决策树 机器学习，西瓜书p80 表4.2 使用信息增益生成决策树及后剪枝
使用信息增益构造决策树，完成后剪枝
1 构造决策树
1 根结点的选择
色泽 信息增益
根据色泽划分为 青绿，乌黑，浅白 三个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{2}{4} log_2 \frac{2}{4}+\frac{2}{4} log_2 \frac{2}{4})=1 \\
Ent(D^2) &= -(\frac{1}{4} log_2 \frac{1}{4}+\frac{3}{4} log_2 \frac{3}{4})=0.811 \\
Ent(D^3)&= -(\frac{2}{2} log_2 \frac{2}{2}+\frac{0}{2} log_2 \frac{0}{2})=0 \\
Ent(D)&= -(\frac{5}{10} log_2 \frac{5}{10}+\frac{5}{10} log_2 \frac{5}{10})=1 \\
Gani(D,色泽)&=Ent(D)-\sum_{v=1}^3 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{4}{10}\times 1+\frac{4}{10} \times 0.811+\frac{2}{10}\times0) \\
&= 0.2756\end{aligned}
\]
根蒂 信息增益
根据根蒂划分为 蜷缩 稍蜷 硬挺 三个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{2}{5} log_2 \frac{2}{5}+\frac{3}{5} log_2 \frac{3}{5})=0.971 \\
Ent(D^2) &= -(\frac{2}{4} log_2 \frac{2}{4}+\frac{2}{4} log_2 \frac{2}{4})=1 \\
Ent(D^3)&= -(\frac{1}{1} log_2 \frac{1}{1}+\frac{0}{1} log_2 \frac{0}{1})=0 \\
Ent(D)&= -(\frac{5}{10} log_2 \frac{5}{10}+\frac{5}{10} log_2 \frac{5}{10})=1 \\
Gani(D,根蒂)&=Ent(D)-\sum_{v=1}^3 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{5}{10}\times 0.971+\frac{4}{10} \times 1+\frac{1}{10}\times0) \\
&= 0.1145\end{aligned}
\]
敲声 信息增益
根据色泽划分为 浊响，沉闷，清脆 三个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{2}{6} log_2 \frac{2}{6}+\frac{4}{6} log_2 \frac{4}{6})=0.918 \\
Ent(D^2) &= -(\frac{2}{3} log_2 \frac{2}{3}+\frac{1}{3} log_2 \frac{1}{3})=0.918 \\
Ent(D^3)&= -(\frac{1}{1} log_2 \frac{1}{1}+\frac{0}{1} log_2 \frac{0}{1})=0 \\
Ent(D)&= -(\frac{5}{10} log_2 \frac{5}{10}+\frac{5}{10} log_2 \frac{5}{10})=1 \\
Gani(D,敲声)&=Ent(D)-\sum_{v=1}^3 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{6}{10}\times 0.918+\frac{3}{10} \times 0.918+\frac{1}{10}\times0) \\
&=0.2346\end{aligned}
\]
纹理 信息增益
根据 纹理 划分为 清晰 稍糊 模糊 三个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{2}{6} log_2 \frac{2}{6}+\frac{4}{6} log_2 \frac{4}{6})=0.918 \\
Ent(D^2) &= -(\frac{2}{3} log_2 \frac{2}{3}+\frac{1}{3} log_2 \frac{1}{3})=0.918 \\
Ent(D^3)&= -(\frac{1}{1} log_2 \frac{1}{1}+\frac{0}{1} log_2 \frac{0}{1})=0 \\
Ent(D)&= -(\frac{5}{10} log_2 \frac{5}{10}+\frac{5}{10} log_2 \frac{5}{10})=1 \\
Gani(D,纹理)&=Ent(D)-\sum_{v=1}^3 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{6}{10}\times 0.918+\frac{3}{10} \times 0.918+\frac{1}{10}\times0) \\
&= 0.2346\end{aligned}
\]
脐部 信息增益
根据色泽划分为 凹陷，稍凹，平坦 三个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{4} log_2 \frac{1}{4}+\frac{3}{4} log_2 \frac{3}{4})=0.811 \\
Ent(D^2) &= -(\frac{2}{4} log_2 \frac{2}{4}+\frac{2}{4} log_2 \frac{2}{4})=1 \\
Ent(D^3)&= -(\frac{2}{2} log_2 \frac{2}{2}+\frac{0}{2} log_2 \frac{0}{2})=0 \\
Ent(D)&= -(\frac{5}{10} log_2 \frac{5}{10}+\frac{5}{10} log_2 \frac{5}{10})=1 \\
Gani(D,脐部)&=Ent(D)-\sum_{v=1}^3 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{4}{10}\times 0.811+\frac{4}{10} \times 1+\frac{2}{10}\times0) \\
&= 0.2756\end{aligned}
\]
触感 信息增益
根据色泽划分为 硬滑，软粘 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{3}{6} log_2 \frac{3}{6}+\frac{3}{6} log_2 \frac{3}{6})=1 \\
Ent(D^2) &= -(\frac{2}{4} log_2 \frac{2}{4}+\frac{2}{4} log_2 \frac{2}{4})=1 \\
Ent(D)&= -(\frac{5}{10} log_2 \frac{5}{10}+\frac{5}{10} log_2 \frac{5}{10})=1 \\
Gani(D,触感)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{6}{10}\times 1 +\frac{4}{10} \times 1 \\
&= 0\end{aligned}
\]
选择根结点构建决策树
\[\begin{aligned}
Gain(D,色泽)=0.2756 \ Gain(D,根蒂)=0.1145 \ Gain(D,敲声)=0.2346 \\
Gain(D,纹理)=0.2346 \ Gain(D,脐部)=0.2756 \ Gain(D,触感)=0\end{aligned}
\]
比较六个属性的信息增益大小，选择脐部作为根结点
则数据集被划分为
2 对分支结点\({1,2,3,14}\)进行划分
色泽 信息增益
根据色泽划分为 青绿，乌黑，浅白 三个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{0}{1} log_2 \frac{0}{1}+\frac{1}{1} log_2 \frac{1}{1})=0 \\
Ent(D^2) &= -(\frac{0}{2} log_2 \frac{0}{2}+\frac{2}{2} log_2 \frac{2}{2})=0 \\
Ent(D^3)&= -(\frac{1}{1} log_2 \frac{1}{1}+\frac{0}{1} log_2 \frac{0}{1})=0 \\
Ent(D)&= -(\frac{1}{4} log_2 \frac{1}{4}+\frac{3}{4} log_2 \frac{3}{4})=0.811 \\
Gani(D,色泽)&=Ent(D)-\sum_{v=1}^3 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 0.811 – (\frac{1}{4}\times 0+\frac{2}{4} \times 0 +\frac{1}{4}\times 0) \\
&= 0.811\end{aligned}
\]
根蒂 信息增益
根据根蒂划分为 蜷缩 稍蜷 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{0}{3} log_2 \frac{0}{3}+\frac{3}{3} log_2 \frac{3}{3})=0 \\
Ent(D^2) &= -(\frac{1}{1} log_2 \frac{1}{1}+\frac{0}{1} log_2 \frac{0}{1})=0 \\
Ent(D)&= -(\frac{1}{4} log_2 \frac{1}{4}+\frac{3}{4} log_2 \frac{3}{4})= 0.811\\
Gani(D,根蒂)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 0.811 – (\frac{3}{4}\times 0 +\frac{1}{4} \times 0) \\
&= 0.811\end{aligned}
\]
敲声 信息增益
根据色泽划分为 浊响，沉闷 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{0}{2} log_2 \frac{0}{2}+\frac{2}{2} log_2 \frac{2}{2})=0 \\
Ent(D^2) &= -(\frac{1}{2} log_2 \frac{1}{2}+\frac{1}{2} log_2 \frac{1}{2})=1 \\Ent(D)&= -(\frac{1}{4} log_2 \frac{1}{4}+\frac{3}{4} log_2 \frac{3}{4})=0.811 \\
Gani(D,敲声)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 0.811 – (\frac{2}{4}\times 0 +\frac{2}{4} \times 1 ) \\
&=0.311\end{aligned}
\]
纹理 信息增益
根据 纹理 划分为 清晰 稍糊 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{0}{3} log_2 \frac{0}{3}+\frac{3}{3} log_2 \frac{3}{3})=0 \\
Ent(D^2) &= -(\frac{1}{1} log_2 \frac{1}{1}+\frac{0}{1} log_2 \frac{0}{1})=0 \\Ent(D)&= -(\frac{1}{4} log_2 \frac{1}{4}+\frac{3}{4} log_2 \frac{3}{4})=0.811 \\
Gani(D,纹理)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 0.811 – (\frac{3}{4}\times 0+\frac{1}{4} \times 0 ) \\
&= 0.811\end{aligned}
\]
触感 信息增益
根据触感划分为 硬滑 一个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{4} log_2 \frac{1}{4}+\frac{3}{4} log_2 \frac{3}{4})=0.811 \\
Ent(D)&= -(\frac{1}{4} log_2 \frac{1}{4}+\frac{3}{4} log_2 \frac{3}{4})= 0.811 \\
Gani(D,触感)&=Ent(D)-\sum_{v=1}^1 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 0.811 – (\frac{4}{4}\times 0.811 ) \\
&= 0\end{aligned}
\]
选择分类结点构建决策树
\[\begin{aligned}
Gain(D,色泽)=0.811 \ Gain(D,根蒂)=0.811 \ Gain(D,敲声)=0.311 \\
Gain(D,纹理)=0.811 \ \ \ Gain(D,触感)=0\end{aligned}
\]
不妨选择色泽作为分类依据
形成的决策树
3 对分支 \({6,7,15,17}\)进行划分
色泽 信息增益
根据色泽划分为 青绿，乌黑 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{2} log_2 \frac{1}{2}+\frac{1}{2} log_2 \frac{1}{2})=1 \\
Ent(D^2) &= -(\frac{1}{2} log_2 \frac{0}{2}+\frac{1}{2} log_2 \frac{1}{2})=1 \\Ent(D)&= -(\frac{2}{4} log_2 \frac{2}{4}+\frac{2}{4} log_2 \frac{2}{4})= 1 \\
Gani(D,色泽)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{2}{4}\times 1+\frac{2}{4} \times 1 ) \\
&= 0\end{aligned}
\]
根蒂 信息增益
根据根蒂划分为 蜷缩 稍蜷 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{3} log_2 \frac{1}{3}+\frac{2}{3} log_2 \frac{2}{3})=0。918\\
Ent(D^2) &= -(\frac{1}{1} log_2 \frac{1}{1}+\frac{0}{1} log_2 \frac{0}{1})=0 \\
Ent(D)&= -(\frac{2}{4} log_2 \frac{2}{4}+\frac{2}{4} log_2 \frac{2}{4})= 1\\
Gani(D,根蒂)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{3}{4}\times 0.918 +\frac{1}{4} \times 0) \\
&= 0.3115\end{aligned}
\]
敲声 信息增益
根据色泽划分为 浊响，沉闷 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{3} log_2 \frac{1}{3}+\frac{2}{3} log_2 \frac{2}{3})=0.918 \\
Ent(D^2) &= -(\frac{1}{1} log_2 \frac{1}{1}+\frac{0}{1} log_2 \frac{0}{1})= 0 \\Ent(D)&= -(\frac{2}{4} log_2 \frac{2}{4}+\frac{2}{4} log_2 \frac{2}{4})=1 \\
Gani(D,敲声)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{3}{4}\times 0.918 +\frac{1}{4} \times 0 ) \\
&=0.3115\end{aligned}
\]
纹理 信息增益
根据 纹理 划分为 清晰 稍糊 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{2} log_2 \frac{1}{2}+\frac{1}{2} log_2 \frac{1}{2})=1 \\
Ent(D^2) &= -(\frac{1}{2} log_2 \frac{1}{2}+\frac{1}{2} log_2 \frac{1}{2})=1 \\Ent(D)&= -(\frac{2}{4} log_2 \frac{2}{4}+\frac{2}{4} log_2 \frac{2}{4})=1 \\
Gani(D,纹理)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{2}{4}\times 1+\frac{2}{4} \times 1 ) \\
&= 0\end{aligned}
\]
触感 信息增益
根据触感划分为 硬滑，软粘 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{3} log_2 \frac{1}{3}+\frac{2}{3} log_2 \frac{2}{3})=0.918 \\
Ent(D^2) &= -(\frac{1}{1} log_2 \frac{1}{1}+\frac{0}{1} log_2 \frac{0}{1})=0 \\Ent(D)&= -(\frac{2}{4} log_2 \frac{2}{4}+\frac{2}{4} log_2 \frac{2}{4})=1 \\
Gani(D,触感)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{3}{4}\times 0.918+\frac{1}{4} \times 0 ) \\
&= 0.2295\end{aligned}
\]
选择分类结点构建决策树
\[\begin{aligned}
Gain(D,色泽)=0 \ Gain(D,根蒂)=0.3115 \ Gain(D,敲声)=0.3115 \\
Gain(D,纹理)=0 \ \ \ Gain(D,触感)=0.2295\end{aligned}
\]
不妨选择根蒂作为分类依据
此时决策树为
4 对分支\({6,7,15}\)进行划分
色泽 信息增益
根据色泽划分为 青绿，乌黑 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{0}{1} log_2 \frac{0}{1}+\frac{1}{1} log_2 \frac{1}{1})=0 \\
Ent(D^2) &= -(\frac{1}{2} log_2 \frac{1}{2}+\frac{1}{2} log_2 \frac{1}{2})=1 \\Ent(D)&= -(\frac{1}{3} log_2 \frac{1}{3}+\frac{2}{3} log_2 \frac{2}{3})= 0.918 \\
Gani(D,色泽)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 0.918 – (\frac{1}{3}\times 0+\frac{2}{3} \times 1 ) \\
&= 0.252\end{aligned}
\]
敲声 信息增益
根据色泽划分为 浊响 一个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{3} log_2 \frac{1}{3}+\frac{2}{3} log_2 \frac{2}{3})=0.918 \\Ent(D)&= -(\frac{2}{3} log_2 \frac{2}{3}+\frac{2}{3} log_2 \frac{2}{3})=0.918 \\
Gani(D,敲声)&=Ent(D)-\sum_{v=1}^1 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 0.918 – (\frac{3}{3}\times 0.918 ) \\
&=0\end{aligned}
\]
纹理 信息增益
根据 纹理 划分为 清晰 稍糊 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{2} log_2 \frac{1}{2}+\frac{1}{2} log_2 \frac{1}{2})=1 \\
Ent(D^2) &= -(\frac{0}{1} log_2 \frac{0}{1}+\frac{1}{1} log_2 \frac{1}{1})=0\\Ent(D)&= -(\frac{1}{3} log_2 \frac{1}{3}+\frac{2}{3} log_2 \frac{2}{3})=0.918 \\
Gani(D,纹理)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 0.918 – (\frac{2}{3}\times 1+\frac{1}{3} \times 0 ) \\
&= 0.252\end{aligned}
\]
触感 信息增益
根据触感划分为 软粘 一个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{3} log_2 \frac{1}{3}+\frac{2}{3} log_2 \frac{2}{3})=0.918 \\Ent(D)&= -(\frac{1}{3} log_2 \frac{1}{3}+\frac{2}{3} log_2 \frac{2}{3})=0.918\\
Gani(D,触感)&=Ent(D)-\sum_{v=1}^1 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 0.918 – (\frac{3}{3}\times 0.918 ) \\
&= 0\end{aligned}
\]
选择分类结点构建决策树
\[\begin{aligned}
Gain(D,色泽)=0 .252 \ \ Gain(D,敲声)=0 \\
Gain(D,纹理)=0.252 \ \ \ Gain(D,触感)=0\end{aligned}
\]
不妨选择色泽作为分类依据
此时决策树为
5 对分支\({7,15}\)进行划分
敲声 信息增益
根据色泽划分为 浊响 一个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{2} log_2 \frac{1}{2}+\frac{1}{2} log_2 \frac{1}{2})=1 \\Ent(D) &= -(\frac{1}{2} log_2 \frac{1}{2}+\frac{1}{2} log_2 \frac{1}{2})=1\\
Gani(D,敲声)&=Ent(D)-\sum_{v=1}^1 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{2}{2}\times 0.918 ) \\
&= 0\end{aligned}
\]
纹理 信息增益
根据 纹理 划分为 清晰 稍糊 两个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{0}{1} log_2 \frac{0}{1}+\frac{1}{1} log_2 \frac{1}{1})=0 \\
Ent(D^2) &= -(\frac{1}{1} log_2 \frac{1}{1}+\frac{0}{1} log_2 \frac{0}{1})=0\\Ent(D)&= -(\frac{1}{2} log_2 \frac{1}{2}+\frac{1}{2} log_2 \frac{1}{2})=1 \\
Gani(D,纹理)&=Ent(D)-\sum_{v=1}^2 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{1}{2}\times 0+\frac{1}{2} \times 0 ) \\
&= 1\end{aligned}
\]
触感 信息增益
根据触感划分为 软粘 一个子集
计算信息熵
\[\begin{aligned}
Ent(D^1) &= -(\frac{1}{2} log_2 \frac{1}{2}+\frac{1}{2} log_2 \frac{1}{2})=1 \\Ent(D) &= -(\frac{1}{2} log_2 \frac{1}{2}+\frac{1}{2} log_2 \frac{1}{2})=1\\
Gani(D,触感)&=Ent(D)-\sum_{v=1}^1 \frac{|D^v|}{|D|}Ent(D^v) \\
&= 1 – (\frac{2}{2}\times 0.918 ) \\
&= 0\end{aligned}
\]
选择分类结点构建决策树
\[\begin{aligned}
\ \ Gain(D,敲声)=0 \ Gain(D,纹理)=1 \ \ \ Gain(D,触感)=0\end{aligned}
\]
选择纹理作为分类依据
此时决策树为
2 决策树后剪枝
1 考虑结点\(7,15\)
原分支（剪枝前），有三个样本被正确分类 验证集精度为 42.8%
剪枝后的决策树
此时验证集有四个样本被正确分类，精度为57.1%
于是后剪枝策略决定剪枝，得到上图的决策树
2 考虑结点\(6,715\)色泽=？
由上图，决策树精度为57.1%
剪去结点后的决策树为
此时验证集有四个样本被正确分类，精度为57.1%
与未剪枝时的精度相同，西瓜书中采用了不剪枝的策略。在这里我们不妨采用剪枝的策略，于是得到上图的决策树
3 考虑结点\(1,2,3,14\)色泽=？
在上图基础上来考虑剪去结点\(1,2,3,14\)色泽=？ ，剪枝后的决策树为
此时的决策树正确分类的样本5个，精度为71.4%
根据后剪枝策略，进行剪枝，得到上图的决策树
4考虑 \(6,7，15,17\)根蒂=？
剪枝后的决策树为
此时的决策树的精度仍然为71.4%
与未剪枝时的精度相同，西瓜书中采用了不剪枝的策略。在这里我们不妨采用剪枝的策略，于是得到上图的决策树
最终得到上图的决策树
