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# Action Rules Discovery Systems: DEAR1, DEAR2, ARED, ….. by Zbigniew W. Ra ś - PowerPoint PPT Presentation

Action Rules Discovery Systems: DEAR1, DEAR2, ARED, ….. by Zbigniew W. Ra ś. Y = {x 2 , x 4 } Z = {x 1 ,x 2 ,x 3 ,x 4 ,x 5 ,x 7 }. LERS. (a, a 1 ) (a, a 2 ) (b, b 1 ) (b,b 2 ) ……….. (d,d 1 ) (d,d 2 ). atomic terms. Decision System S. r = [[(a, a 2 )*(b, b 1 )] → (d, d 1 )]

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## PowerPoint Slideshow about ' Action Rules Discovery Systems: DEAR1, DEAR2, ARED, ….. by Zbigniew W. Ra ś' - dara-conrad

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by

ZbigniewW. Raś

Y = {x2, x4}

Z = {x1,x2,x3,x4,x5,x7}

LERS

(a, a1)

(a, a2)

(b, b1)

(b,b2)

………..

(d,d1)

(d,d2)

atomic terms

Decision System S

r = [[(a, a2)*(b, b1)] → (d, d1)]

w  Y → w  Z

rule

sup(r) = 2

conf(r) = 2/2 = 1

Support:

Confidence:

Partition decision table S

Stable:{ a, b}

Flexible: {c, e, f}

Reclassification direction:

2 1 or 3 1

Splitting the node using the stable attribute

Dom(a) = {1,2,3} & Dom(b) = {1,2,3,4,5}

a = 1

a = 3

a = 2

All objects have the same value 8 for attribute f, so it is crossed out from the sub-table ( this condition is used for stable attributes as well)

T1

T3

All objects have the same decision value, so this sub-table is not analyzed any further

None of the objects contain the desired class “1”, so this sub-table stops splitting any further

T2

b = 5

b = 1

T5

T4

All the flexible values are the same for both objects , therefore this sub-table is not analyzed any further

Stable Attribute: {a, c}

Flexible Attribute: b

Decision Attribute: d

a = ?

a = 0

Table: Set of rules R with supporting objects

c = ?

c = ?

c = 1

a = 2

c = 0

a = ?

T6

T4

T5

c = ?

c = 2

Figure of (d, L)-tree T2

T3

(T3, T1) : (a = 2)  (b, 21) ( d, L  H)

(a = 2)  (b, 31) ( d, L  H)

T1

T2

Figure of (d, H)-tree T1

b = 1

b = 2

b = 3

Set of rules R with supporting objects

Stable Attribute: b

Flexible Attribute: {a, c}

Decision Attribute: d

d = L

d = H

(b = 1)  (a, 02) ( d, L  H)

(b = 1)  (c, 02) ( d, L  H)

(b = 1)  (c, 12) ( d, L  H)

Action rule r:

[(b1, v1→ w1)  (b2, v2→ w2)  … ( bp, vp→ wp)](x)  (d, k1→ k2)(x)

The cost of r in S:

costS(r) = {S(vi , wi) : 1  i  p}

Action rule r is feasible in S, if costS(r) < S(k1 , k2).

For any feasible action rule r, the cost of the conditional

part of r is lower than the cost of its decision part.

Example:

r = [(b1, v1 → w1) … (bj, vj → wj) …  ( bp, vp → wp)](x) 

(d, k1 → k2)(x)

In RS[(bj, vj → wj)] we find

r1= [(bj1, vj1 → wj1)  (bj2, vj2 → wj2)  … ( bjq, vjq → wjq)](x)

(bj, vj → wj)(x)

Then, we can compose r with r1 and the same replace

term (bj, vj → wj) by term from the left hand side of r1:

[(b1, v1 → w1)  … [(bj1, vj1 → wj1)  (bj2, vj2 → wj2)  … 

( bjq, vjq → wjq)] … ( bp, vp → wp)](x)  (d, k1 → k2)(x)

(a, a1 →a1)

(a, a2 → a2)

(b, b1 → b1)

(b, b2 → b2)

………..

(d, d1 → d1)

(d, d2 → d2)

Y = {x2, x4}

Z = {x1,x2,x3,x4,x5,x7}

Decision System S

atomic action terms

r=[(a, a2→ a2)*(b, b1→ b1)] → (d, d1→ d1)

(w, w)  (Y, Y ) → (w,w)  (Z, Z)

action rule

Support:

Confidence:

sup(r) = 2

conf(r) = 2/2 = 1

(a, a1 →a1)

(a, a2 → a2)

(b, b1 → b1)

(b, b2 → b2)

………..

(d, d1 → d1)

(d, d2 → d2)

Y = {x2, x4}

Z = {x1,x2,x3,x4,x5,x7}

Decision System S

atomic action terms

r=[(a, a2→ a1)*(b, b1→ b1)] → (d, d1→ d2)

(w1, w2)  (Y1, Y2) → (w1,w2)  (Z1, Z2)

Y=(Y1,Y2), Z=(Z1,Z2)

w = (w1,w2)

action rule

Support:

Confidence:

sup(r) = ?

conf(r) = ?

(a, a1 →a1)

(a, a1 → a2)

(b, b1 → b2)

(b, b2 → b2)

………..

(d, d1 → d1)

(d, d2 → d2)

atomic action terms

Decision System S

Y1→ Z1, Y2→ Z2

r=[(a, a2→ a1)*(b, b1→ b1)] → (d, d1→ d2)

(Y1, Y 2) (Z1, Z2)

action rule

sup(r) = 2

conf(r) = 2/2 = 1

Support:

Confidence:

Y1 = {x2, x4}

Z1 = {x1,x2,x3,x4,x5,x7}

Y2 = {x1, x6}

Z2 = { x6}

(a, a1 →a1)

(a, a1 → a2)

(b, b1 → b2)

(b, b2 → b2)

………..

(d, d1 → d1)

(d, d2 → d2)

atomic terms

Decision System S

r=[(a, a2→ a1)*(b, b1→ b1)] → (d, d1→ d2)

(Y1, Y 2) (Z1, Z2)

rule

sup(r) = 2

conf(r) = 2/2 = 1

Y1 = {x2, x4}

Z1 = {x1,x2,x3,x4,x5,x7}

Y2 = {x1, x6}

Z2 = { x6}

(a, a1 →a1)

(a, a1 → a2)

(b, b1 → b2)

(b, b2 → b2)

………..

(d, d1 → d1)

(d, d2 → d2)

atomic terms

Decision System S

r=[(a, a2→ a1)*(b, b1→ b1)] → (d, d1→ d2)

(Y1, Y 2) (Z1, Z2)

rule

sup(r) = 2

conf(r) = 2/2 = 1

Y1 = {x2, x4}

Z1 = {x1,x2,x3,x4,x5,x7}

Y2 = {x1, x6}

Z2 = { x6}

λ1 - minimum support, λ2 - minimum confidence

Object reclassification from class d1 to d2

λ1=2, λ2=1/4

Meaning of (d,d1  d2) in S:

NS(d,d1 d2)=[{x1,x2, x3, x4, x5, x7}, {x6}]

Atomic classification terms:

(b,b1b1), (b,b2b2), (b,b3b3)

(a,a1a2), (a,a1a1), (a,a2a2), (a,a2a1)

(c,c1c2), (c,c2c1), (c,c1c1), (c,c2c2)

stable attribute

flexible attributes

λ1 - minimum support, λ2 - minimum confidence

Object reclassification from class d1 to d2

λ1=2, λ2=1/4

Notation:

t1=(b,b1b1), t2=(b,b2b2), t3=(b,b3b3),

t4=(a,a1a2), t5=(a,a1a1), t6=(a,a2a2), t7=(a,a2a1),

t8=(c,c1c2), t9=(c,c2c1), t10=(c,c1c1),

t11=(c,c2c2),

t12 = (d,d1 d2).

stable attribute

flexible attributes

λ1=2, λ2=1/4

For decision attribute in S:

NS(d,d1 d2)=[{x1,x2, x3, x4, x5, x7}, {x6}]

For classification attribute in S:

Not marked λ1=3

NS(t1) = NS(b,b1b1) = [{x1,x2, x4, x6}, {x1,x2, x4,x6}]

Mark “-” λ2=0

NS(t2) = NS(b,b2b2) = [{x3,x7, x8}, {x3,x7, x8}]

Mark “-” λ1=1

NS(t3) = NS(b,b3b3) = [{x5}, {x5}]

Mark “-” λ2=0

NS(t4) = NS (a,a1a2) = [{x1,x6, x7, x8}, {x2,x3, x4,x5}]

λ1=2, λ2=1/4

For decision attribute in S:

NS(d,d1 d2)=[{x1,x2, x3, x4, x5, x7}, {x6}]

For classification attribute in S:

Not marked λ1=2

NS(t5) = NS(a,a1a1) = [{x1,x6, x7, x8}, {x1,x6, x7,x8}]

Mark “-” λ2= 0

NS(t6)= NS(a,a2a2) = [{x2,x3, x4, x5}, {x2,x3, x4,x5}]

Mark “+” λ1=4,λ2=1/4

NS(t7)= NS(a,a2a1) = [{x2,x3, x4, x5}, {x1,x6, x7,x8}]

λ1=2, λ2=1/4

For decision attribute in S:

NS(t12)=[{x1,x2, x3, x4, x5, x7}, {x6}]

For classification attribute in S:

Not marked λ1=3

NS(t1)=[{x1,x2, x4, x6}, {x1,x2, x4,x6}]

Marked “-” λ2=0

NS(t2)=[{x3,x7, x8}, {x3,x7, x8}]

Marked “-” λ1=1

NS(t3)=[{x5}, {x5}]

Marked “-” λ2=0

NS(t4)=[{x1,x6, x7, x8}, {x2,x3, x4,x5}]

Not marked λ1=2

NS(t5)=[{x1,x6, x7, x8}, {x1,x6, x7,x8}]

Marked “-” λ2=0

NS(t6)=[{x2,x3, x4, x5}, {x2,x3, x4,x5}]

NS(t7)=[{x2,x3, x4, x5}, {x1,x6, x7,x8}]

Mark “+” λ1=4, λ2=1/4

r = [t7 t1]

λ1=2, λ2=1/4

For decision attribute in S:

NS(t12)=[{x1,x2, x3, x4, x5, x7}, {x6}]

For classification attribute in S:

conf = 2/3 *1/5 <λ2

Not marked

NS(t8)= NS(c,c1c2) = [{x1,x4, x8}, {x2, x3, x5,x6, x7}]

Marked “-”

NS(t9) = NS(c,c2c1) = [{x2, x3, x5, x6, x7}, {x1, x4, x8}]

Marked “-”

NS(t10) = NS(c,c1c1) = [{x1, x4, x8}, {x1, x4,x8}]

Not marked

NS(t11) = NS (c,c2c2)=

[{x2, x3, x5, x6, x7}, {x2, x3, x5,x6, x7}]

λ1=2, λ2=1/4

Now action terms of length 2

from unmarked action terms of length 1

For decision attribute in S:

NS(t12)=[{x1,x2, x3, x4, x5, x7}, {x6}]

For classification attribute in S:

NS(t1)=[{x1,x2, x4, x6}, {x1,x2, x4,x6}] , NS(t5)=[{x1,x6, x7, x8}, {x1,x6, x7,x8}],

NS(t8)=[{x1,x4, x8}, {x2, x3, x5,x6, x7}], NS(t11)= [{x2, x3, x5, x6, x7}, {x2, x3, x5,x6, x7}].

Marked “-”,λ1=1

NS(t1*t5)=[{x1, x6}, {x1, x6}]

Marked “+”

NS(t1*t8)=[{x1, x4}, {x2, x6}]

Rule r = [t1*t8→t12], conf = 1/2 ≥ λ2, sup=2 ≥ λ1

Marked “-”,λ1=1

NS(t1*t11)=[{x2, x6}, {x2, x6}]

Marked “-”,λ1=1

NS(t5*t8)=[{x1, x8}, {x6, x7}]

Marked “-”,λ1=1

NS(t5*t11)=[{x6, x7}, {x6, x7}]

Marked “-”

NS(t8*t11)=[Ø, {x2, x3, x5, x6, x7}]

Object reclassification from class d1 to d2

λ1=2, λ2=1/4

For decision attribute in S:

NS(t12)=[{x1,x2, x3, x4, x5, x7}, {x6}]

For classification attribute in S:

Action rules:

[[(b,b1→b1)*(c,c1→c2)] → (d, d1→d2)]

[[(a,a2→a1] → (d, d1→d2)]

Thank You