Machine_Learning_2019_Task 8 决策树

Machine_Learning_2019_Task 8 决策树

• ID3（基于信息增益）
• C4.5（基于信息增益比）
• CART（Gini指数）

扩充：学习CART的生成(回归树模型)【参考统计学习方法】

熵：$H(x)=-{\sum }_{i=1}^n{p}_i\log p_i$

条件熵: $H(X|Y)=\sum P(X|Y)\log P(X|Y)$

信息增益 : $g(D,A)=H(D)-H(D|A)$

信息增益比: $g_R(D,A)=\frac{g(D,A)}{H(A)}$

Gini 指数:$Gini(D)={\sum }_{k=1}^K{p}_k\log p_k=1-{\sum }_{k=1}^K{p}_k^2$

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
%matplotlib inline

from sklearn.datasets import load_iris
from sklearn.model_selection import train_test_split

from collections import Counter
import math
from math import log

import pprint

生成数据集

def create_data():
    datasets = [['青年', '否', '否', '一般', '否'],
               ['青年', '否', '否', '好', '否'],
               ['青年', '是', '否', '好', '是'],
               ['青年', '是', '是', '一般', '是'],
               ['青年', '否', '否', '一般', '否'],
               ['中年', '否', '否', '一般', '否'],
               ['中年', '否', '否', '好', '否'],
               ['中年', '是', '是', '好', '是'],
               ['中年', '否', '是', '非常好', '是'],
               ['中年', '否', '是', '非常好', '是'],
               ['老年', '否', '是', '非常好', '是'],
               ['老年', '否', '是', '好', '是'],
               ['老年', '是', '否', '好', '是'],
               ['老年', '是', '否', '非常好', '是'],
               ['老年', '否', '否', '一般', '否'],
               ]
    labels = [u'年龄', u'有工作', u'有自己的房子', u'信贷情况', u'类别']
    # 返回数据集和每个维度的名称
    return datasets, labels

datasets, labels = create_data()

train_data = pd.DataFrame(datasets, columns=labels)

train_data

	年龄	有工作	有自己的房子	信贷情况	类别
0	青年	否	否	一般	否
1	青年	否	否	好	否
2	青年	是	否	好	是
3	青年	是	是	一般	是
4	青年	否	否	一般	否
5	中年	否	否	一般	否
6	中年	否	否	好	否
7	中年	是	是	好	是
8	中年	否	是	非常好	是
9	中年	否	是	非常好	是
10	老年	否	是	非常好	是
11	老年	否	是	好	是
12	老年	是	否	好	是
13	老年	是	否	非常好	是
14	老年	否	否	一般	否

# 熵
def calc_ent(datasets):
    data_length = len(datasets)
    label_count = {}
    for i in range(data_length):
        label = datasets[i][-1]
        if label not in label_count:
            label_count[label] = 0
        label_count[label] += 1
    ent = -sum([(p/data_length)*log(p/data_length, 2) for p in label_count.values()])
    return ent

# 条件熵
def cond_ent(datasets, axis=0):
    data_length = len(datasets)
    feature_sets = {}
    for i in range(data_length):
        feature = datasets[i][axis]
        if feature not in feature_sets:
            feature_sets[feature] = []
        feature_sets[feature].append(datasets[i])
    cond_ent = sum([(len(p)/data_length)*calc_ent(p) for p in feature_sets.values()])
    return cond_ent

# 信息增益
def info_gain(ent, cond_ent):
    return ent - cond_ent

def info_gain_train(datasets):
    count = len(datasets[0]) - 1
    ent = calc_ent(datasets)
    best_feature = []
    for c in range(count):
        c_info_gain = info_gain(ent, cond_ent(datasets, axis=c))
        best_feature.append((c, c_info_gain))
        print('特征({}) - info_gain - {:.3f}'.format(labels[c], c_info_gain))
    # 比较大小
    best_ = max(best_feature, key=lambda x: x[-1])
    return '特征({})的信息增益最大，选择为根节点特征'.format(labels[best_[0]])

info_gain_train(np.array(datasets)) # 将dataset转换为ndarray类型

特征(年龄) - info_gain - 0.083
特征(有工作) - info_gain - 0.324
特征(有自己的房子) - info_gain - 0.420
特征(信贷情况) - info_gain - 0.363

'特征(有自己的房子)的信息增益最大，选择为根节点特征'

利用ID3算法生成决策

# 定义节点类
class Node:
    def __init__(self, root=True, label=None, feature_name=None, feature=None):
        self.root = root
        self.label = label
        self.feature_name = feature_name
        self.feature = feature
        self.tree = {}
        self.result = {'label:': self.label, 'feature': self.feature, 'tree': self.tree}

    def __repr__(self):
        return '{}'.format(self.result)

    def add_node(self, val, node):
        self.tree[val] = node

    def predict(self, features):
        if self.root is True:
            return self.label
        return self.tree[features[self.feature]].predict(features)

# 定义二叉树
class DTree:
    def __init__(self, epsilon=0.1):
        self.epsilon = epsilon
        self._tree = {}

    # 熵
    @staticmethod
    def calc_ent(datasets):
        data_length = len(datasets)
        label_count = {}
        for i in range(data_length):
            label = datasets[i][-1]
            if label not in label_count:
                label_count[label] = 0
            label_count[label] += 1
        ent = -sum([(p/data_length)*log(p/data_length, 2) for p in label_count.values()])
        return ent

    # 条件熵
    def cond_ent(self, datasets, axis=0):
        data_length = len(datasets)
        feature_sets = {}
        for i in range(data_length):
            feature = datasets[i][axis]
            if feature not in feature_sets:
                feature_sets[feature] = []
            feature_sets[feature].append(datasets[i])
        cond_ent = sum([(len(p)/data_length)*self.calc_ent(p) for p in feature_sets.values()])
        return cond_ent

    # 信息增益
    @staticmethod
    def info_gain(ent, cond_ent):
        return ent - cond_ent

    def info_gain_train(self, datasets):
        count = len(datasets[0]) - 1
        ent = self.calc_ent(datasets)
        best_feature = []
        for c in range(count):
            c_info_gain = self.info_gain(ent, self.cond_ent(datasets, axis=c))
            best_feature.append((c, c_info_gain))
        # 比较大小
        best_ = max(best_feature, key=lambda x: x[-1])
        return best_

    def train(self, train_data):
        """
        input:数据集D(DataFrame格式)，特征集A，阈值eta
        output:决策树T
        """
        _, y_train, features = train_data.iloc[:, :-1], train_data.iloc[:, -1], train_data.columns[:-1]
        # 1,若D中实例属于同一类Ck，则T为单节点树，并将类Ck作为结点的类标记，返回T
        if len(y_train.value_counts()) == 1:
            return Node(root=True,
                        label=y_train.iloc[0])

        # 2, 若A为空，则T为单节点树，将D中实例树最大的类Ck作为该节点的类标记，返回T
        if len(features) == 0:
            return Node(root=True, label=y_train.value_counts().sort_values(ascending=False).index[0])

        # 3,计算最大信息增益 Ag为信息增益最大的特征
        max_feature, max_info_gain = self.info_gain_train(np.array(train_data))
        max_feature_name = features[max_feature]

        # 4,Ag的信息增益小于阈值eta,则置T为单节点树，并将D中是实例数最大的类Ck作为该节点的类标记，返回T
        if max_info_gain < self.epsilon:
            return Node(root=True, label=y_train.value_counts().sort_values(ascending=False).index[0])

        # 5,构建Ag子集
        node_tree = Node(root=False, feature_name=max_feature_name, feature=max_feature)

        feature_list = train_data[max_feature_name].value_counts().index
        for f in feature_list:
            sub_train_df = train_data.loc[train_data[max_feature_name] == f].drop([max_feature_name], axis=1)

            # 6, 递归生成树
            sub_tree = self.train(sub_train_df)
            node_tree.add_node(f, sub_tree)

        return node_tree

    def fit(self, train_data):
        self._tree = self.train(train_data)
        return self._tree

    def predict(self, X_test):
        return self._tree.predict(X_test)

datasets, labels = create_data()
data_df = pd.DataFrame(datasets, columns=labels)
dt = DTree()
tree = dt.fit(data_df)

tree

{'label:': None, 'feature': 2, 'tree': {'否': {'label:': None, 'feature': 1, 'tree': {'否': {'label:': '否', 'feature': None, 'tree': {}}, '是': {'label:': '是', 'feature': None, 'tree': {}}}}, '是': {'label:': '是', 'feature': None, 'tree': {}}}}

dt.predict(['老年', '否', '否', '一般'])

'否'

sklearn.tree.DecisionTreeClassifier

criterion : string, optional (default=”gini”)

用于测量分割质量的函数.

Supported criteria are “gini” for the Gini impurity and “entropy” for the information gain.

def create_data():
    iris = load_iris()
    df = pd.DataFrame(iris.data, columns=iris.feature_names)
    df['label'] = iris.target
    df.columns = ['sepal length', 'sepal width', 'petal length', 'petal width', 'label']
    data = np.array(df.iloc[:100, [0, 1, -1]])
    print(data)
    return data[:,:2], data[:,-1]

X, y = create_data()
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3)

[[5.1 3.5 0. ]
 [4.9 3.  0. ]
 [4.7 3.2 0. ]
 [4.6 3.1 0. ]
 [5.  3.6 0. ]
 [5.4 3.9 0. ]
 [4.6 3.4 0. ]
 [5.  3.4 0. ]
 [4.4 2.9 0. ]
 [4.9 3.1 0. ]
 [5.4 3.7 0. ]
 [4.8 3.4 0. ]
 [4.8 3.  0. ]
 [4.3 3.  0. ]
 [5.8 4.  0. ]
 [5.7 4.4 0. ]
 [5.4 3.9 0. ]
 [5.1 3.5 0. ]
 [5.7 3.8 0. ]
 [5.1 3.8 0. ]
 [5.4 3.4 0. ]
 [5.1 3.7 0. ]
 [4.6 3.6 0. ]
 [5.1 3.3 0. ]
 [4.8 3.4 0. ]
 [5.  3.  0. ]
 [5.  3.4 0. ]
 [5.2 3.5 0. ]
 [5.2 3.4 0. ]
 [4.7 3.2 0. ]
 [4.8 3.1 0. ]
 [5.4 3.4 0. ]
 [5.2 4.1 0. ]
 [5.5 4.2 0. ]
 [4.9 3.1 0. ]
 [5.  3.2 0. ]
 [5.5 3.5 0. ]
 [4.9 3.6 0. ]
 [4.4 3.  0. ]
 [5.1 3.4 0. ]
 [5.  3.5 0. ]
 [4.5 2.3 0. ]
 [4.4 3.2 0. ]
 [5.  3.5 0. ]
 [5.1 3.8 0. ]
 [4.8 3.  0. ]
 [5.1 3.8 0. ]
 [4.6 3.2 0. ]
 [5.3 3.7 0. ]
 [5.  3.3 0. ]
 [7.  3.2 1. ]
 [6.4 3.2 1. ]
 [6.9 3.1 1. ]
 [5.5 2.3 1. ]
 [6.5 2.8 1. ]
 [5.7 2.8 1. ]
 [6.3 3.3 1. ]
 [4.9 2.4 1. ]
 [6.6 2.9 1. ]
 [5.2 2.7 1. ]
 [5.  2.  1. ]
 [5.9 3.  1. ]
 [6.  2.2 1. ]
 [6.1 2.9 1. ]
 [5.6 2.9 1. ]
 [6.7 3.1 1. ]
 [5.6 3.  1. ]
 [5.8 2.7 1. ]
 [6.2 2.2 1. ]
 [5.6 2.5 1. ]
 [5.9 3.2 1. ]
 [6.1 2.8 1. ]
 [6.3 2.5 1. ]
 [6.1 2.8 1. ]
 [6.4 2.9 1. ]
 [6.6 3.  1. ]
 [6.8 2.8 1. ]
 [6.7 3.  1. ]
 [6.  2.9 1. ]
 [5.7 2.6 1. ]
 [5.5 2.4 1. ]
 [5.5 2.4 1. ]
 [5.8 2.7 1. ]
 [6.  2.7 1. ]
 [5.4 3.  1. ]
 [6.  3.4 1. ]
 [6.7 3.1 1. ]
 [6.3 2.3 1. ]
 [5.6 3.  1. ]
 [5.5 2.5 1. ]
 [5.5 2.6 1. ]
 [6.1 3.  1. ]
 [5.8 2.6 1. ]
 [5.  2.3 1. ]
 [5.6 2.7 1. ]
 [5.7 3.  1. ]
 [5.7 2.9 1. ]
 [6.2 2.9 1. ]
 [5.1 2.5 1. ]
 [5.7 2.8 1. ]]

from sklearn.tree import DecisionTreeClassifier
from sklearn.tree import export_graphviz

clf = DecisionTreeClassifier()
clf.fit(X_train, y_train,)

DecisionTreeClassifier(class_weight=None, criterion='gini', max_depth=None,
            max_features=None, max_leaf_nodes=None,
            min_impurity_decrease=0.0, min_impurity_split=None,
            min_samples_leaf=1, min_samples_split=2,
            min_weight_fraction_leaf=0.0, presort=False, random_state=None,
            splitter='best')

clf.score(X_test, y_test)

0.9333333333333333

CART树

决策树生成

决策树剪枝

回归树：平方误差最小化

分类树：Gini Index

回归树的生成

设Y是连续变量，给定训练数据集$D=\left\{\left(x_1,y_1\right),\left(x_2,y_2\right),\cdots ,\left(x_N,y_N\right)\right\}$

假设已将输入空间划分为M各单元R1,R2…Rm,并且每个单元Rm上有一个固定的输出Cm，回归树表示为

$f(x)={\sum }_{m=1}^M{c}_{m}I\left(x\in R_m\right)$

平方误差来表示预测误差，用平方误差最小准则求解每个单元上的最优输出值

${\sum }_{x\in R_n}{\left(y_i-f\left(x_i\right)\right)}^2$

Rm上的Cm的最优值${\hat{c}}_m=\mathrm{ave}\left(y_i|x_i\in R_m\right)$

如何对输入空间进行划分？

启发式：选择第j个变量x(j)和它取的值s，作为切分变量和切分点，定义两个区域

$R_1(j,s)=\left\{x|x^{(j)}⩽s\right\}$
和
$R_2(j,s)=\left\{x|x^{(j)}>s\right\}$

然后寻找最优切分变量和切分点

${\min }_{j,s}\left[{\min }_{a_1}{\sum }_{x\in R_1(y,s)}{\left(y_i-c_1\right)}^2+{\min }_{c_2\in R_2(j,s)}{\left(y_i-c_2\right)}^2\right]$

同时满足

${\hat{c}}_1=\mathrm{ave}\left(y_i|x_i\in R_1(j,s)\right)$

${\hat{c}}_2=\mathrm{ave}\left(y_i|x_i\in R_2(j,s)\right)$；

再对两个区域重复上述划分，直至满足停止条件。

最小二乘回归树生成算法

输入：训练数据集D

输出：回归树f(x)

在训练数据集所在的输入空间中，递归地将每个区域划分为两个子区域并决定每个子区域上的输出值，构建二叉决策树

（1）选择最优切分变量j与切分点s，求解

${\min }_{j,s}\left[{\min }_{a_1}{\sum }_{x\in R_1(y,s)}{\left(y_i-c_1\right)}^2+{\min }_{c_2\in R_2(j,s)}{\left(y_i-c_2\right)}^2\right]$

遍历变量j，对固定的切分变量j扫描切分点s，选择使上式达到最小值的对(j, s)

（2）用选定的对(j, s)划分区域并决定相应的输出值

$R_1(j,s)=\left\{x|x^{(j)}⩽s\right\},R_2(j,s)=\left\{x|x^{(j)}>s\right\}$

${\hat{c}}_m=\frac{1}{N_m}{\sum }_{x_i\in R_n(j,s)}{y}_i,x\in R_m,m=1,2$

（3）继续对两个子区域调用步骤（1）（2），直至满足条件；

（4）将输入空间划分为M个区域R1，R2，…，RM，生成决策树

$f(x)={\sum }_{m=1}^M{\hat{c}}_{m}I\left(x\in R_m\right)$