Interpretation of output from Weka for Apriori Algorithm

Our Dataset:


Best rules found in Weka using Apriori:

1. item5=t 2 ==> item1=t 2    <conf:(1)> lift:(1.5) lev:(0.07) [0] conv:(0.67)
2. item4=t 2 ==> item2=t 2    <conf:(1)> lift:(1.29) lev:(0.05) [0] conv:(0.44)
3. item5=t 2 ==> item2=t 2    <conf:(1)> lift:(1.29) lev:(0.05) [0] conv:(0.44)
4. item2=t item5=t 2 ==> item1=t 2    <conf:(1)> lift:(1.5) lev:(0.07) [0] conv:(0.67)
5. item1=t item5=t 2 ==> item2=t 2    <conf:(1)> lift:(1.29) lev:(0.05) [0] conv:(0.44)
6. item5=t 2 ==> item1=t item2=t 2    <conf:(1)> lift:(2.25) lev:(0.12) [1] conv:(1.11)
7. item1=t item4=t 1 ==> item2=t 1    <conf:(1)> lift:(1.29) lev:(0.02) [0] conv:(0.22)
8. item3=t item5=t 1 ==> item1=t 1    <conf:(1)> lift:(1.5) lev:(0.04) [0] conv:(0.33)
9. item3=t item5=t 1 ==> item2=t 1    <conf:(1)> lift:(1.29) lev:(0.02) [0] conv:(0.22)
10. item2=t item3=t item5=t 1 ==> item1=t 1    <conf:(1)> lift:(1.5) lev:(0.04) [0] conv:(0.33)

Rule 1: item5=t 2 ==> item1=t 2    <conf:(1)> lift:(1.5) lev:(0.07) [0] conv:(0.67)

INTERPRETATION OF RULES:
item5=t 2
Meaning: item5 is in two transactions.

item1=t 2
Meaning: item1 is in two transactions.

Confidence(X -> Y) = P(Y | X) = (# txn with both X and Y) / (# txn with X) 
Lift = P(A and B) / (P(A) * P(B))
A -> item5
B -> item1
= (2/9) / ((2/9) * (6/9))
= 1.5

Conviction(X -> Y) = (1 - supp(Y)) / (1 - conf(X -> Y))

Support(X) => Support(item5) = 2/9
Support(Y) => Support(item1) = 6/9
Support(x -> Y) = 2 / 9 
Confidence(item5 -> item1) = 1 

Conviction(X -> Y) = (1 - (6/9)) / (1 - (1)) = Division by zero error

Coverage (also called cover or LHS-support) is the support of the left-hand-side of the rule X => Y, i.e., supp(X). It represents a measure of to how often the rule can be applied. Coverage can be quickly calculated from the rule's quality measures (support and confidence) stored in the quality slot.

Leverage computes the difference between the observed frequency of A and C appearing together and the frequency that would be expected if A and C were independent. A leverage value of 0 indicates independence.

Leverage(A -> C) = support(A -> C) - support(A) * support(C)
Range: [-1,1]

Leverage(item5 -> item1) = support(item5 -> item1) - support(item5)*support(item5)

Leverage(item5 -> item1) = (2/9) - (2/9)*(6/9) = 0.074

Rule 2: item4=t 2 ==> item2=t 2    <conf:(1)> lift:(1.29) lev:(0.05) [0] conv:(0.44)

Confidence(X -> Y) = P(Y | X) 
(# of txns with Y and X) / (# of txns with X)

X = item4
Y = item2

(# of txns with Y and X) = 2
(# of txns with item4) = 2

Confidence = 2/2 = 1

Lift = P(A and B) / (P(A) * P(B))

A = item4
B = item2

P(A and B) = (2/9) / ((2/9) * (7/9))
(P(A) * P(B)) = (2/9) * (7/9)
Lift = (2/9) / ((2/9) * (7/9)) = 1.285


Rule 10: item2=t item3=t item5=t 1 ==> item1=t 1    <conf:(1)> lift:(1.5) lev:(0.04) [0] conv:(0.33)

item2=t item3=t item5=t 1
Meaning: Items 2, 3 and 5 are appearing together in one transaction.

item2=t item3=t item5=t 1 ==> item1=t 1
Meaning: Items 1, 2, 3 and 5 are appearing together in one transaction.

Confidence(X -> Y) = P(Y | X) 
RHS = (# txn with both X and Y) / (# txn with X) = 1 / 1 = 1

Lift = P(A and B) / (P(A) * P(B))

= (1/9) / ((1/9) * (6/9)) = 1.5

Leverage(A -> C) = support(A -> C) - support(A) * support(C)

support(Items 2, 3, 5) = 1/9
support(Item 1) = 6/9
support(Items 1, 2, 3, 5) = 1/9
Leverage(A -> C) = (1/9) - ((1/9) * (6/9)) = 0.037

Tags: Technology,Machine Learning,Weka,