## February 15, 2019

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Question:
For the following vectors x and y, calculate the cosine similarity and euclidean distance measures:
x =(4,4,4,4), y=(2,2,2,2)

Solution:

Cosine
x ● y = 4*2 + 4*2 + 4*2 + 4*2 = 32
||x|| = sqrt(4*4 + 4*4 + 4*4 + 4*4) = sqrt (64)   = 8
||y|| = sqrt(2*2 + 2*2 + 2*2 + 2*2) = sqrt (16) = 4
cos(x,y) = (x ● y) /  (||x||*||y||) = (32)/ (8*4)
cos(x,y) = 1

Euclidean
d(x, y) = sqrt((4-2)^2 + (4-2)^2 + (4-2)^2 + (4-2)^2)
Euclidean distance = 4

Question:
Consider the one-dimensional data set shown on the below table

 X 0.6 3.2 4.5 4.6 4.9 5.2 5.6 5.8 7.1 9.5 Y - - + + + - - + - -

Classify the data point x=5.0 according to its 3- and 9- nearest neighbors (Using majority Vote)

We need to first find the difference of each data set with respect to x=5.0, Refer the below table for the same.

 x X Difference (x & X) Y 5.0 0.6 4.4 − 5.0 3.2 1.8 − 5.0 4.5 0.5 + 5.0 4.6 0.4 + 5.0 4.9 0.1 + 5.0 5.2 0.2 − 5.0 5.6 0.6 − 5.0 5.8 0.8 + 5.0 7.1 2.1 − 5.0 9.5 4.5 −

Using 3- nearest neighbors method, 3 Closest points to the point x=5.0 will be the one who has least difference among them - > 4.9, 5.2, 4.6
Classes ->   +
Using Majority Vote, 3-nearest neighbor: +

Using 9- nearest neighbors method, 9 Closest points to the point x=5.0 will be the one who has least difference among them - > 4.9, 5.2, 4.6, 4.5, 5.6, 5.8, 3.2, 7.1, 0.6
Classes -> +  + +  +
Using Majority Vote, 9-nearest neighbor:

Question:
Suppose a group of 12 sales price records has been sorted as follows:
5; 10; 11; 13; 15; 35; 50; 55; 72; 90; 204; 215:
Partition them into three bins by each of the following methods.
(a) equal-frequency partitioning
(b) equal-width partitioning
(c) clustering

(a) equal-frequency (equidepth) partitioning:
Partition the data into equidepth bins of depth 4: [given as n=4]
Bin 1: 5, 10, 11, 13
Bin 2: 15, 35, 50, 55
Bin 3: 72, 90, 204, 215

(b) equal-width partitioning:
Partitioning the data into 3 equi-width bins will require the width to be (215−5)/3 = 70.
We get interval like- (1,70),(71,140),(141,210),(211,280)
Bin 1: 5, 10, 11, 13, 15, 35, 50, 55
Bin 2:72, 90
Bin 3: 204
Bin 4: 215

(c) clustering:
Using K-means clustering to partition the data into three bins we get
Bin 1: 5, 10, 11, 13, 15, 35
Bin 2: 50, 55, 72, 90
Bin 3: 204, 215

Question:

a. How do you evaluate a classifier when there is a class imbalance?
Answer: In normal case, accuracy and error rate can help. In case of class imbalance, we need specificity and sensitivity.

b. Assume that a search for ‘computer programming’ gave you a result of 100 web pages. Give 5-6 factors that would have been used by the search engine to determine the order in which the result pages are listed.
Frequency of the search terms in the pages, Place of occurrence – title/tags/paragraphs etc., no. of users visiting the page, no. of links to the page, geographical location, search-click history of user, domain’s importance, search term appearing in the domain etc.

c. How do clustering tendency, cluster validity help in data mining?
Clustering makes sense only if the data is non-random. Clustering tendency measures such as Hopkin statistic help. Cluster validity helps in evaluating clusters using unsupervised, supervised, or relative measures.

Question:
1.      A database has five transactions. Let min sup = 60% and min conf = 80%.       (5+2 marks)

 TID items bought T100 Bread,Butter,Beans,Potato,Jam, Milk T200 Bread,Butter,Shampoo,Potato,Jam, Milk T300 Beans,Soap,Butter, Bread T400 Beans, Onion, Apple, Butter, Milk T500 Apple, Banana, Jam, Bread,Butter
(a) Find all frequent itemsets using FP-growth algorithm.
(b) List all of the strong association rules (with support s and confidence c) matching the following buys(X; item1) ^ buys(X; item2) => buys(X; item3) [s; c]

Solution:NOTE: To solve this question, we will first go through 1 question below for your practice and then you can do it by yourself.

Reference question:

A database has five transactions. Let min sup = 60% and min conf = 75%.
 TID items bought T100 M, O, N, N, K, E, Y, Y T200 D, D, O, N, K, E, Y T300 M, M, A, K, E, E T400 M, U, C, C, Y, C, E, O T500 C, O, O, K, I, I, E

(a) Find all frequent itemsets using Apriori method.

Solution:
Database is scanned once to generate frequent 1-itemsets. To do this, I use absolute support, where duplicate values are counted only once per TID. The total number of TID is 5, so minimum support of 60% is equivalent to 3/5. Thus itemsets with 1 or 2 support counts are eliminated.

Table 1a. 1-itemset results, raw

Table 1b. 1-itemset results, consolidated

Now, database is scanned second time to generate frequent 2-itemsets. The possible combinations are 5!/(3!2!) = 10. Using absolute support, each combination is counted per TID, and combinations that are below support value of 3 are eliminated.
Table 2a. 2-itemset results, raw
Table 2a. 2-itemset results, consolidated

I proceed to scan the database again to generate frequent 3-itemsets. Sets {E, K}, {K, O}, {E, O} make {E, K, O} possible. Likewise, {E, O}, {E, Y}, {O, Y} make {E, O, Y}.
Table 3a. 3-itemset results

Frequent 4-itemsets cannot be generated, because sets {K, O, Y} and {E, K, Y} are missing. So, all frequent itemsets have been found.

(b) List all of the strong association rules (with support s=60% and confidence c=75%) matching the following metarule, where X is a variable representing customers, and itemi denotes variables representing items (e.g., “A”, “B”, etc.): buys(X; item1) and buys(X; item2) ) => buys(X; item3)
[s; c]

Solution:
The highest itemsets are {E, K, O} and {E, O, Y}. Thus, there can be 2(3!/(1!2!)) = 6 total possible association rules following the metarule of selecting 2 inputs for testing association with 1 output.
Association rules from {E, K, O}:
R1. E ∩ K -> O confidence = #{E, K, O} / #{E, K} = 3 / 4 = 75% Therefore, R1 is a strong association rule.
R2. E ∩ O -> K confidence = #{E, K, O} / #{E, O} = 3 / 4 = 75% Therefore, R2 is a strong association rule.
R3. K ∩ O -> E confidence = #{E, K, O} / #{K, O} = 3 / 3 = 100% Therefore, R3 is a strong association rule.
Association rules from {E, O, Y}:
R4. E ∩ O -> Y confidence = #{E, O, Y} / #{E, O} = 3 / 4 = 75% Therefore, R4 is a strong association rule.
R5. E ∩ Y -> O confidence = #{E, O, Y} / #{E, Y} = 3 / 3 = 100% Therefore, R5 is a strong association rule.
R6. O ∩ Y -> E confidence = #{E, O, Y} / #{O, Y} = 3 / 3 = 100% Therefore, R6 is a strong association rule.
In this case, all 6 association rules are strong, meaning that customers who purchase any of the two products among E, K, O are likely to purchase the remaining one, and customers who purchase two items among E, O, Y are likely to purchase the remaining one.

Question: Give appropriate solutions for the following 3+3=6 Marks

a. Suppose that the data for analysis includes the attribute age. The age values for the data tuples are 70, 20, 16, 16, 52, 15, 20, 21, 22, 25, 22, 30, 25, 46, 25, 33, 36, 35, 40, 35, 35, 33, 35, 45, 25, 19, 13.Use smoothing by bin means to smooth the above data, using a bin depth of 3. Illustrate your
steps.

Step 1: Sort the data. 13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30, 33, 33, 35, 35, 35,
35, 36, 40, 45, 46, 52, 70
Step 2: Partition the data into equal-frequency bins of size 3.
Bin 1: 13, 15, 16
Bin 2: 16, 19, 20
Bin 3: 20, 21, 22
Bin 4: 22, 25, 25
Bin 5: 25, 25, 30
Bin 6: 33, 33, 35
Bin 7: 35, 35, 35
Bin 8: 36,40, 45
Bin 9: 46, 52, 70
Step 3: Calculate the arithmetic mean of each bin.
Step 4: Replace each of the values in each bin by the arithmetic mean calculated for the bin.

Bin 1: 14.6, 14.6, 14.6
Bin 2: 18.3, 18.3, 18.3
Bin 3: 21, 21, 21
Bin 4: 24, 24, 24
Bin 5: 26.6, 26.6, 26.6
Bin 6: 33.6, 33.6, 33.6
Bin 7: 35, 35, 35
Bin 8: 40.3,40.3, 40.3
Bin 9: 56, 56, 56

b. Outliers are often discarded as noise. However, one person’s garbage could be another’s treasure. For example, exceptions in credit card transactions can help us detect the fraudulent use of credit cards. Taking fraudulence detection as an example, propose two methods that can be used to detect outliers and discuss which one is more reliable.
-> Using clustering techniques: After clustering, the different clusters represent the different kinds of data (transactions). The outliers are those data points that do not fall into any cluster. Among the various kinds of clustering methods, density-based clustering may be the most effective.
-> Using prediction (or regression) techniques: Constructed a probability (regression) model based on all of the data. If the predicted value for a data point differs greatly from the given value, then the given value may be consider an outlier.

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