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Action Rules Discovery Systems: DEAR1, DEAR2, ARED, ….. by Zbigniew W. Ra ś

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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Action Rules Discovery Systems: DEAR1, DEAR2, ARED, ….. by Zbigniew W. Ra ś

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  1. Action Rules Discovery Systems: DEAR1, DEAR2, ARED, ….. by ZbigniewW. Raś

  2. 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:

  3. Action Rules Discovery (Preprocessing) 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

  4. System DEAR1 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

  5. System DEAR2 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)

  6. Cost of Action Rule 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.

  7. Cost of Action Rule 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)

  8. ARED (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

  9. ARED (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) = ?

  10. ARED (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}

  11. ARED (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}

  12. ARED (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}

  13. ARED 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

  14. ARED 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

  15. Object reclassification from class d1 to d2 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}]

  16. Object reclassification from class d1 to d2 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}]

  17. 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: 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]

  18. 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: 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}]

  19. Object reclassification from class d1 to d2 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}]

  20. ARED Algorithm 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)]

  21. Questions? Thank You

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