From 464f107928902c76d449a034a7f6817fa9af1702 Mon Sep 17 00:00:00 2001
From: benjas <909336740@qq.com>
Date: Sun, 18 Apr 2021 20:58:48 +0800
Subject: [PATCH] Add. Build the decision tree

---
 .../5.决策树——每次选一边.md       |  21 ++++++++++++++++++
 .../assets/1618749843112.png                  | Bin 0 -> 1310 bytes
 .../assets/1618749968444.png                  | Bin 0 -> 1605 bytes
 .../assets/1618750408613.png                  | Bin 0 -> 4516 bytes
 4 files changed, 21 insertions(+)
 create mode 100644 机器学习算法理论及应用/李航——统计学习方法/assets/1618749843112.png
 create mode 100644 机器学习算法理论及应用/李航——统计学习方法/assets/1618749968444.png
 create mode 100644 机器学习算法理论及应用/李航——统计学习方法/assets/1618750408613.png

diff --git a/机器学习算法理论及应用/李航——统计学习方法/5.决策树——每次选一边.md b/机器学习算法理论及应用/李航——统计学习方法/5.决策树——每次选一边.md
index b1d513d..06036f4 100644
--- a/机器学习算法理论及应用/李航——统计学习方法/5.决策树——每次选一边.md
+++ b/机器学习算法理论及应用/李航——统计学习方法/5.决策树——每次选一边.md
@@ -213,3 +213,24 @@ Information gain ratio
 
 如上面的年龄，有3个类（青年、中年、老年），![1618749717493](assets/1618749717493.png)
 
+
+
+信息增益比和信息增益的区别就是除以![1618749843112](assets/1618749843112.png)
+
+
+
+### 决策树的构建
+
+Build the decision tree
+
+ID3算法：
+
+- 输入：训练数据集D，特征A，阈值ε ；
+- 输出：决策树T
+  1. 若D中所有实例属于同一类![1618749968444](assets/1618749968444.png)，则T为单节点数，并将类![1618749968444](assets/1618749968444.png)作为该节点的类标记，返回T；
+  2. 若A = Ø，则T为单节点树，并将D中实例数最大的类![1618749968444](assets/1618749968444.png)作为该节点的类标记，返回T；
+  3. 否则，按算法计算A中各特征对D的信息增益，选择信息增益最大的特征Ag；
+  4. 如果Ag的信息增益小于阈值ε，则置T为单节点树，并将D中实例数最大的类![1618749968444](assets/1618749968444.png)作为该节点的类标记，返回T；
+  5. 否则，对Ag的每一个可能值ai，依![1618750408613](assets/1618750408613.png)将D分割为若干非空子集Di，将Di中实例最大的类作为标记，构建子节点，由节点及其子节点构成树T，返回T；
+  6. 对第i个子节点，以Di为训练集，以A - {Ag}为特征集，递归地调用1~5步，得到树Ti，返回Ti。
+
diff --git a/机器学习算法理论及应用/李航——统计学习方法/assets/1618749843112.png b/机器学习算法理论及应用/李航——统计学习方法/assets/1618749843112.png
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