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一、（30分）设最小支持度阈值为0.2500, 最小置信度为0.6500。对于下面的规则模板和信息表找出R中的所有强关联规则：

S∈R，P（S，x ）∧ Q（S，y ）==> Gpa（S，w ） [ s, c ]
其中，P，Q ∈{ Major, Status ，Age }.

Major	Status	Age	Gpa	Count
Arts	Graduate	Old	Good	50
Arts	Graduate	Old	Excellent	150
Arts	Undergraduate	Young	Good	150
Appl_	science	Undergraduate	Young	Excellent
Science	Undergraduate	Young	Good	100

解答：
样本总数为500，最小支持数为500*0.25 = 125。
在Gpa取不同值的情形下，分别讨论。
（1）Gpa = Good，

Major	Status	Age	Count
Arts	Graduate	Old	50
Arts	Undergraduate	Young	150
Science	Undergraduate	Young	100

频繁1项集L1 = {Major= Arts:200; Status=Undergraduate: 250; Age = Young:250} -----10分
频繁2项集的待选集C2={Major= Arts，Status= Undergraduate:150; Major= Arts，Age=Young:150；Status=Undergraduate, Age=Young:250 }
频繁2项集L2=C2

(2) Gpa = Excellent

Major	Status	Age	Count
Arts	Graduate	Old	150
Appl_science	Undergraduate	Young	50

频繁1项集L1 = {Major= Arts:150; Status=Graduate: 150; Age = Old:250}
频繁2项集的待选集C2={Major= Arts，Status= Graduate:150; Major= Arts，Age=Old:150；Status=Graduate, Age=Old:150 }
频繁2项集L2=C2

考察置信度：
Major(S,Arts)^Status(S,Undergraduate)=>Gpa(S,Good) [s=150/500=0.3000, c=150/150=1.0000]
Major(S, Arts)^Age(S,Young)=>Gpa(S, Good)[s=150/500=0.3000, c=150/150=1.0000]
Status(S,Undergraduate)^Age(S,Young)=>Gpa(S,Good) [s=250/500=0.5000, c=250/300=0.8333]
Major(S, Arts)^Status(S,Graduate)=>Gpa(S, Excellent)[s=150/500=0.3000, c=150/200=0.7500]
Major(S, Arts)^Age(S,Old)=>Gpa(S, Excellent)[s=150/500=0.3000, c=150/200=0.7500]
Status(S,Graduate)^Age(S,Old)=>Gpa(S,Excellent) [s=150/500=0.3000, c=150/200=0.7500]

因此，所有强关联规则是：
Major(S,Arts)^Status(S,Undergraduate)=>Gpa(S,Good) [s=150/500=0.3000, c=150/150=1.0000]
Major(S, Arts)^Age(S,Young)=>Gpa(S, Good)[s=150/500=0.3000, c=150/150=1.0000]
Status(S,Undergraduate)^Age(S,Young)=>Gpa(S,Good) [s=250/500=0.5000, c=250/300=0.8333]
Major(S, Arts)^Status(S,Graduate)=>Gpa(S, Excellent)[s=150/500=0.3000, c=150/200=0.7500]
Major(S, Arts)^Age(S,Old)=>Gpa(S, Excellent)[s=150/500=0.3000, c=150/200=0.7500]
Status(S,Graduate)^Age(S,Old)=>Gpa(S,Excellent) [s=150/500=0.3000, c=150/200=0.7500]

二、（30分）设类标号属性 Gpa 有两个不同的值（ 即{ Good, Excellent } ）, 基于信息增益，利用判定树进行归纳分类。

解答：
定义P: Gpa = Good
N: Gpa = Excellent
任何分割进行前,样本集的熵为:

p	n	I(p,n)
300	200	0.97095

I(p,n)=-0.6log2(0.6) –0.4log2(0.4)
= 0.97095

考虑按属性Major分割后的样本的熵

Major	pi	ni	I(pi,ni)
Arts	200	150	0.98523
Appl_science	0	50	0
Science	100	0	0

E(Major) = 350/500*0.98523 = 0.68966

I(p,n)=-(4/7)log2(4/7) –(3/7)log2(3/7) =0.98523

考虑按属性Status分割后的样本的熵

Status	pi	ni	I(pi,ni)
Graduate	50	150	0.81128
Undergraduate	250	50	0.65002

E(Status) = 200/5000.81128+300/5000.65002 = 0.71452

考虑按属性Age分割后的样本的熵

Age	pi	ni	I(pi,ni)
Old	50	150	0.81128
Young	250	50	0.65002

E(Age) = E(Status) = 0.71452

各属性的信息增益如下:
Gain(Major) =0.97095-0.68966 = 0.28129
Gain(Status) =Gain(Age) =0.97095-0.71452 = 0.25643

比较后,由于Gain(Major)的值最大,按照最大信息增益原则,按照属性Major的不同取值进行第一次分割.
分割后,按照Major的不同取值,得到下面的3个表:

(1)Major = Arts

Status	Age	Gpa	Count
Graduate	Old	Good	50
Graduate	Old	Excellent	150
Undergraduate	Young	Good	150

考虑按属性Status分割后的样本的熵

Status	pi	ni	I(pi,ni)
Graduate	50	150	0.81128
Undergraduate	150	0	0

E(Status) = 200/350*0.81128= 0.46359

考虑按属性Age分割后的样本的熵

Status	pi	ni	I(pi,ni)
Old	50	150	0.81128
Young	150	0	0

E(Age) = E(Status)= 0.46359

由于E(Age) = E(Status)，可按照属性Status的不同取值进行第二次分割。分割后，按照Status的不同取值，得到下面的2个表：

(1.1) Status =Graduate

Age	Gpa	Count
Old	Good	50
Old	Excellent	150

由于表中属性Age的取值没有变化，停止分割。按照多数投票原则，该分支可被判定为Gpa=Excellent。
（1.2）Status = Undergraduate

Status	Age	Gpa	Count
Undergraduate	Young	Good	150

在这种情形下,所有样本的Gpa属性值都相同.停止分割.
（2）Major= Appl_Science

Status	Age	Gpa	Count
Undergraduate	Young	Excellent	50

在这种情形下,所有样本的Gpa属性值都相同.停止分割.
（3）Major=Science

Status	Age	Gpa	Count
Undergraduate	Young	Good	100

在这种情形下,所有样本的Gpa属性值都相同.停止分割.
综合以上分析,有以下的判定树:
Major--------- Arts ----------Status-------Graduate ------Excellent
\ ______Undergraduate______Good
_______Appl_Science_______________________Excellent

__________Science______________________Good

小 tricks

计算信息熵的代码

import mathdef entropy(probabilities):total = sum(probabilities)probabilities= [p / total for p in probabilities]entropy = 0for p in probabilities:if p > 0:entropy -= p * math.log2(p)return entropyprobabilities = [100,100,150]#计算100 100 150的信息熵result = entropy(probabilities)
print("信息熵:", result)