apriori principle example question and answer
DESCRIPTION
Apriori Principle example question and answer which is related to data mining algorithms. Answer is not guranteed. Refer on your own risk.TRANSCRIPT
1) a) Define what is “Apriori principle” and briefly discuss why Apriori principle is useful in
association rule mining.
Apriori Principle:-
If an item set is frequent, then all of its subsets must also be frequent,
Or
If an item set is infrequent, then all of its supersets must be infrequent.
Apriori principle reduces the number of candidate item sets in an association rule mining process by
eliminating the candidates that are infrequent and leaving only those that are frequent.
b) Compare and contrast FP-Growth algorithm with Apriori algorithm.
Apriori Algorithm FP-Growth Algorithm
Use Apriori property and join and prune property.
It constructs conditional frequent pattern tree and conditional pattern base from database which satisfy minimum support.
Due to large number of candidates are generated require large memory space.
Due to compact structure and no candidate generation require less memory.
Multiple scans for generating candidate sets. Database scanning happens twice only.
Execution time is higher than FP-Growth algorithm as time is wasted in producing candidates every time.
Execution time is less than Apriori algorithm.
2) Consider the market basket transactions given in the following table. Let min_sup = 40% and
min_conf = 40%.
a) Find all the frequent item sets using Apriori algorithm.
Minimum Support = 40%
Minimum Confidence = 40%
Transaction ID Items Bought
T1 A,B,C
T2 A,B,C,D,E
T3 A,C,D
T4 A,C,D,E
T5 A,B,C,D
C1
Item Number of Transactions Minimum Support
A 5 5/5=100%
B 3 3/5=60%
C 5 5/5=100%
D 4 4/5=80%
E 2 2/5=40%
L1
Item Number of Transactions
A 5
B 3
C 5
D 4
E 2
C2
Item Pairs Number of Transactions Minimum Support
A,B 3 3/5=60%
A,C 5 5/5=100%
A,D 4 4/5=80%
A,E 2 2/5=40%
B,C 3 3/5=60%
B,D 2 2/5=40%
B,E 1 1/5=20%
C,E 2 2/5=40%
C,D 4 4/5=80%
E,D 2 2/5=40%
L2
Item Pairs No of Transactions
A,B 3
A,C 5
A,D 4
A,E 2
B,C 3
B,D 2
C,D 4
C,E 2
E,D 2
AB & AC => ABC AB & AD =>ABD AB & AE => ABE AC & AD =>ACD AC & AE => ACE AD & AE => ADE
BC & BD =>BCD
CD & CE =>CDE
C3
Item Set Number of Transactions Minimum Support
A,B,C 3 3/5=60%
A,B,D 2 2/5=40%
A,B,E 1 1/5=20%
A,C,D 4 4/5=80%
A,C,E 2 2/5=40%
A,D,E 2 2/5=40%
B,C,D 2 2/5=40%
C,D,E 2 2/5=40%
L3
Item set Number of Transactions
A,B,C 3
A,B,D 2
A,C,D 4
A,C,E 2
A,D,E 2
B,C,D 2
C,D,E 2
ABC & ABD => ABCD ACD & ACE => ACDE
Item Set Number of Transactions
A, B,C, D 2
A, C, D, E 2
Sets of {A, B, C, D} & {A, C, D, E} are bought together most frequently.
b) Obtain significant decision rules.
Subsets of {A, B, C, D}
{A} {B} {C} {D} {A, B} {A, C} {A, D}
{B, C} {B, D} {C, D} {A, B, C} {A, C, D} {A, B, D} {B, C, D}
{A} => {B, C, D}
C= σ{A, B, C, D}/ σ{A}
=2/5 = 40% Confidence
{B} => {A, C, D}
C= {A, B, C, D}/ {B}
=2/3 = 66.66% Confidence
{C} => {A, B, D}
C= σ{A, B, C, D}/σ {C}
=2/5=40% Confidence
{D} => {A, B, C}
C=σ {A, B, C, D}/ σ{D}
=2/4=50% Confidence
{A, B} => {C, D}
C= σ{A, B, C, D}/ σ{A, B}
=2/3=66.66% Confidence
{A, C} => {B, D}
C= σ{A, B, C, D}/σ{A, C}
=2/5=40% Confidence
{A, D} => {B, C}
C= σ{A, B, C, D}/σ{A, D}
=2/4=50% Confidence
{B, C} => {A, D}
C=σ {A, B, C, D}/σ{B, C}
=2/3=66.66% Confidence
{B, D} => {A, C}
C= σ{A, B, C, D}/σ{B, D}
=2/2=100% Confidence
{C, D} => {A, B}
C= σ{A, B, C, D}/σ{C, D}
=2/4=50% Confidence
{A, B, C} => {D}
C= {A, B, C, D}/{A, B, C}
=2/3=66.66% Confidence
{A, C, D} => {B}
C= σ{A, B, C, D}/σ{A, C, D}
=2/4=50% Confidence
{A, B, D} => {C}
C= σ{A, B, C, D}/σ{A, B, D}
=2/2=100% Confidence
{B, C, D} => {A}
C=σ {A, B, C, D}/σ{B, C, D}
=2/2=100% Confidence
Subsets of {A, C, D, E}
{A} {C} {D} {E} {A, C} {A, D} {A, E}
{C, D} {C, E} {D, E} {A, C, D} {A, D, E} {A, C, E} {C, D, E}
{A} => {C, D, E}
C=σ{A, C , D, E}/σ{A}
=2/5=40% Confidence
{C} => {A, D, E}
C=σ{A, C, D, E}/σ{C}
=2/5=40% Confidence
{D} => {A, C, E}
C=σ{A, C, D, E}/σ{D}
=2/4=50% Confidence
{E} => {A, C, D}
C=σ{A, C, D, E}/σ{E}
=2/2=100% Confidence
{A, C} => {D, E}
C= σ{A, C, D, E}/σ{A, C}
=2/5=40% Confidence
{A, D} => {C, E}
C=σ{A, C, D, E}/σ{A, D}
=2/4=50% Confidence
{A, E} => {C, D}
C=σ{A, C, D, E}/σ{A, E}
=2/2=100% Confidence
{C, D} => {A, E}
C= σ{A, C, D, E}/ σ {C, D}
=2/4=50% Confidence
{C, E} => {A, D}
C= σ {A, C, D, E}/ σ {C, E}
=2/2=100% Confidence
{D, E} => {A, C}
C= σ {A, C, D, E}/ σ {D, E}
=2/2=100% Confidence
{A, C, D} => {E}
C= σ {A, C, D, E}/ σ {A, C, D}
=2/4=50% Confidence
{A, D, E} => {C}
C= σ {A, C, D, E}/ σ {A, D, E}
=2/2=100% Confidence
{A, C, E} => {D}
C= σ {A, C, D, E}/ σ {A, C, E}
=2/2=100% Confidence
{C, D, E} => {A}
C= σ {A, C, D, E}/ σ {C, D, E}
=2/2=100% Confidence
c) Derive the FP-Tree for the above transaction table.
Step 01
Support for each item.
A=5/5=100%
B=3/5=60%
C=5/5=100%
D=4/5=80%
E=2/5=40%
Transaction ID Items Bought
T1 A,C,B
T2 A,C,D,B,E
T3 A,C,D
T4 A,C,D,E
T5 A,C,D,B
TID:1 =>
TID:2 =>
NULL
A:1
C:1
B:1
A:2
C:2
B:1
NULL
D:1
B:1
E:1
TID:3 =>
TID:4 =>
TID:5 =>
A:3
C:3
B:1
NULL
E:1
D:2
B:1
A:4
C:4
B:1
NULL
E:1
D:3
B:1
E:1
A:5
C:5
B:1
NULL
E:1
D:4
B:2
E:1