1、數(shù)據(jù)挖掘第二次作業(yè)第一題：1. a) Compute the Information Gain for Gender, Car Type and Shirt Size.b) Construct a decision tree with Information Gain.答案：a) 因為class分為兩類：C0和C1，其中C0的頻數(shù)為10個，C1的頻數(shù)為10，所以class元組的信息增益為Info(D)=-1020*log21020-1020*log2(10/20)=11.按照Gender進行分類：Infogender(D)=1020*-410*log2410-610*log2610+1020*

2、(-410*log2410-610*log2610)=0.971Gain(Gender)=1-0.971=0.0292.按照Car Type進行分類InfocarType(D)=420*-14*log214-34*log234+820*-78*log278-18*log218+820* -88*log288-08*log208=0.314Gain(Car Type)=1-0.314=0.6863.按照Shirt Size進行分類：InfoshirtSize(D)= 520*-35*log235-25*log225+720*-47*log247-37*log237+420* -24*log224

3、-24*log224+420*-24*log224-24*log224=0.988Gain(Shirt Size)=1-0.988=0.012b) 由a中的信息增益結(jié)果可以看出采用Car Type進行分類得到的信息增益最大，所以決策樹為：Car Type?medium,large, extra largesmallC1C0C0luxurySportfamilyShirt Size?C1第二題：2. (a) Design a multilayer feed-forward neural network (one hidden layer) for the data set in Q1. Labe

4、l the nodes in the input and output layers.(b) Using the neural network obtained above, show the weight values after one iteration of the back propagation algorithm, given the training instance “(M, Family, Small). Indicate your initial weight values and biases and the learning rate used.a)b) 由a可以設(shè)每

5、個輸入單元代表的屬性和初始賦值X11X12X21X22X23X31X32X33X34FMFamilySportsLuxurySmallMediumLargeExtra Large011001000由于初始的權(quán)重和偏倚值是隨機生成的所以在此定義初始值為：W1,10W1,11W2,10W2,11W3,10W3,11W4,10W4,11W5,10W5,110.20.2-0.2-0.10.40.3-0.2-0.10.1-0.1W6,10W6,11W7,10W7,11W8,10W8,11W9,10W9,11W10,12W11,120.1-0.2-0.40.20.20.2-0.10.3-0.3-0.110

6、1112-0.20.20.3凈輸入和輸出：單元 j凈輸入 Ij輸出Oj100.10.52110.20.55120.0890.48每個節(jié)點的誤差表：單元jErrj100.0089110.003012-0.12權(quán)重和偏倚的更新：W1,10W1,11W2,10W2,11W3,10W3,11W4,10W4,11W5,10W5,110.2010.198-0.211-0.0990.40.308-0.202-0.0980.101-0.100W6,10W6,11W7,10W7,11W8,10W8,11W9,10W9,11W10,12W11,120.092-0.211-0.4000.1980.2010.190-

7、0.1100.300-0.304-0.099101112-0.2870.1790.344第三題：3.a) Suppose the fraction of undergraduate students who smoke is 15% and the fraction of graduate students who smoke is 23%. If one-fth of the college students are graduate students and the rest are undergraduates, what is the probability that a studen

8、t who smokes is a graduate student?b) Given the information in part (a), is a randomly chosen college student more likely to be a graduate or undergraduate student?c) Suppose 30% of the graduate students live in a dorm but only 10% of the undergraduate students live in a dorm. If a student smokes an

9、d lives in the dorm, is he or she more likely to be a graduate or undergraduate student? You can assume independence between students who live in a dorm and those who smoke.答：a) 定義：A=A1 ,A2其中A1表示沒有畢業(yè)的學(xué)生，A2表示畢業(yè)的學(xué)生，B表示抽煙則由題意而知：P(B|A1)=15% P(B|A2)=23% P(A1)=4/5 P(A2)=1/5 則問題則是求P(A2|B)由則b) 由a可以看出隨機抽取一個抽

10、煙的大學(xué)生，是畢業(yè)生的概率是0.277，未畢業(yè)的學(xué)生是0.723，所以有很大的可能性是未畢業(yè)的學(xué)生。c) 設(shè)住在宿舍為事件C則P(C|A2)=30% P(C|A1)=10% =0.4所以由上面的結(jié)果可以看出是畢業(yè)生的概率大一些第四題：4. Suppose that the data mining task is to cluster the following ten points (with(x, y, z) representing location) into three clusters:A1(4,2,5), A2(10,5,2), A3(5,8,7), B1(1,1,1), B2(2

11、,3,2), B3(3,6,9), C1(11,9,2), C2(1,4,6), C3(9,1,7), C4(5,6,7)The distance function is Euclidean distance. Suppose initially we assign A1, B1, C1 as the center of each cluster, respectively. Use the K-Means algorithm to show only(a) The three cluster center after the first round execution(b) The fina

12、l three clusters答：a) 各點到中心點的歐式距離第一輪：A1B1C1A2549817A34110162B2146165B33393122C21434141C33010093C4217770從而得到的三個簇為：A1, A3,B3,C2, C3, C4 B1,B2 C1,A2所以三個簇新的中心為：(4.5,4.5,6.83)，(1.5,2,1.5)，(10.5,7,2)第二輪：新的簇均值為：(4.5,4.5,6.83)，(1.5,2,1.5)，(10.5,7,2)(4.5,4.5,6.83)(1.5,2,1.5)C1(10.5,7,2)A19.18.576.25A253.86111

13、81.54.25A312.5277878.556.25B158.527781.5127.25B231.861111.588.25B39.74.5106.25C185.86111139.54.25C213.1944424.5115.25C332.5277887.563.25C42.58.556.25所以得到的新的簇為：A1, A3,B3,C2, C3, C4 B1,B2 C1,A2得到的新的簇跟第一輪結(jié)束得到的簇的結(jié)果相同，不再變化，所以上面的簇是最終的結(jié)果。Part II: LabQuestion 1 Assume this supermarket would like to promote

14、milk. Use the data in “transactions” as training data to build a decision tree (C5.0 algorithm) model to predict whether the customer would buy milk or not. 1. Build a decision tree using data set “transactions” that predicts milk as a function of the other fields. Set the “type” of each field to “F

15、lag”, set the “direction” of “milk” as “out”, set the “type” of COD as “Typeless”, select “Expert” and set the “pruning severity” to 65, and set the “minimum records per child branch” to be 95. Hand-in: A figure showing your tree.2. Use the model (the full tree generated by Clementine in step 1 abov

16、e) to make a prediction for each of the 20 customers in the “rollout” data to determine whether the customer would buy milk. Hand-in: your prediction for each of the 20 customers.3. Hand-in: rules for positive (yes) prediction of milk purchase identified from the decision tree (up to the fifth level

17、. The root is considered as level 1). Compare with the rules generated by Apriori in Homework 1, and submit your brief comments on the rules (e.g., pruning effect)答：1生成的決策樹為：生成的決策樹模型為：juices = 1 Mode: 1 water = 1 Mode: 1 = 1 water = 0 Mode: 0 pasta = 1 Mode: 1 = 1 pasta = 0 Mode: 0 tomato souce = 1

18、Mode: 1 = 1 tomato souce = 0 Mode: 0 biscuits = 1 Mode: 1 = 1 biscuits = 0 Mode: 0 = 0 juices = 0 Mode: 0 yoghurt = 1 Mode: 1 water = 1 Mode: 1 = 1 water = 0 Mode: 0 biscuits = 1 Mode: 1 = 1 biscuits = 0 Mode: 0 brioches = 1 Mode: 1 = 1 brioches = 0 Mode: 0 beer = 1 Mode: 1 = 1 beer = 0 Mode: 0 = 0

19、yoghurt = 0 Mode: 0 beer = 1 Mode: 0 biscuits = 1 Mode: 1 = 1 biscuits = 0 Mode: 0 rice = 1 Mode: 1 = 1 rice = 0 Mode: 0 coffee = 1 Mode: 1 water = 1 Mode: 1 = 1 water = 0 Mode: 0 = 0 coffee = 0 Mode: 0 = 0 beer = 0 Mode: 0 frozen vegetables = 1 Mode: 0 biscuits = 1 Mode: 1 pasta = 1 Mode: 1 = 1 pas

20、ta = 0 Mode: 0 = 0 biscuits = 0 Mode: 0 oil = 1 Mode: 1 = 1 oil = 0 Mode: 0 brioches = 1 Mode: 0 water = 1 Mode: 1 = 1 water = 0 Mode: 0 = 0 brioches = 0 Mode: 0 = 0 frozen vegetables = 0 Mode: 0 pasta = 1 Mode: 0 mozzarella = 1 Mode: 1 = 1 mozzarella = 0 Mode: 0 water = 1 Mode: 1 biscuits = 1 Mode:

21、 1 = 1 biscuits = 0 Mode: 0 brioches = 1 Mode: 1 = 1 brioches = 0 Mode: 0 coffee = 1 Mode: 1 = 1 coffee = 0 Mode: 0 = 0 water = 0 Mode: 0 coke = 1 Mode: 0 coffee = 1 Mode: 1 = 1 coffee = 0 Mode: 0 = 0 coke = 0 Mode: 0 = 0 pasta = 0 Mode: 0 water = 1 Mode: 0 coffee = 1 Mode: 1 = 1 coffee = 0 Mode: 0

22、= 0 water = 0 Mode: 1 rice = 1 Mode: 0 = 0 rice = 0 Mode: 1 tunny = 1 Mode: 0 biscuits = 1 Mode: 1 = 1 biscuits = 0 Mode: 0 = 0 tunny = 0 Mode: 1 brioches = 1 Mode: 0 = 0 brioches = 0 Mode: 1 coke = 1 Mode: 0 = 0 coke = 0 Mode: 1 coffee = 1 Mode: 0 = 0 coffee = 0 Mode: 1 biscuits = 1 Mode: 0 = 0 bis

23、cuits = 0 Mode: 1 oil = 1 Mode: 0 = 0 oil = 0 Mode: 1 tomato souce = 1 Mode: 0 = 0 tomato souce = 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 = 12按照1中生成的據(jù)冊數(shù)進行預(yù)測的結(jié)果：4. 生成的關(guān)聯(lián)規(guī)則為：Question 2: Churn ManagementThe