【数据挖掘】国科大刘莹老师数据挖掘课程作业 —— 第二次作业

​σc,i​1​exp(−2σc,i2​(xi​−μc,i2​)​)(6)
对于样本 Z = (Height = Short, Hair = blond, Eye = brown) 而言，先计算先验概率

P

(

C

1

)

P(rm C1)

P(C1) 和

P

(

C

2

)

P(rm C2)

P(C2)，

P

(

C

1

)

=

5

9

P({rm C1}) = frac{5}{9}

P(C1)=95​，

P

(

C

2

)

=

4

9

P({rm C2}) = frac{4}{9}

P(C2)=94​。针对属性 Height，

P

(

H

e

i

g

h

t

=

S

h

o

r

t

∣

C

1

)

=

2

5

P({rm Height = Short} mid {rm C1}) = frac{2}{5}

P(Height=Short∣C1)=52​，

P

(

H

e

i

g

h

t

=

S

h

o

r

t

∣

C

2

)

=

1

4

P({rm Height = Short} mid {rm C2}) = frac{1}{4}

P(Height=Short∣C2)=41​；针对属性 Hair，

P

(

H

a

i

r

=

b

l

o

n

d

∣

C

1

)

=

2

5

P({rm Hair = blond}mid {rm C1}) = frac{2}{5}

P(Hair=blond∣C1)=52​，

P

(

H

a

i

r

=

b

l

o

n

d

∣

C

2

)

=

1

2

P({rm Hair = blond}mid {rm C2}) = frac{1}{2}

P(Hair=blond∣C2)=21​；针对属性 Eye，

P

(

E

y

e

=

b

r

o

w

n

∣

C

1

)

=

3

5

P({rm Eye = brown}mid {rm C1}) = frac{3}{5}

P(Eye=brown∣C1)=53​，

P

(

E

y

e

=

b

r

o

w

n

∣

C

2

)

=

0

P({rm Eye= brown}mid {rm C2}) = 0

P(Eye=brown∣C2)=0。

因此，

P

(

C

1

∣

Z

)

=

P

(

C

1

)

P

(

H

e

i

g

h

t

=

S

h

o

r

t

∣

C

1

)

P

(

H

a

i

r

=

b

l

o

n

d

∣

C

1

)

P

(

E

y

e

=

B

r

o

w

n

∣

C

1

)

=

0.0533

P({rm C1}mid {rm Z}) = P({rm C1})P({rm Height = Short} mid {rm C1})P({rm Hair = blond} mid {rm C1})P({rm Eye = Brown} mid {rm C1}) = 0.0533

P(C1∣Z)=P(C1)P(Height=Short∣C1)P(Hair=blond∣C1)P(Eye=Brown∣C1)=0.0533​；

P

(

C

2

∣

Z

)

=

P

(

C

2

)

P

(

H

e

i

g

h

t

=

S

h

o

r

t

∣

C

2

)

P

(

H

a

i

r

=

b

l

o

n

d

∣

C

2

)

P

(

E

y

e

=

B

r

o

w

n

∣

C

2

)

=

0

P({rm C2}mid {rm Z}) = P({rm C2})P({rm Height = Short} mid {rm C2})P({rm Hair = blond} mid {rm C2})P({rm Eye = Brown} mid {rm C2}) = 0

P(C2∣Z)=P(C2)P(Height=Short∣C2)P(Hair=blond∣C2)P(Eye=Brown∣C2)=0；

在不考虑平滑的前提下，

P

(

E

y

e

=

b

r

o

w

n

∣

C

2

)

=

0

P({rm Eye= brown}mid {rm C2}) = 0

P(Eye=brown∣C2)=0 导致

P

(

C

2

∣

Z

)

P(rm C2mid Z)

P(C2∣Z) 为

0

0

0。所以样本 Z 被分类为 C1。

Lab Part

假设一家超市想推销意大利面。使用“Transactions.txt”中的数据作为训练数据来构建基于 C5.0 算法的决策树模型，以预测客户是否会购买意大利面。

1. 使用数据集 “Transactions.txt” 构建决策树，利用其它字段来预测 “pasta” 字段。使用 Field Ops 中的 Type 模块，将除了 COD 字段外的每个字段的 “type” 设置为 “Flag”，将 COD 字段的 “type“ 设置为 “Typeless”，将 “pasta” 字段的 “direction” 属性设置为 “out”。使用 Modeling 中的 C5.0 模块，选择 “Expert” 并将 “Pruning severity” 设置为

65

65

65，将 “Minimum records per child branch” 设置为

95

95

95。

图

5

5

5 为 Clementine 的使用截图。使用数据集 “Transaction.txt” 构建的决策树如图

6

6

6 所示。

图 5    Clementine 使用截图

图 6    决策树

虽然横向显示决策树会比较美观，但是缩放严重出现失真，故还是选择了纵向显示。

2. 使用上面创建好的模型对 “rollout.txt” 数据中的

20

20

20 位客户中的每一位进行预测，以确定客户是否会购买意大利面。

图

7

7

7 和图

8

8

8 分别展示了数据类型配置和对 “rollout.txt” 的预测结果。

图 7    rollout 数据类型配置

图 8    决策树预测结果

前五层的预测规则如下：

tomato souce = 1 [ Mode: 1 ] 
	tunny = 1 [ Mode: 1 ] => 1 
	tunny = 0 [ Mode: 1 ] 
		rice = 1 [ Mode: 1 ] => 1 
		rice = 0 [ Mode: 0 ] 
			brioches = 1 [ Mode: 1 ] => 1 
			brioches = 0 [ Mode: 0 ] 
				frozen vegetables = 1 [ Mode: 1 ] => 1 
				frozen vegetables = 0 [ Mode: 0 ] 
					coffee = 1 [ Mode: 1 ] => 1 
					coffee = 0 [ Mode: 0 ] => 0 
tomato souce = 0 [ Mode: 0 ] 
	rice = 1 [ Mode: 0 ] 
		coffee = 1 [ Mode: 1 ] => 1 
		coffee = 0 [ Mode: 0 ] 
			biscuits = 1 [ Mode: 1 ] => 1 
			biscuits = 0 [ Mode: 0 ] 
				coke = 1 [ Mode: 1 ] => 1 
				coke = 0 [ Mode: 0 ] => 0 
	rice = 0 [ Mode: 0 ] 
		tunny = 1 [ Mode: 0 ] => 0 
		tunny = 0 [ Mode: 0 ] 
			oil = 1 [ Mode: 0 ] => 0 
			oil = 0 [ Mode: 0 ] 
				water = 1 [ Mode: 0 ] => 0 
				water = 0 [ Mode: 0 ] 
					milk = 1 [ Mode: 0 ] => 0 
					milk = 0 [ Mode: 0 ] 
						yoghurt = 1 [ Mode: 0 ] => 0 
						yoghurt = 0 [ Mode: 0 ] 
							coke = 1 [ Mode: 0 ] => 0 
							coke = 0 [ Mode: 0 ] 
								biscuits = 1 [ Mode: 0 ] => 0 
								biscuits = 0 [ Mode: 0 ] 
									brioches = 1 [ Mode: 0 ] => 0 
									brioches = 0 [ Mode: 1 ] 
										coffee = 1 [ Mode: 0 ] => 0 
										coffee = 0 [ Mode: 1 ] 
											frozen vegetables = 1 [ Mode: 0 ] => 0 
											frozen vegetables = 0 [ Mode: 1 ] 
												beer = 1 [ Mode: 0 ] => 0 
												beer = 0 [ Mode: 1 ] 
													juices = 1 [ Mode: 0 ] => 0 
													juices = 0 [ Mode: 1 ] 
														mozzarella = 1 [ Mode: 0 ] => 0 
														mozzarella = 0 [ Mode: 1 ] 
															crackers = 1 [ Mode: 0 ] => 0 
															crackers = 0 [ Mode: 1 ] 
																frozen fish = 1 [ Mode: 0 ] => 0 
																frozen fish = 0 [ Mode: 1 ] => 1 

通过对某在线培训系统的标注数据集进行建模，预测其它会员期末考试的结果。数据集来自在线培训系统的日志，数据包括每个会员的在线学习行为。请尝试多种不同的模型、不同的参数，建立高质量的预测模型。

训练集有

873

873

873 条记录，测试集有

461

461

461 条记录。训练集和测试集包含如下变量：

1. 对训练数据集进行决策树分类。将除 “人员 ID” 之外的字段设置为输入。将 “Class” 的 “direction” 设置为 “out”，“type” 设置为 ”Flag“。自定义 “pruning severity” 和 “minimum records per child branch”，然后勾选“use global pruning”。

尝试了多组参数如图

9

9

9 所示。其中，PS 为 pruning severity，MRPCB 为 minimum records per child branch。可见，“最佳”参数组合为 PS=5，MRPCB=5。

图 9    决策树混淆矩阵

2. 使用默认设置的神经网络处理训练数据集。设置同上。

默认设置的神经网络对应的混淆矩阵如图

10

10

10 所示。

图 10    神经网络混淆矩阵

3. 使用默认设置的逻辑回归模型处理训练数据集。设置同上。

默认设置的逻辑回归模型对应的混淆矩阵如图

11

11

11 所示。

图 11    逻辑回归模型混淆矩阵

4. 分析上面三个模型生成的混淆矩阵，评估模型质量。

对比 PS=5, MRPCB=5 的决策树、默认设置的神经网络和逻辑回归模型，决策树在准确率（accuracy）、召回率（recall）还是精度（precision）指标上的效果均优于其它两个模型，但这并不意味着决策树模型更适合这个数据集。

REF

Clementine教程 – 知乎

使用clementine得到混淆矩阵 – CSDN