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