5002 assignment1

2.1 Wavelet Transform

Question 1

step1: Calculate the average of input samples

step2: Calculate the difference of input samples

step3: Regard averages as input sequence. Repeat step1 and step2 until only one output average.

Quention 2

Original Signal	1	4	2	3	-2	-1	2	1
1.Iteration	3.53	3.53	-2.12	2.12	-2.12	-0.71	-0.71	0.71
2.Iteration	5.00	0.00			0.00	-3.00
3.Iteration	3.54				3.54
Haar wavelet transformation	3.54	3.54	0.00	-3.00	-2.12	-0.71	-0.71	0.71

2.2 Principal Components Analysis

import numpy as np

dimension = 3
data_num = 7
data = [[-1, -2, -3, 1, 2, 3, 1], [-1, -1, -2, 1, 1, 2, 2],
        [1, 4, -2, 1, 2, 1, 4]]

data_mean = [np.mean(data[i]) for i in range(dimension)]

# normalize data by average
for i in range(0, dimension):
    for j in range(0, len(data[0])):
        data[i][j] -= data_mean[i]

# calculate covariance matrix according to definition
cov = []
for i in range(dimension):
    for j in range(dimension):
        sumOfRowMulti = 0
        for k in range(0, data_num):
            sumOfRowMulti += data[i][k] * data[j][k]
        cov.append(sumOfRowMulti / (data_num - 1))

cov_matrix = np.asarray(cov).reshape(dimension, dimension)
print 'covariance matrix is:'
print cov_matrix

# calculate eigen value and eigen vector using numpy function
eig_val, eig_vec = np.linalg.eig(cov_matrix)

for i in range(dimension):
    print 'eigen value', i, eig_val[
        i], ' corresponding eigen vector:', eig_vec[:, i]

# sort eigen value in descending order
eig_val_sorted = np.sort(eig_val)[::-1]

# calcuate propotion
# propotion=sum of eigen value of first k components/sum of all n eigen values
propotion = (eig_val_sorted[0] + eig_val_sorted[1]) / np.sum(eig_val_sorted)

print 'proportion of total population variance explained by the first two components: %.4f' % propotion

The result is:

image.png

3 Pattern Discovery

Question 1

Step 1: Discretization

Web page

1. Sorted data for web pages : 7,9,9,11,13,14,18,35,36,45
2. Partition into 4 equal-frequency (equal-depth) bins:

• web1: 7,9,9
• web2: 11,13
• web3: 14,18,35
• web4: 36,45

Session length

1. Find the max and min value of session length: 267,2017
2. Calculate the width of intervals: width = ( 2017 – 267 )/ 3≈583.4
3. Partition into 3 equal-frequency (equal-width) bins:

• len1 (267~850.4): 267,598,672
• len2 (850.3~1433.8): 898,998,1213,1345
• len3(1433.8~2017.2): 1543,1702,2017

Transactions table after discretization

Session ID	Country	Session Length	Web Pages	Buy
1	NA	len2	web1	no
2	AS	len3	web2	yes
3	EU	len1	web3	yes
4	EU	len2	web4	no
5	NA	len1	web1	no
6	AS	len2	web3	yes
7	AS	len3	web3	yes
8	AS	len1	web1	no
9	NA	len3	web2	no
10	EU	len2	web4	yes

Step 2: Apriori algorithm

1st scan

generate length 1 candidate itemsets

itemset	sup
NA	0.3
AS	0.4
EU	0.3
len1	0.3
len2	0.4
len3	0.3
web1	0.3
web2	0.2
web3	0.3
web4	0.2
yes	0.5
no	0.5

prune infrequent itemsets

itemset	sup
NA	0.3
AS	0.4
EU	0.3
len1	0.3
len2	0.4
len3	0.3
web1	0.3
web3	0.3
yes	0.5
no	0.5

2nd scan

generate length 2 candidate itemsets

itemset	sup	itemset	sup
NA,len1	0.1	EU,yes	0.2
NA,len2	0.1	EU,no	0.1
NA,len3	0.1	len1,web1	0.2
AS,len1	0.1	len2,web1	0.1
AS,len2	0.1	len3,web1	0
AS,len3	0.2	len1,web3	0.1
EU,len1	0.1	len2,web3	0.1
EU,len2	0.2	len3,web3	0.1
EU,len3	0	len1,yes	0.1
NA,web1	0.2	len2,yes	0.2
NA,web3	0	len3,yes	0.2
AS,web1	0.1	len1,no	0.2
AS,web3	0.2	len2,no	0.2
EU,web1	0	len3,no	0.1
EU,web3	0.1	web1,yes	0
NA,yes	0	web1,no	0.3
NA,no	0.3	web3,yes	0.3
AS,yes	0.3	web3,no	0
AS,no	0.1

prune infrequent itemsets

itemset	sup
NA,no	0.3
AS,yes	0.3
web1,no	0.3
web3,yes	0.3

3rd scan

generate length 3 candidate itemsets
no length 3 candidate

Step 3: Get frequent itemsets

{web3} (support = 0.3)
{web1} (support = 0.3)
{len3} (support = 0.3)
{len1} (support = 0.3)
{NA} (support = 0.3)
{EU} (support = 0.3)
{len2} (support = 0.4)
{AS} (support = 0.4)
{yes} (support = 0.5)
{no} (support = 0.5)
{web3,yes} (support = 0.3)
{web1,no} (support = 0.3)
{NA,no} (support = 0.3)
{AS,yes} (support = 0.3)

Question 2

Step 1: Discretization

same as question question 1

Step 2: Deduce the ordered frequent items

item	sup
no	0.5
yes	0.5
AS	0.4
len2	0.4
EU	0.3
NA	0.3
len1	0.3
len3	0.3
web1	0.3
web3	0.3

ID	Frequent Items
1	no,len2,NA,web1
2	yes,AS,len3
3	yes,EU,len1,web3
4	no,len2,EU
5	no,NA,len1,web1
6	yes,AS,len2,web3
7	yes,AS,len3,web3
8	no,AS,len1,web1
9	no,NA,len3
10	yes,len2,EU

Step 3: Construct FP tree

FP-tree

Step 4: construct conditional FP-tree for each item

Cond. FP-tree on "web3"

{web3} (support = 0.3)

web3

Cond. FP-tree on "web1"

{web1} (support = 0.3)

web1

Cond. FP-tree on "len3"

{len3} (support = 0.3)

len3

Cond. FP-tree on "len1"

{len1} (support = 0.3)

len1

Cond. FP-tree on "NA"

{NA} (support = 0.3)

NA

Cond. FP-tree on "EU"

{EU} (support = 0.3)

EU

Cond. FP-tree on "len2"

{len2} (support = 0.4)

len2

Cond. FP-tree on "AS"

{AS} (support = 0.4)

AS

Cond. FP-tree on "yes"

{yes} (support = 0.5)

yes

Cond. FP-tree on "no"

{no} (support = 0.5)

no

Step 5: Determine frequent patterns

{web3} (support = 0.3)
{web3,yes} (support = 0.3)
{web1} (support = 0.3)
{web1,no} (support = 0.3)
{len3} (support = 0.3)
{len1} (support = 0.3)
{NA} (support = 0.3)
{NA,no} (support = 0.3)
{EU} (support = 0.3)
{len2} (support = 0.4)
{AS} (support = 0.4)
{AS,yes} (support = 0.3)
{yes} (support = 0.5)
{no} (support = 0.5)

Question 3

closed frequent patterns

{web3,yes}
{web1,no}
{len3}
{len1}
{NA,no}
{EU}
{len2}
{AS}
{AS,yes}
{yes}
{no}

maximal frequent patterns

{web3,yes}
{web1,no}
{len3}
{len1}
{NA,no}
{EU}
{len2}
{AS,yes}