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2306687144@qq.com:数据挖掘复习题和答案

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一、

考虑表中二元分类问题的训练样本集

1. 2. 3. 4. 5. 6.

整个训练样本集关于类属性的熵是多少? 关于这些训练集中 a1,a2 的信息增益是多少? 对于连续属性 a3,计算所有可能的划分的信息增益。 根据信息增益,a1,a2,a3 哪个是最佳划分? 根据分类错误率,a1,a2 哪具最佳? 根据 gini 指标,a1,a2 哪个最佳?

答 1.

P(+) = 4/9 and P(?) = 5/9 ?4/9 log2(4/9) ? 5/9 log2(5/9) = 0.9911.

1 / 14

答 2:

(估计不考)

答 3:

2 / 14

答 4: According to information gain, a1 produces the best split. 3 / 14

答 5:

For attribute a1: error rate = 2/9. For attribute a2: error rate = 4/9. Therefore, according to error rate, a1 produces the best split.

答 6:

4 / 14

二、考虑如下二元分类问题的数据集

5 / 14

1. 计算 a.b 信息增益,决策树归纳算法会选用哪个属性

2. 计算 a.b gini 指标,决策树归纳会用哪个属性?

这个答案没问题

6 / 14

3. 从图 4-13 可以看出熵和 gini 指标在[0,0.5]都是单调递增,而[0.5,1]之间单调递减。有没有可能信息增益和 gini 指标增益支持不同的属性?解释你的理由
Yes, even though these measures have similar range and monotonous behavior, their respective gains, Δ, which are scaled differences of the measures, do not necessarily behave in the same way, as illustrated by the results in parts (a) and (b). 贝叶斯分类

7 / 14

1. P(A = 1|?) = 2/5 = 0.4, P(B = 1|?) = 2/5 = 0.4, P(C = 1|?) = 1, P(A = 0|?) = 3/5 = 0.6, P(B = 0|?) = 3/5 = 0.6, P(C = 0|?) = 0; P(A = 1|+) = 3/5 = 0.6, P(B = 1|+) = 1/5 = 0.2, P(C = 1|+) = 2/5 = 0.4, P(A = 0|+) = 2/5 = 0.4, P(B = 0|+) = 4/5 = 0.8, P(C = 0|+) = 3/5 = 0.6.

2. 3. P(A = 0|+) = (2 + 2)/(5 + 4) = 4/9, P(A = 0|?) = (3+2)/(5 + 4) = 5/9, P(B = 1|+) = (1 + 2)/(5 + 4) = 3/9, P(B = 1|?) = (2+2)/(5 + 4) = 4/9, P(C = 0|+) = (3 + 2)/(5 + 4) = 5/9, P(C = 0|?) = (0+2)/(5 + 4) = 2/9. 4. Let P(A = 0,B = 1, C = 0) = K

8 / 14

5. 当的条件概率之一是零,则估计为使用 m-估计概率的方法的条件概率是更好的,因为我们不希望整个表达
式变为零。

1. P(A = 1|+) = 0.6, P(B = 1|+) = 0.4, P(C = 1|+) = 0.8, P(A = 1|?) = 0.4, P(B = 1|?) = 0.4, and P(C = 1|?) = 0.2
2.

Let R : (A = 1,B = 1, C = 1) be the test record. To determine its class, we need to compute P(+|R) and P(?|R). Using Bayes theorem, P(+|R) = P(R|+)P(+)/P(R) and P(?|R) = P(R|?)P(?)/P(R). Since P(+) = P(?) = 0.5 and P(R) is constant, R can be classified by comparing P(+|R) and P(?|R). For this question, P(R|+) = P(A = 1|+) × P(B = 1|+) × P(C = 1|+) = 0.192 P(R|?) = P(A = 1|?) × P(B = 1|?) × P(C = 1|?) = 0.032 Since P(R|+) is larger, the record is assigned to (+) class. 3. P(A = 1) = 0.5, P(B = 1) = 0.4 and P(A = 1,B = 1) = P(A) × 9 / 14

P(B) = 0.2. Therefore, A and B are independent. 4.
P(A = 1) = 0.5, P(B = 0) = 0.6, and P(A = 1,B = 0) = P(A =1)× P(B = 0) = 0.3. A and B are still independent.

5. Compare P(A = 1,B = 1|+) = 0.2 against P(A = 1|+) = 0.6 and P(B = 1|Class = +) = 0.4. Since the product between P(A = 1|+) and P(A = 1|?) are not the same as P(A = 1,B = 1|+), A and B are not conditionally independent given the class.

三、 使用下表中的相似度矩阵进行单链和全链层次聚类。绘制树状况显示结果,树状图应该清楚地显示合并的次序。

10 / 14

There are no apparent relationships between s1, s2, c1, and c2.

11 / 14

12 / 14

13 / 14

A2: Percentage of frequent itemsets = 16/32 = 50.0% (including the null set).

A4: False alarm rate is the ratio of I to the total number of itemsets. Since the count of I = 5, therefore the false alarm rate is 5/32 = 15.6%.

14 / 14


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