|
|
@@ -91,7 +91,7 @@ $$\begin{aligned}
|
|
|
## 4.7
|
|
|
$$T_a={\lbrace{\frac{a^i+a^{i+1}}{2}|1\leq{i}\leq{n-1}}\rbrace}$$
|
|
|
[解析]：这个公式所表达的思想很简单，就是以每两个相邻取值的中点作为划分点，下面以西瓜书中表4.3中西瓜数据集3.0为例来说明此公式的用法。对于“密度”这个连续属性，已观测到的可能取值为$\{0.243,0.245,0.343,0.360,0.403,0.437,0.481,0.556,0.593,0.608,0.634,0.639,0.657,0.666,0.697,0.719,0.774\}$共17个值，根据公式（4.7）可知，此时$i$依次取1到16，那么“密度”这个属性的候选划分点集合为
|
|
|
-$T_{a} = \{\frac{(0.243+0.245)}{2}, \frac{(0.245+0.343)}{2},\frac{(0.343+0.360)}{2},\frac{(0.360+0.403)}{2},\frac{(0.403+0.437)}{2},\frac{(0.437+0.481)}{2},\frac{(0.481+0.556)}{2},\frac{(0.556+0.593)}{2},\frac{(0.593+0.608)}{2},\frac{(0.608+0.634)}{2},\\\frac{(0.634+0.639)}{2},\frac{(0.639+0.657)}{2},\frac{(0.657+0.666)}{2},\frac{(0.666+0.697)}{2},\frac{(0.697+0.719)}{2},\frac{(0.719+0.774)}{2}\}$
|
|
|
+$T_{a} = \{\frac{(0.243+0.245)}{2}, \frac{(0.245+0.343)}{2},\frac{(0.343+0.360)}{2},\frac{(0.360+0.403)}{2},\frac{(0.403+0.437)}{2},\frac{(0.437+0.481)}{2},\frac{(0.481+0.556)}{2},\\\frac{(0.556+0.593)}{2},\frac{(0.593+0.608)}{2},\frac{(0.608+0.634)}{2},\frac{(0.634+0.639)}{2},\frac{(0.639+0.657)}{2},\frac{(0.657+0.666)}{2},\frac{(0.666+0.697)}{2},\frac{(0.697+0.719)}{2},\frac{(0.719+0.774)}{2}\}$
|
|
|
|
|
|
## 4.8
|
|
|
$$\begin{aligned}
|
Sm1les