]\n",
"# 每类样本及其对应数目的字典\n",
"count_dict = {'前往': df[df['是否前往游乐场']=='1'].shape[0], '不前往': df[df['是否前往游乐场']=='1'].shape[1]}\n",
"# 计算信息熵\n",
"entropy = calc_entropy(total_num, count_dict)\n",
"entropy\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<center><video src=\"http://files.momodel.cn/decision_tree_build.mp4\" controls=\"controls\" width=800px></center>"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**计算天气状况所对应的信息熵**： \n",
"天气状况的三个属性记为 $a_0=“晴”$ ，$a_1=“多云”$ ，$a_2=“雨”$ ， \n",
"属性取值为 $a_i$ 对应分支节点所包含子样本集记为 $D_i$ ，该子样本集包含样本数量记为 $|D_i|$ 。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"|天气属性取值$a_i$|“晴”|“多云”|“雨”|\n",
"|:--:|:--:|:--:|:--:|\n",
"|对应样本数$|D_i|$|5|4|5|\n",
"|正负样本数量|（2+，3-）|（4+，0-）|（3+，2-）|\n",
"\n",
"计算天气状况每个属性值的信息熵：\n",
"\n",
"$“晴”：E(D_0)=-(\\frac{2}{5}×log_{2}{\\frac{2}{5}}+\\frac{3}{5}×log_{2}{\\frac{3}{5}})=0.971$\n",
"\n",
"$“多云”：E(D_1)=-(\\frac{4}{4}×log_{2}{\\frac{4}{4}})=0$\n",
"\n",
"$“雨”：E(D_2)=-(\\frac{3}{5}×log_{2}{\\frac{3}{5}}+\\frac{2}{5}×log_{2}{\\frac{2}{5}})=0.971$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"现在，我们来编写代码进行完成上面的计算。\n",
"\n",
"首先，我们可以使用下面的写法，对 Dataframe 进行多个条件的筛选。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# 筛选出 天气为晴并且去游乐场的样本数据\n",
"df[(df['天气']=='晴') & (df['是否前往游乐场']=='1')]\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# 天气为晴的总天数\n",
"total_num_sun = df[df['天气']=='晴'].shape[0]\n",
"\n",
"# 天气为晴时，去游乐场和不去游乐场的人数\n",
"count_dict_sun = {'前往':df[(df['天气']=='晴') & (df['是否前往游乐场']=='1')].shape[0],\n",
" '不前往':df[(df['天气']=='晴') & (df['是否前往游乐场']=='0')].shape[0]}\n",
"print(count_dict_sun)\n",
"\n",
"# 计算天气-晴 的信息熵\n",
"ent_sun = calc_entropy(total_num_sun, count_dict_sun)\n",
"print('天气-晴 的信息熵为：%s' % ent_sun)\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# 天气为多云的总天数\n",
"total_num_cloud = df[df['天气']=='多云'].shape[0]\n",
"\n",
"# 天气为多云时，去游乐场和不去游乐场的人数\n",
"count_dict_cloud = {'前往':df[(df['天气']=='多云') & (df['是否前往游乐场']=='1')].shape[0],\n",
" '不前往':df[(df['天气']=='多云') & (df['是否前往游乐场']=='0')].shape[0]}\n",
"print(count_dict_cloud)\n",
"\n",
"# 计算天气-多云 的信息熵\n",
"ent_cloud = calc_entropy(total_num_cloud, count_dict_cloud)\n",
"print('天气-多云 的信息熵为：%s' % ent_cloud)\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# 天气为雨的总天数\n",
"total_num_rain = df[df['天气']=='雨'].shape[0]\n",
"\n",
"# 天气为雨时，去游乐场和不去游乐场的人数\n",
"count_dict_rain = {'前往':df[(df['天气']=='雨') & (df['是否前往游乐场']=='1')].shape[0],\n",
" '不前往':df[(df['天气']=='雨') & (df['是否前往游乐场']=='0')].shape[0]}\n",
"print(count_dict_rain)\n",
"\n",
"# 计算天气-雨 的信息熵\n",
"ent_rain = calc_entropy(total_num_rain, count_dict_rain)\n",
"print('天气-雨 的信息熵为：%s' % ent_rain)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"计算天气状况的信息增益: \n",
"$$Gain(D,A)=E(D)-\\sum_{i}^{n}\\frac{|D_i|}{D}E(D)$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"其中，$A=“天气状况”$。于是天气状况这一气象特点的信息增益为：\n",
"$$Gain(D,天气)=0.940-（\\frac{5}{14}×0.971+\\frac{4}{14}×0+\\frac{5}{14}×0.971）=0.246$$\n",
"\n",
"同理可以计算温度高低、湿度大小、风力强弱三个气象特点的信息增益。 \n",
"通常情况下，某个分支的信息增益越大，则该分支对样本集划分所获得的“纯度”越大，信息不确定性减少的程度越大。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**动手练**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"使用上面的公式计算信息增益。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# todo 计算按天气状况分割的信息增益\n",
"gain = \n",
"gain\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 扩展内容\n",
"\n",
"**基尼指数**\n",
"\n",
"除了使用信息增益以外，我们也可以使用基尼指数来构建决策树。\n",
"\n",
"分类问题中，假设有 $K$ 个类，样本点属于第 $k$ 类的概率为 $p_{k}$，则概率分布的基尼指数定义为：\n",
"\n",
"$$\\operatorname{Gini}(p)=\\sum_{k=1}^{K} p_{k}\\left(1-p_{k}\\right)=1-\\sum_{k=1}^{K} p_{k}^{2}$$\n",
"\n",
"\n",
"对于给定的样本集合 $D$，其基尼指数为\n",
"\n",
"$$\\operatorname{Gini}(D)=1-\\sum_{k=1}^{K}\\left(\\frac{\\left|C_{k}\\right|}{|D|}\\right)^{2}$$\n",
"\n",
"这里，$C_{k}$ 是 $D$ 中属于第 $k$ 类的样本子集，$K$ 是类的个数。\n",
"\n",
"如果样本集合 $D$ 根据特征 $A$ 是否取某一可能值 $a$ 被分割为 $D_{1}$ 和 $D_{2}$ 两部分，即\n",
"\n",
"$$D_{1}=\\{(x, y) \\in D | A(x)=a\\}, \\quad D_{2}=D-D_{1}$$\n",
"\n",
"则在特征 $A$ 的条件下，集合 $D$ 的基尼指数定义为\n",
"\n",
"$$\\operatorname{Gini}(D, A)=\\frac{\\left|D_{1}\\right|}{|D|}\n",
"\\operatorname{Gini}\\left(D_{1}\\right)+\\frac{\\left|D_{2}\\right|}{|D|} \\operatorname{Gini}\\left(D_{2}\\right)$$\n",
"\n",
"基尼指数 $Gini(D)$ 表示集合 $D$ 的不确定性，基尼指数 $Gini(D, A)$ 表示经过分割后集合 $D$ 的不确定性。基尼指数值越大，样本集合的不确定性也就越大，这一点与信息熵相似。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**想一想**："
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"对于二分类问题，若样本点属于第 1 个类的概率是 $p$，则概率分布的基尼指数是多少？"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 思考与练习 "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"1. 分别将天气状况、温度高低、湿度大小、风力强弱作为分支点来构建决策树，查看信息增益。\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# todo 以温度 26 为温度高低的分界线，计算信息增益\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# todo 以湿度 75 为湿度大小的分界线，计算信息增益\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# todo 以有风无风划分，计算信息增益\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"2. 每朵鸢尾花有萼片长度、萼片宽度、花瓣长度、花瓣宽度四个特征。现在需要根据这四个特征将鸢尾花分为杂色鸢尾花、维吉尼亚鸢尾和山鸢尾三类，试构造决策树进行分类。\n",
"\n",
"|序号|萼片长度|萼片宽度|花瓣长度|花瓣宽度|种类|\n",
"|:--:|:--:|:--:|:--:|:--:|:--:|\n",
"|1|5.0|2.0|3.5|1.0|杂色鸢尾|\n",
"|2|6.0|2.2|5.0|1.5|维吉尼亚鸢尾|\n",
"|3|6.0|2.2|4.0|1.0|杂色鸢尾|\n",
"|4|6.2|2.2|4.5|1.5|杂色鸢尾|\n",
"|5|4.5|2.3|1.3|0.3|山鸢尾|"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"观察上表中的五笔数据，我们可以看到 杂色鸢尾 和 维吉尼亚鸢尾 的花瓣宽度明显大于山鸢尾，所以可以通过判断花瓣宽度是否大于 0.7，来将山鸢尾区从其他两种鸢尾中区分出来。\n",
"\n",
"同时，杂色鸢尾 和 维吉尼亚鸢尾 的花瓣长度明显大于山鸢尾，所以也可以通过判断花瓣长度是否大于 2.4，来将山鸢尾区从其他两种鸢尾中区分出来。\n",
"\n",
"然后我们观察到 维吉尼亚鸢尾的花瓣长度明显大于杂色鸢尾，所以可以通过判断花瓣长度是否大于 4.75，来将杂色鸢尾和维吉尼亚鸢尾区分出来。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"实际上是否如此呢？你能否想到其他的切分方式？"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"上面的表格只是 Iris 数据集的一小部分，完整的数据集包含 150 个数据样本，分为 3 类，每类 50 个数据，每个数据包含 4 个属性。即花萼长度，花萼宽度，花瓣长度，花瓣宽度4个属性。\n",
"\n",
"我们使用 sklearn 工具包来构建决策树模型，先导入数据集。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"from sklearn.datasets import load_iris\n",
"# 加载数据集\n",
"iris = load_iris()\n",
"# 查看 label\n",
"print(list(iris.target_names))\n",
"# 查看 feature\n",
"print(iris.feature_names)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"setosa 是山鸢尾，versicolor是杂色鸢尾，virginica是维吉尼亚鸢尾。\n",
"\n",
"sepal length， sepal width，petal length，petal width 分别是萼片长度，萼片宽度，花瓣长度，花瓣宽度。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"然后进行训练集和测试集的切分。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"from sklearn.model_selection import train_test_split\n",
"# 载入数据\n",
"X, y = load_iris(return_X_y=True)\n",
"# 切分训练集合测试集\n",
"X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.33, random_state=42)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"接下来，我们在训练集数据上训练决策树模型。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"from sklearn import tree\n",
"from sklearn.tree import DecisionTreeClassifier\n",
"# 初始化模型，可以调整 max_depth 来观察模型的表现\n",
"# 也可以调整 criterion 为 gini 来使用 gini 指数构建决策树\n",
"clf = tree.DecisionTreeClassifier(criterion='entropy', max_depth=2)\n",
"# 训练模型\n",
"clf = clf.fit(X_train, y_train)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"我们可以使用 graphviz 包来展示构建好的决策树。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"import graphviz\n",
"feature_names = ['萼片长度','萼片宽度','花瓣长度','花瓣宽度']\n",
"target_names = ['山鸢尾', '杂色鸢尾', '维吉尼亚鸢尾']\n",
"# 可视化生成的决策树\n",
"dot_data = tree.export_graphviz(clf, out_file=None,\n",
" feature_names=feature_names,\n",
" class_names=target_names,\n",
" filled=True, rounded=True,\n",
" special_characters=True)\n",
"graph = graphviz.Source(dot_data)\n",
"graph\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"我们看模型在测试集上的表现。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"from sklearn.metrics import accuracy_score\n",
"y_test_predict = clf.predict(X_test)\n",
"accuracy_score(y_test,y_test_predict)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 实践与体验\n",
"\n",
"**计算文章的信息熵**\n",
"\n",
"收集中英文对照的短文，在计算短文内中文单词和英文单词出现概率基础上，计算该两篇短文的信息熵，比较中文短文信息熵和英文短文信息熵的大小。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"首先定义一个方法来辅助读取文件的内容。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"def read_file(path):\n",
" \"\"\"\n",
" 读取某文件的内容\n",
" :param path: 文件的路径\n",
" :return: 文件的内容\n",
" \"\"\"\n",
" contents = \"\"\n",
" with open(path) as f:\n",
" # 读取每一行的内容\n",
" for line in f.readlines():\n",
" contents += line\n",
" return contents\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"使用上面定义的方法读取英文短文及其对应的中文短文。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# 读取英文短文\n",
"en_essay = read_file('essay3_en.txt')\n",
"# 读取中文短文\n",
"ch_essay = read_file('essay3_ch.txt')\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"处理文本，统计单词出现的概率，并计算信息熵。"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"from collections import Counter\n",
"import re\n",
"\n",
"\n",
"def cal_essay_entropy(essay, split_by=None):\n",
" \"\"\"\n",
" 计算文章的信息熵\n",
" :param essay: 文章内容\n",
" :param split_by: 切分方式，对于中文文章，不需传入，按字符切分，\n",
" 对于英文文章，需传入空格字符来进行切分\n",
" :return: 文章的信息熵\n",
" \"\"\"\n",
" # 把英文全部转为小写\n",
" essay = essay.lower()\n",
" # 去除标点符号\n",
" essay = re.sub(\n",
" \"[\\f+\\n+\\r+\\t+\\v+\\?\\.\\!\\/_,$%^*(+\\\"\\']+|[+——！，。？、~@#《》￥%……&*（）]\", \"\",\n",
" essay)\n",
" # print(essay)\n",
" # 把文本分割为词\n",
" if split_by:\n",
" word_list = essay.split(split_by)\n",
" else:\n",
" word_list = list(essay)\n",
" # 统计总的单词数\n",
" word_number = len(word_list)\n",
" print('此文章共有 %s 个单词' % word_number)\n",
" # 得到每个单词出现的次数\n",
" word_counter = Counter(word_list)\n",
" # print('每个单词出现的次数为：%s' % word_counter)\n",
" # 使用信息熵公式计算信息熵\n",
" ent = -sum([(p / word_number) * log(p / word_number, 2) for p in\n",
" word_counter.values()])\n",
" print('信息熵为：%.2f' % ent)\n",
" return ent\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"ent = cal_essay_entropy(ch_essay)\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"ent = cal_essay_entropy(en_essay, split_by = ' ')\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**问题 1**: 你在上面的试验中观察到了什么？请在下面写下你观察到的现象，并尝试分析其原因。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**答案 1**：（在此处填写你的答案。）"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 扩展阅读\n",
"\n",
"1. [决策树与随机森林](https://www.bilibili.com/video/av26086646?from=search&seid=6716049859412037731)\n",
"2. [从决策树到随机森林：树型算法的原理与实现](https://www.jiqizhixin.com/articles/2017-07-31-3)\n",
"3. [sklearn 决策树](https://scikit-learn.org/stable/modules/generated/sklearn.tree.DecisionTreeClassifier.html)"
]
}
],
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