Cho bang quyet dinh to, D=Hl£,R) va cae ngtKJng minsupp, mine:onf, t1m cae lu~t ke't h<;1pco d~ng r: S~{ e}. voi c EC va S cH. Co th~ dl{aVaGlu~t
54
ke't hQpnay lam cae lu~t phan lOp dii'li~u. rho bang quye't d!nh (0, D=Hl£.R)
va cae ngu'<Jngminsupp, mineonf fun cac lu~t ke't h<;1pco dl:lng r: S~{e}. vdi ceC va S cR. Theo dinh nghi'a dQ tin c~y eua lu~t ke't hQp r: S~{e} la :
CF(r) IP(S)~~({C}) I va peS) Ia t~p cac d6i tu'Qngco ehua cae thuQc tinh trong
S, p({e}) la ~p cae d6i tu'QngthuQelOpc do do p(S)np({c}} se xae dinh cae d6i
tu'<;1ngthuQe Idp e va co chua cae thuQc nnh trong S. Ne'u e la ldp e2 thi Ip(S)()p({e2})1=TP, peS)=TP uTN hay Ip(S)1=ITPI+ITNIvi TPnTN=0. Noi cach khae:
ITNI
CF(S~{el })= ITP I+1TN I
ITPI
CF(S~{e2})= ITP I+1TN I
(1.7)
(LX)\ \
Nhqn xii: Co thE sa d~ng dQtin e~y cua lu~t ke't hQpd~ daub gia dQ chinh de
eua ham phan ldp
Vi d~ 1.10. Vdi bang quy~t dinh nb! phan trong bang 1.11, se co cae lu~t ke't h~p
.~'
thee ngtttJng ph6 bie'n t6i thi~u minsupp=OJ2 va nglliJng tin e~y t6i thiEu mineonf=O.7 rl:{dl} -> {ell; SP= 0.25 CF= 1.00 r2: {d3}-> {e2}; SP= 0.38 CF= 1.00 r3:{d4} -> {e2}; SP= 0.38 CF= 0.75 r4:{d5}-> {ell; SP= 0.38 CF= 0.75 r5:{d2,dS}-> {c1};SP= 0.25 CF= 1.00 r6:{d3,d4}-> {e2};SP= 0.25 CF= 1.00
55
1.8.4. Uimg Iu~t ke"t h(jp d~ md rqng h~ s6 ph~ thuQc thuqc tinh trong Iy
thuye't t~p tho
1.8.4.1. Cae khai ni?m cd ban trong Ii thuylt tqp tho
Ph~n nay sii' d~ng cac djnh ngma cd ban cua 1:9thuyet t~p tho lam cd sa
xiiy dlfng h~ s6 phl;1thuQcthuQctinh ma rQng [33],[79].
Dinh nghia 1.24: H~ th6ng thong tin
Cho t~p h<;1p0 hii'uh~n, khac r6ng cac t~p d6i ut<;1ngva A la t~p hii'u h.,n
khac r5ng cac thuQc tinh roi r~c. GQidom(a;) Iii ffii~ngia tri cua thuQc tmh aiEA
RAIl
va V=Udom(a;), ham is: O~AxV xac dinh ghi teiciia cac doi ttf<;1ngU'ngvoi cac
1=1
thuQc tinh cua A. H~ th6ng thong tin Iii bQ ba (O,A,fs). (adsbygoogle = window.adsbygoogle || []).push({});
Bang 1.12 MQt vi d~ v~ h~ thong thong tin
\
'.z~
Bang LI2.la mQt vi d1,lv~ h~ thong thong tin vdi O={01,02,03,04.o5,06, 07, 08} va A={a.b.c}.
Cho h~ th6ng thong tin (O,A,fs). BcA, ky hi~u neB) 130gici tri thuQctinh cua t~p thuQc tinh B U'ngvoi d6i tu'<;1ngu. M5i doi ttf(1ngCEO se U'ngvdi ffiQt vector d~c tntng cho doi ttf<;1ngc6 thanh ph§n Iii cac c~p <a,v> voi a E A va v=o( {a}). E>6itu'<;1ngI trong bang 1.12 tu'dng U'ngvoi vector d~c trung cho d6i tu'<;1ng«a,l>,<b,4>,<c,6». O/A a b c 01 1 4 6 02 2 4 7 03 3 4 7 04 1 5 6 ',05 2 5 6 06 3 5 7 07. 2 5 6 08 3 4 7
56
Dink ngkia 1.25. Quan h~ bit kha phan va phan ho~ch t~p d6i tu<;1ng
Cho h~ th6ng thong tin (O,A,fs), BcA, quail h~ bit kha phan ind(B) tren
t~p dO'i ttf<;1ng0 du'<;1cd!nh nghla nhu'sau:
'if B c A , 'if u, V EO, U ind(B) v ~ u(B) =v(B) (1.9)
Quan h~ bit kha phan ind(B) xac dinh hai d6i tu<;1ngu va v co cling gia tIi thuQctinh dO'ivoi tit d cae thuQetinh trong B ( u(B)=v(B » .
Cho BcA, co th~ ki€m ITaquail h~ bit kha phan ind(B) Ia mQt quail h~ tu'dng du'dng. Quan h~ bit kha phan ind(B) xae dinh mQt phan ho~eh t~p dO'i
tu'<;1ng0 thanh cae lop ttfdng du'dng. Vdi u E 0, k9 hi~u [U]ind(B) 130lOp ttfdng
du'dng eila u theo quail h~ ind(B) va O/B Ia phan ho<:1ehdu'<;1c1<:10tll quail h~
ind(B). M6i phgn tli eila phan ho~ch O/B du'<;1cgQiIa IDQlt~p co sa hay IDQtIdp
tu'dngduong.
Vi dlJ1.11: Vdi bang dii'Ii~u a bang 1.11 va B= {e} se co cae lop tu'dng du'ong:
. (jng vdi <e,6>
.,
[ol]ind(B)=[04]ind(B) =[~~1jnd(B)= [07]ind(B) = {ol ,04,05,07} e (j ng vdi <e,7>
[02]ind(B)=[03]ind(B) =[06]ind(B)= [08]ind(B) = {02, 03, 06, 08}
Dink ngkia 1.26: Bang quy€t dinh trong 19thuy€t t~p tho
Cho h~ thO'ngthong tin (O,A,fs), gQi HR va CR la cae t~p con khac r6ng
eila A sao cho A=HRuCR va HRi1CR=0, (0, A=HRuCR, fs» du'<;1cgQi hi mQt bang quy€t dinh trong 19 thuy€t t~p tho. T~p HR du<JcgQi la t~p cae thuQetinh di~u ki~n va CR la t~p cae thuQc tinh quy€t dinh. Bang 1.12. Ia IDQtvi d~lv~ bang quy€t d!nh trang 19thuy€tt~ptho vdi H={ a,b} va C={c}.
57
Dink ngkia 1.27. Xa'p xl t~p h<;fp
Cho h~ th6ng thong tin (O,A,fs), X, la cac t~p can khac r6ng cua 0, XcO va B la t~p con khac r6ng cua A, BcA. -BE 1!oe Iu'<;fngt~p X cae d6i tu'<;fngqua t?P B cac thuQc tinh, Z.Pawlak dung khai ni~m xa'p xi du'oi eua X qua B ky hi~u
la B.(Xr va xa'pxi tren eua X quaB kYhi~uIa B*(X)[79]. Cae xa'pxi du'oiva
tren B.(X) va B.(X) dtr<;fCdinhnghia nhu'sau:
B.(X) ={u EO I[U]ind(B)C X} . (adsbygoogle = window.adsbygoogle || []).push({});
B (X)= {U E o([U]ind(B) II X* 0 }
(1.10)
Dink nghia 1.28. H~ so' ph1,1thuQc thuQc tlnh
Cho tru'dc hai ~p con khac r6ng U, V cua ~p thuQc tlnh A, h~ sO' ph1,1 thuQc thuQc tinh cua t~p thuQc tmh V VaGt~p thuQc tinh U du'<;fCsa d1,1ngdE khao sat s1,1'ph1,1thuQc cua t~p thuQc tinh V VaGt~p thuQc tlnh U va du'<;fcdinh nghIa nhasau:
y(U,V) = LIU.(X)IIIOI
XeOIV (1.11)
-t.
Ph1,1thuQc thuQc"tihh cua V VaG U du'<;fCkj hi~u la: U~V , k. Voi k =1, t?P thuQc tlnh V bean loan ph1,1thuQC VaGt~p thuQc tlnh U. Voi k<I: V phtJ. thuQc mQt ph~n VaGU; Voi k=0: V bean loan khong ph1,1thuQc VaG U.
H~ so' ph1,1thuQc thuQc tinh y(U,V) du'<;fCsu-d1,1ngdE phan anh mti'c dQ ph1,1 thuQccua hai t~p thuQctinh [79].
Vi dl} 1.12. Vdi h~ th6ng thong tin d bang dii'li~u 3.2, rho: U={a, b} va V={c;},
hay tinh Y(U,V)?
a) V8i U={a, b }se eo cae 18p ttfdng dtfdng:
. {<a,I> ; <b,4>}: UI=[ol]ind(U)=[oI]
{<a,2> ; <b, 4>}: U2=[ 02]ind(U)=[02] .
58
. {<a,3>; ,<b,4>} : U3=[ 03]ind(U)=[08]ind(U)={03,08 }
. {<a,l >; <b,5>} : U4=[04]ind(U)=[04]
. {<a,2>; <b,5>}: U5=[05]ind(U)=[07]ind(U)= {05,07} . {<a,3>; <b,5> }: U5=[06]ind(U)={06}
b)V8i V= {c}se co cae 18p tudng dudng:
. (fng vdi <c, 6>
XI= [ol]ind(V)=[04]ind(V) =[05]ind(V)= [07]ind(V)= {01,04,05, 07}
. (fng vdi <c,7>
X2= [02] ind(V)=[03]ind(V)=[06]ind(V)= [08]ind(V)= {02, 03,06,08}
Bi tinh h~ s6 pht;1thuQccua thuQc tinh cua V vao U b~ng c6ng thU'c1.11,
dn tinhU*(X)vdix eON.
. VdiXl={01,04,05, 07}, U*(Xl)={01,04,05, 07}
. Vdi X2= {02, 03, o~, 08}, U*(X2)={ 02, 03, 06, 08}
y(U,V)= 2)u.(X)I/IOI-lu.(Xl)I+IU.(X2)1-
XeDif' 8 - 1,0
~f ,~' (adsbygoogle = window.adsbygoogle || []).push({});
V~y h~ 86 pht;1thuQc thuQc tinh cua V vao U la 1,0 hay V pht;1thue}choan toan vao U.