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Update chapter4.md to v2
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Sm1les authored Oct 28, 2024
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所以可进一步推得式(4.5)

$$
\operatorname{Gini}(D)=\sum_{k=1}^{|\mathcal{Y}|} \sum_{k^{\prime} \neq k} p_k p_{k^{\prime}}=1-\sum_{k=1}^{\mid \mathcal{Y |}} p_k^2
$$

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### 4.4.1 式(4.7)的解释

此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$\{0.243,0.245,0.343,\linebreak0.360,0.403,0.437,0.481,0.556,0.593,0.608,0.634,0.639,0.657,0.666,0.697,0.719,0.774\}$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
此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

$$
\begin{aligned}
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