{ "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", "\n", "\n", "预测时，在树的内部节点处用某一属性值进行判断，根据判断结果决定进入哪个分支节点，\n", "\n", "直到到达叶节点处，得到分类结果。\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "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。\n", "\n", "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", "\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "首先判断贷款用户是否拥有房产，\n", "如果用户拥有房产，则说明该用户具有偿还贷款的能力；\n", "否则需要判断该用户是否结婚，\n", "如果已经结婚则具有偿还贷款的能力；\n", "否则需要判断该用户的收入大小，\n", "如果该用户月收入小于 4K 元，\n", "则该用户不具有偿还贷款的能力，\n", "否则该用户是具有偿还能力的。\n", "\n", "现在有一个贷款用户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", "很显然，他是具有偿还贷款能力的。\n", "\n", "那么，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？\n", "\n", "学完本章即可知道答案。" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 2.2.1 决策树分类概念" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "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", "" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "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": [ "由上面的例子可以看到，构建决策树的过程就是：\n", "1. 选择一个属性值；\n", "2. 基于该属性对样本集进行划分；\n", "3. 重复步骤 1 和 2 直到最后所得划分结果中每个样本为同一类别。" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "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", "import math\n", "from math import log\n", "import warnings\n", "warnings.filterwarnings(\"ignore\")" ] }, { "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": [ "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 2.2.2 构建决策树 " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**信息增益**用来衡量样本集合复杂度（不确定性）所减少的程度。 \n", "\n", "**信息熵**用来度量信息量的大小。从信息论的角度来看，对信息的度量等于计算信息不确定性的多少。 " ] }, { "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", "这 $K$ 个信息的信息熵： \n", "$$E(D)=-\\sum_{k=1}^{K}p_k log_{2} p_k$$\n", "\n", "需要指出：**所有 $p_k$ 累加起来的和为1**。" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "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", " # 使用信息熵公式计算\n", " ent = -sum([(p / total_num) * log(p / total_num, 2) for p in count_dict.values() if p != 0])\n", " # 避免 print 显示异常\n", " if not ent:\n", " ent = 0\n", " # 返回信息熵，精确到小数点后 3 位\n", " return round(ent, 3)\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "\n", "现在用**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", "\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", "\n", "# 计算信息熵\n", "entropy = calc_entropy(total_num, count_dict)\n", "entropy\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "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": "markdown", "metadata": {}, "source": [ "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "然后，我们是使用上面的筛选方法，分别计算不同天气下的信息熵。" ] }, { "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": [ "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "得到不同天气下的信息熵后，我们可以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": [ "我们来编写代码，使用上面的公式计算信息增益。" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# 信息增益\n", "gain = entropy - (total_num_sun/total_num*ent_sun +\n", " total_num_cloud/total_num*ent_cloud +\n", " total_num_rain/total_num*ent_rain)\n", "gain\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "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？\n", "\n", "$$\\text { Gini }(p)=2 p(1-p)$$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### 思考与练习 " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "1. 在上面的游客去往游乐场的例子中，分别将温度高低、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": "markdown", "metadata": {}, "source": [ "计算按温度高低进行切分的信息增益。" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# 温度 >26 的信息熵\n", "total_num_temp_high = df[df['温度']=='>26'].shape[0]\n", "count_dict_temp_high = {'前往':df[(df['温度']=='>26') & (df['是否前往游乐场']=='1')].shape[0],\n", " '不前往':df[(df['温度']=='>26') & (df['是否前往游乐场']=='0')].shape[0]}\n", "print(count_dict_temp_high)\n", "ent_temp_high = calc_entropy(total_num_temp_high, count_dict_temp_high)\n", "print('温度 >26 的信息熵为：%s' % ent_temp_high)\n", "\n", "# 温度 <=26 的信息熵\n", "total_num_temp_low = df[df['温度']=='<=26'].shape[0]\n", "count_dict_temp_low = {'前往':df[(df['温度']=='<=26') & (df['是否前往游乐场']=='1')].shape[0],\n", " '不前往':df[(df['温度']=='<=26') & (df['是否前往游乐场']=='0')].shape[0]}\n", "print(count_dict_temp_low)\n", "ent_temp_low = calc_entropy(total_num_temp_low, count_dict_temp_low)\n", "print('温度 <=26 的信息熵为：%s' % ent_temp_low)\n", "\n", "# 如果按照温度高低进行划分，则对应的信息增益为\n", "gain = entropy - (total_num_temp_high/total_num*ent_temp_high +\n", " total_num_temp_low/total_num*ent_temp_low)\n", "print('按照温度高低进行划分的信息增益为：%s' % gain)\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "计算按湿度高低进行切分的信息增益。" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# 湿度 >75 的信息熵\n", "total_num_hum_high = df[df['湿度']=='>75'].shape[0]\n", "count_dict_hum_high = {'前往':df[(df['湿度']=='>75') & (df['是否前往游乐场']=='1')].shape[0],\n", " '不前往':df[(df['湿度']=='>75') & (df['是否前往游乐场']=='0')].shape[0]}\n", "print(count_dict_hum_high)\n", "ent_hum_high = calc_entropy(total_num_hum_high, count_dict_hum_high)\n", "print('湿度 >75 的信息熵为：%s' % ent_hum_high)\n", "\n", "# 湿度 <=75 的信息熵\n", "total_num_hum_low = df[df['湿度']=='<=75'].shape[0]\n", "count_dict_hum_low = {'前往':df[(df['湿度']=='<=75') & (df['是否前往游乐场']=='1')].shape[0],\n", " '不前往':df[(df['湿度']=='<=75') & (df['是否前往游乐场']=='0')].shape[0]}\n", "print(count_dict_hum_low)\n", "ent_hum_low = calc_entropy(total_num_hum_low, count_dict_hum_low)\n", "print('湿度 <=75 的信息熵为：%s' % ent_hum_low)\n", "\n", "# 如果按照湿度高低进行划分，则对应的信息增益为\n", "gain = entropy - (total_num_hum_high/total_num*ent_hum_high +\n", " total_num_hum_low/total_num*ent_hum_low)\n", "print('按照湿度高低进行划分的信息增益为：%s' % gain)\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "计算按风力强弱进行切分的信息增益。" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# 有风 的信息熵\n", "total_num_wind = df[df['是否有风']=='是'].shape[0]\n", "count_dict_wind = {'前往':df[(df['是否有风']=='是') & (df['是否前往游乐场']=='1')].shape[0],\n", " '不前往':df[(df['是否有风']=='是') & (df['是否前往游乐场']=='0')].shape[0]}\n", "print(count_dict_wind)\n", "ent_wind = calc_entropy(total_num_wind, count_dict_wind)\n", "print('有风 的信息熵为：%s' % ent_wind)\n", "\n", "# 无风 的信息熵\n", "total_num_nowind = df[df['是否有风']=='否'].shape[0]\n", "count_dict_nowind = {'前往':df[(df['是否有风']=='否') & (df['是否前往游乐场']=='1')].shape[0],\n", " '不前往':df[(df['是否有风']=='否') & (df['是否前往游乐场']=='0')].shape[0]}\n", "print(count_dict_nowind)\n", "ent_nowind = calc_entropy(total_num_nowind, count_dict_nowind)\n", "print('无风 的信息熵为：%s' % ent_nowind)\n", "\n", "# 如果按照是否有风进行划分，则对应的信息增益为\n", "gain = entropy - (total_num_wind/total_num*ent_wind +\n", " total_num_nowind/total_num*ent_nowind)\n", "print('按照是否有风进行划分的信息增益为：%s' % gain)\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] }, { "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** 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 **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": [ "
" ] }, { "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": [ "
" ] }, { "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('essay1_en.txt')\n", "# 读取中文短文\n", "ch_essay = read_file('essay1_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": null, "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": [ "## 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)" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.5.2" } }, "nbformat": 4, "nbformat_minor": 2 }