nghiên cứu và phát triển một số thuật giải, mô hình ứng dụng khai thác dữ liệu 3

51

Vdi t~p phdbi€n {iI, i 2 , i3}, eo th€ t~o lu~t k€t h<;ipeo d~ng:

Co 50% khdch hang mua MANY(nhi~u) {il,i2},mua MANY (nhi~u) (ill.

1.7.4. Tun lu~t ke"th(/p cae ngii' canh khai thac du Ii~u mit [7]

GQi FFS(O,I,RF,{/-li};r,minsupp)la t~p h<;ipcae t?P ph6 bien cua ngii'dnh

khai iliac dii' li~u mo ung vdi bt) ham thanh VieD{J-li},gia tIi nguong chuy€n d6i

figii' dnh 1: va nguong minsupp. Vdi ba bt) ham tMnh VieD

{J..liMANY},{J-liAVER},{/-liFEw}cho tUng m~t hang iEI, co th€ t~o ra ba ngii' dnh khai

thac dii' li~u mo khac nhau va ttrdo sii' dt,mgcac thu~t giiii £lmt?P phd bi€n da

trlnh bay d cac ph~n tren d€ tlm cac t?p:

. FFSl= FFS(O,I,RF,{J-liMANY};r,mimsupp) ti'ng vdi bt) ham MANY

. FFSz=. FFS(O,I,RF,{/-liAVER};r,mimsupp)ti'ngvdi bt) ham AVERAGE

. FFS3= FS(O,I,RF,{J..liFEW};r,mimsupp) ti'ngvdi bi) ham FEW.

TIm SFl E FFSj, SFzEFF~z saDcho SFI2=SF1nSF27:0 va phan ra SF12

thanh cae t?P con X, Y khac r6ng cua SFI2saDcho SFI2=XuY va XnY=0 d€

t~o Iu?t ket h<;ipx~ Y giii'acac ngii'dnh khac nhau. N€u lu~t nay co dt) tin c~y

vu<;itngu'Ongminconf, thi co th€ cq.,caelu~t k€t h<;ipeo d~ng:
,,"",,9

. C6 56% khdch hang mua MANY(nhi~u) m(ithang X, thi se mua FEW (it) m(it

hang Y

1.8. DUNG LV! T KET H(1P DE PHAN LOP DU LltV VA M<1RQNG Ht

s6 PHt} THVQC THUQC TINH TRONG LY THVYET T~P THO [9]

1.8.1. Cae khai ni~m cd ban

lJinh nghia 1.22. Bang quyet dinh nhi phan

Xet ngii' dnh khai iliac dii' li~u (O,D,R) vdi 0 la t~p khae r6ng cac d6i

tu<;ing,D la t~p khac r6ng cac chi M.o ( thut)c tloh nhi phan), cho H va C la cae

t~p con khac r6ng cua D saDcho D=HuC , HnC=0, bi) ba (0, D=HvC,R) (hf\1e

gQi IIImi)t bang quy€t dinh nhi phan.



52

Bang 1.11: MQtvi dl,lv~ bang quye'tdinh nb! phan

Bang 1.11 Ia mOt vi dl,l v~ bang quye't dinh nhi phan voi

H={dl ,d2,d3,d4,d5} va C={cl ,c2}. ThuQc tinh cl xac dioh lOpam; thuQc tinh c2

xac dinh lop dlfdng.

Djnh nghia 1.23. Lu~t pMn lop ireD bang quye't d!nh ohi phao

rho bang quye't dinh nhi phan (0, D=HvC,R), gQiS la cac t~p coo khac,
ding cua H, lu~t pMn lop tren bang quye't dinh nhi phan (;6 d~ng S~ {c} voi

CEc. Ham pMn lop f dU'c;1c~o tit lu~t phan lop co d~ng f =1\ dEH" d va H' c H.

Vi dlJ1.8.MQts6lu~t pMn lop trangbang quye'tdinhnhipMn (jbang L11

RI:{d3,d4}~ {c2li;'R2:{d2,d5}~ {cl};R3: {d5}~ {el}

Cac ham pMo lop tu'ong t1ng Ia fl=d3 " d4; f2 = d2 J\ d5, f3=d5. E>6i tlfc;1ng0

thoa ham phan lop f ne'ua co chtta ta'tca cac chI baa co m~t trang H'.

1.8.2.Dq chinhxac cua ham phan lap

rho bangquye'tdiohnhi phan (0, D=Hr..£,R)trongdo cac d6i tu'c;1ngcua

0 du'c;1cxe'p vao hai lop. GQi0+ la t~p cae d6i tu'c;1ngcua 0 thuQcv~ lop e2 va 0-

la eac t~p cae d6i tu'c;1ngcua 0 thuQc v~ lop cl. rho f la mOtham phan lop, eo

th€ stl' dl,lngcac tieu chu~n san d€ xae dioh dO chinh xae cua ham phan lOp f

[24],[38],[48].

GQi TP={OEO+I f(a) dung}; FP = {oEO+1 f(a)sai}

dl d2 d3 d4 d5 c1 c2
01 1 0 0 1 0 1 0
02 0 1 0 1 0 0 1
03 0 0 1 1 0 0 1
04 1 0 0 0 1 1 0
05 0 1 0 0 1 1 0
06 0 0 1 0 1 0 1

07 0 1 0 0 1 1 0
08 0 0 1 1 0 0 1



53

TN ={ 0 E 0-' reO)dung }; FN ={ 0 E 0-' f(o) sai }

Be>chinh xac ciia phan lop c I dtt<;1ctinh bAng Gong thti'c:

11N!
ITPI+I1N1 (1.5)

Be>chinh xac cua phan lop c2 dtt<;1etinh bAng Gong thti'c

IIPI

I TP I + 11N I (1.6)

Vi dlJ.1.9. Voi bang quye'tdinh nbi phan trong bang 1.11

. Xet lu?t phan lop cl : {d2,d5}~ {c1}voi f= d2 J\ d5

0+ ={02, 03,06, 08} ti'ng voi c2; O. ={oI, 04, 05, 07} ti'ng voi c1

TP ={o E 0+1 reo) dung}=0;

FP= {oEO+1 f(0)sai}={02,03,06,08}

TN ={ 0 E 0.1 reo) dung }=~05, 07};

FN ={ 0 E 0- I f( s) sai } ={0 I, 04 }

B6 chinhxac phan lopc1 11NI I{o5,o7}I =10
. ITPI+I1N1 101+I{o5,o7}1 '

. Xet lu?t phan lop e~'d~ng {d3,d4}~ {e2}voi f =d3 J\ d4:

0+ ={02,03, 06, 08} ung voi c2; O. ={01, 04, 05, 07} ti'ngvoi cl

TP ={ 0 E 0+ I reo) dung}={ 03,08}

FP= {o E 0+ I f(o)sai }={02,06}

TN ={ 0 E 0.1 res) dung }=0

FN={oEO.1 f(s)sai} ={ol, 04,05, 07}

Be>chinh xac phan lop c2 - ITPI = I{oJ.o8}I -1,0
ITP I+! nv I I{oJ,oS}I+ 101

1.8.3. Dung lu~t ke't hc1plam lu~t phan lop dii' Ii~u

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'
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 thuQctinh 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 dQchinh 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=OJ2va 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

Trong doco cae lu~t phan ldp dung 100% If!: rl,r2,r5,r6.



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'uh.,n

khac r5ng cac thuQc tinh roi r~c. GQidom(a;) Iii ffii~ngia tri cua thuQctmh 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).

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 Iiicac c~p <a,v> voi a E A va

v=o( {a}). E>6itu'<;1ngI trong bang 1.12 tu'dngU'ngvoi vector d~c trung cho d6i

tu'<;1ng«a,l>,<b,4>,
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'dngdu'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'<;1cgQihi mQt

bang quy€t dinh trong 19 thuy€t t~p tho. T~p HR du
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€t t~ptho vdi H={a,b} va C={c}.

57

Dink ngkia 1.27. Xa'pxl 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}
.

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
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~u3.2, rho: U={a, b} va V={c;},

hay tinh Y(U,V)?

a) V8i U={a, b }se eo cae 18pttfdng 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 18ptudng 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
,~'

V~y h~ 86 pht;1thuQc thuQc tinh cua V vao U la 1,0 hay V pht;1thue}choan

toan vao U.

1.8.4.2. Mil TQnghi sit ph1;lthuQc thuQc linh [9J

Phin nay trlnh bay cd sd 19 lu~n dE dinh nghia va tinh tmin h~ s6 pht;1

thuQc thue}ctinh md fe}ng.

Dinh nghia 1.29. Ham phan anh muc de}bao ham

Cho ngU'ongdo mue dQ bao ham 8e[0,1], gQi ~(S,T) la ham phan anh

mue dQbao ham cua Strong T, ham ~(S,T) dU<;fC(t!nhnghia nhu san:



59

J.lc (S,T) =IS II T)IIISI (1.12)

Neu J.lc(S,T);::: 8, thi t~p h<;1pS du'
baa ham la 8. Neu 8=1,0 thi S c T

Dtnh nghia 1.30. Xa'pXldu'oimd fQng

Vdi dinh nghla cila ham philo anh mue dQ baa ham, co th~ dinh nghia

Xa'pXlmo fQngB**(X)trong Iy thuyet t~p tho nhu'sau:

B**(X) ={ u E 0 I J.lc([U]ind(B),X);::: 8 J\ U EX} (1.13)

Dtnh nghia 1.31. H~ s6 ph\! thuQcthuQctfnhmd fQng

H~ s6 ph\! thuQcmo fQngdu'<;1cdinh nghla qua ham phan anh mue dQbaa

ham. Cho hai t~p thuQctinh U va t~p thuQctinh V, M s6 ph\! thuQcthuQctinh mo

fQngcila V vao U du'<;1cky hi~u Ia '¥ (U,V)va du'<;1cd!nh nghia nhu'sau:

'¥ (U,V) = II U..(X)l1!0 I
XeO/V

(1.14)

Vi dl,lI.13 sail day neu leDkha Dangphan ldp cila h~ s6 ph\! thuQcthuQc

tinh md fQng.
~{'

Vi dl} 1.13: Xet bang quyet dinh 1.12,cho U={b} va V={c}, ta co:

. Voi U={b} se co cae lop tu'dngdu'dng:

[01]ind(U)=[02]ind(U)=[03]ind(U)=[08]ind(U)={01,02,03,08}

[04]ind(U)=[05]ind(U)=[06]ind(U)=[07]ind(U)={04,05,06,07}

. Voi V= {c}se eo cae lop tu'dngdu'dng:-

[ol]ind(B)=[04]ind(B) =[05]ind(B)= [07]ind(B)= {ol,04.05, 07}

[02]ind(B)=[03]ind(B)=[06]ind(B)= [08]ind(B)= {o2,03, 06, 08}

Dung h~ s6 ph\! thuQcthuQctinh truy~n th6ng y(U,V)= II U.(X) 1/101 =0
'eO/!



60

Trong 1:9thuytt t~p tho khi y(U,V)=Oco nghla l?iV khong ph\,!thuQcVaG

U, nhung theo yeu cftu cua pIlau lap gftndung v~n co th8 suy fa duQCV tIcU.

Tit hai lu~t phan ldp :

<b, 4> ~ <c,7>, dQchfnh xac cua pMn lap =0,75

<b, 5>~ <C,6>, dQchinh xaccua pMn Idp =0,75

D\fa VaGnh~n xet tren, lu~n an md fQngkhai ni~m xa'p XlduOicua t~p tho

nh~m (ijnh nghla h~ s6 ph1,1thuQcthuQctinh md fQng \fI(U,V).

Vdi cac t~p cd sd cua phan ho~ch ON va muc dQbaa ham e =0,75:

Vdi Xl= {oI,04.a5, a7}, U..(XI)={a4,05, 07}

Vdi X2= {02,03,06, 08}, U..(X2)={a2, 03, 08}

\fI (U,V) = II U..(X) I/ 10 I = (I {04,05, a7}1+I{02,03,08} I)/101=6/8=0,75
XeOIV ",

Do v~y M s6 ph1,1thuQcthuQctinh md fQngco kha DangpMn ldp t6t hdn

h~ s6 ph1,1thuQcthuQctinh truy~n th6ng, d~cbi~t l?icac pMn lap g~n dung [91.

Nhq.n xet:Khi nguong do mue dQbaa ham 8=1,0 thl '¥ (U,V) =y(U,V).

1.8.4.1. Chuyin tl/Jibang quye'Fi1/nhtTongIi thuylt tljp tho sang bang quylt dink

nhjphlin

IAII

Cho h~ th6ng thong tin (O,A=HRuHC,fs), V=Udom(a,), gQiD Ia t~p h;=1

cac em baa d= <a,v>eAxV va thoa ham is. Tit (O,A=HRuHC,fs) t~o quaDh~ hai

ngoi RcOxD, saDcho 0 R d <=>o(a)=v va d=<a,v>.

Bang 1.1I Ia bang quytt dinh nhi pMn du

dinh truy~n th6ng (bang 1.12)vdi cac chIbaa d nhu san:

dl=<a,l> ; d2=<a,2>; d3 = < a,3>; d4=<b, 4>; d5=<b, 5>; cl=<c,6>; c2=<c,7>

Xet ham attributes duQcdinh nghla nhu san:



61

v SeD, attributes(S)={ a e A I<a,V>-e S } (1.15)

Ham attributes d~ la'yten cac thuQctinh trong t~p con S cac chi baa cua D.

Tinh chat 1.6: Voi c~p ham (p, A) dfi dtnh nghia a ireD, gQi U eA va OIU la

mQtphilo ho~ch o thee quaDh~ ba't kha philo ind(U) va U1,Uz,., Ukla cac ~p cd

sa cua philo ho~ch OIU thi p(A(Uj»=UjVj=I,...,k.

Vi dl} 1.14: Voi U={a, b} va t~p cd sa cua phan hOi;lChOIU ung voi lop tttdng

du'dng U5=[o5]ind(U)=[o7]ind(U)={o5,o7} du'<;1cxac dtnh bai: <a,3> va <b, 5>.

Theo cach ma hoa ireD,hai chi baa tttdng ung la d2=<a,2>; d5=<b, 5>. Dung c~p

ham p,A da du'<;1cdtnh nghia aireD, ta co:

A(05, o7)={d2,d5,cl}; p(A-(o5, 07») = p({d2,dS,cl})={o5,o7}= U5

1.8.4.4. Tinh hf srfphI} thul)c thul)c tinh md rl)ng qua dl) tin cljy va dl)phil bitn

cua luat kit hd,rp "-.. ,
Rtl dl 1.1: Cho SeD va TeD, muc dQcua peS)bao ham trong peT) du'<;1ctlnh:

J.Ic(p(S) ,peT»~ =Ip(S) tlp(T)llIp(S)1 =CF(S-+ T) (1.16)

-.}-
.,~.

Dinh Ii 1.7([9]).Cho (O,A=HRuHC,fs) la bang quye't dtnh va bang chuy~n d6i

quye't dtnh nht philo (O,D=HuC,R) tttdng ung, gQiU va VIa hai t?P h<;1pcon cua

A, Uj la cac t?P cd sa cua philo hOi;lChOIU va X la t?P cd sa cua philo hOi;lCh

ON, J la t~p cac chi s6 sao rho VjeJ, !lc(Uj,X)~ e thi:

'I' (U,V) = I I(CF(A.(Uj)-+A,(X»*SP(A,(Uj)))
XeOlVjeJ

(1.17)

Trang do D la t~p chi baa cua bang quye't dtnh nht phan (O,D,R) dtt<;1c

chuy~n d6i tITbang quye'tdtnh (O,AJs).



62

Chung minh: Gqi J Ia ~p cac chi s6 saGcho 'v'jeJ, J.1c(Uj,X);::e voi l!j Ia ~p cd

sd cua phin ho~ch 01U, co th€ tinh I(U (X»I bhg:

I(U (X»I = IIUj(JXI
jeJ

Do l(Uv cD, A.(X)g), lu~t ke't h<;1pA.(Uj)-+A.(X)di'idu<;1etlnh dQph6

bie'n va dQtin c~y Den CF(A(Uj)-+A,(X»= Ip(A,(Uj»(\ p(A,(X)l/lp(A(Uj»1.Theo

tlnh cha't 1.6 do Uj va X la cac t~p co sd eua phin ho~ch Den p(A(Uj»=Ujva

p(A(X)=X,do v~y Ip(A.(Uj»n p(A.(X)I=IUjn XI = CF(A(Uj)~A(X»* IV). Ngoai

fa, dQ ph6 bie'n cua ~p h<;fpA(Uj)Ia SP(A,(Uj»= Ip(A(Uj))I/IOI=IUpIOI,Den

IUjl=SP(A(Uv)* 101. Tom l~i:IUjn XI =CF(A(Uj)~A(X»* SP(I..(Uj»* 101

Ne'uA.(Uj ) la t~p ph6 bie'n va A(Uj)~A(X) la lu~t ke'th<;fp,co th€ tlnhh~

s6 ph1:lthuQcthuQc tinh md rQng nhu san:

'¥ (U,V)= I I( GtF(A(U)~ A(X»*SP(A(Uj)))
XeD/V jeJ

1.8.4.5 Xliytb!ng thuQ.tgiai dJ!a tren hi siJphlJ. thllQCthuQc tilllz mll TQng

Cho bang quye't dinh (O,A=HRuCR,fs) va nglliJng dQ ehlnh xae cua phin

~.

lOpminprecisione[O,I], fun cae lu~t'phin lop S~T voi S ~HR va TcCR, saGtho

do chlnh xae cua lu~t phin lop S~ V Ion hon ho~c bing minprecision. Cho bang

quye't dinh (O,A=HRuCR,fs), gQi (O,D=HuC,R) la bang quye't djnh nb! phin

dU<;fCehuy~n d6i tU bang quye't djnh (O,A=HRuCR,fs). Cho trUoc cac nglliJng

minsupp, minconf, minprecision. GQi FS(O,D=HuC,R,minsupp) la t~p cae t~p

ph6 bie'n cia (O,D=HuC,R) va R(O,D=HuC,R,minsupp,mincont) la t~p cae lu~t

ke't h<;fpeo d~ng lu~t phin lop S ~ T, saGcho S~H va Tcc.A=Huc.

Thu~t giai 1.11. san dfty sa d1:lngh~ s6 ph1:lthuQcthuQetinh md rQngd~

tlm lu~( phan Idp dli li~u.



63

Thu4t giiii 1.11: TIm lu~t phan lop dt!a tren h~ 56ph1:1thuQcmd rQng

Vao: Bang quy~t djnh (O,A=HR0CR,fs)

NgU'Ongminsupp, mineonf, minpreeision

Ra: T~p cae lu~t phan lop S ~ T, sac cho S c H, T c C, A=HuC, ngU'Qngphan

lOp la minprecision.

BlIUc 1: Chuy~n bang quy~t dtnh (O,A=HRuCR,fs) sang bang quy€t djnh nht

phan (O,D=HuC, R)

BlIf1c 2: Tinh FS(O,D=HuC,R,minsupp) va R(O,D=HuC,R,minsupp,minconf)

thee cae thu~t giai fun t~pph6 bi€n va lu~t k~t h
BlIUc 3: Phan hoi;1cht~p R(O,D=HuC,R,minsupp,mincont) ra cae nhom lu~it

phan lop S ~ T, co cac thuQc tinh trong t~p S gi6ng nhau va cae thuQc tmh

trong t~p T gi5ng nhau, gQiC={G!,Gz,...,Gd la cac nhom lu~t san khi phan lop.

,BlIUc 4: g6m cae b1foegall:

1) For each G E C do

2)

3)

4)

5)

6)

7)

8)

9)

10)

11)

12)

La'y rEG va r = S ~ T

GQi U=Attributes(S) v~V la Attributes(T)
:::::;':;1

/I Tinh '¥(U,V)

Psi=O

For each r: S ~ T va r EG do

Tinh CF(S ~ T) va SP(S) II dung thu~tgiai t1mlu~t ke't hc;1P

Psi=Psi +CF(S~ T)* SpeS)

Endfor /I r

If Psi ~ minprecision

Ghi (U,V) vao t~p KetQua

Endif

13)Endfor /I G



64

Vi dl!-minh h{Ja thuq.t gidi 1.11

Voi bang quytt dinh nhi phan (j bang 1,12, ngU'ong ph6 bitn t6i thi~u

minsupp=O,1. ngu'Ongtin c~y t6i thi~u III minconf=0,75, ngu'ong cmnh xac toi

thi~u Iii minprecision=O,75.Ung dl,mgcac thu~t giai rim Iu~t phan lop tit lu~t ktt

h
Nhom Gl:

. Lu~t ke't h<;1p{dl} 40 {el}

r1:<a, 1> ~ <c,6>

ThuQc tmh vt trai a, thuQc tinh vt phai c,

SP(rl)= 0,25 CF(rl)= 1,00 SP({dl D=0,25

. Lu~t ke't h<;1p{d3} 40 {e2}

r2:<a,3> ~ <c,7>

ThuQctinh vt trai a, thuQctinh,v€ phiii c.

SP= 0,38 CF= 1,00

SP(r2)= 0,38 CF(r2)= 1.00 SP({d3D=0,38

Tinh 'P({a},{C})=CF(rl)*SP( {dl} )+CF(r2)*SP({d3}}=0,63

Nhom G2:

. Lu~tktth
-.J
c~",'

r3:<b,4> ~ <c,7>

ThuQctinh vt trai b, thuQctinh vt phiii c.

SP(r3)= 0,38 CF(r3)= 0,75 SP({d4})=0,5

. Lu~tktth<;1p {dS} 40 {el}

r4:<b, 5> ~ <c,6>

ThuQc tinh vt trai b, thuQc tinh vt phiii c.

SP(r4)= 0,38 CF(r4)= 0,75 SP({d5})=O,5



65

\f'({b},{ c})=

CF(r3)*SP( {d4} )+CF(r4)*SP( {d5} }=0.5*0.75+0.5*0.75=0,75

Nhom G3:

. Lu~t ke't h<;1p{d1,d4} ~ {el }

r5:<a,l> * <b, 4> ~ < c, 6>

ThuQctinh ve tnIi a,b; thuQctinh ve ph:ii c.

SP(r5)= 0,13 CF(r5)= 1,00 SP({d1,d4})=0,125

. Lu~tketh<;1p{dl,d5} ~ {el}

r6:<a,I>*<b, 5> ~ <c,6>

ThuQctinh ve tnIi a, b; thuQctinh ve phai c.

SP(r6)= 0,13 CF(r6)= 1,00 SP( {dl,d5})=0,125

. {d2,d4} ~ {c2}

r7:<a,2>*<b, 4> ~ <c,7> ,
ThuQctinhve tnii a,b; thuQctinh ve phai c.

SP(r7)= 0,13 CF(r7)= 1,00 SP({d2,d4})=O,125

. Lu~tketh<;1p{d2,d5}~ {el}

r8:<a,2> * <b,5> ~ <c,6>

ThuQctinh ve tnii a, b; thuQctinh ve ph:ii c.

SP(r8)= 0,25 CF(r8)= 1,00 SP({d2,d5})=O,25

0 Lu~t ket h<;1p{d3,d4} ~ {c2}

r9:*<a,3>*<b,4> ~ <c,7>

Ten thuQc tinh ve tnii a,b; ten thuQc tinh ve ph:ii c.

SP(r9)= 0,25 CF(r9)= 1,00 SP({d3,d4})=O,25

. Lu~t ket h<;1p{d3,d5} ~ {c2}

rlO:<a,3>*<b, 5> ~. <c,7>

ThuQctinh ve treE a, b ; thuQctinh ve phai c.

66

SP(rlO)= 0,13 CF(rlO)= 1,00 SP( {d3,d5})=0,125

Tinh 'I'({a, b},{c})= CF(r5)*SP({dl,d4})+ CF(r6)*SP({dl,d5))+

CF(r7)*SP( {d2,d4}) + CF(r8)*SP( {d2,d5})+ CF(r9)* SP( {d3,d4})+

CF(rlO)*SP( {d3,d5})= 1,0

1.9.KET LU~N

Chu'c1ngnay phat tri~n cac thu?t giiii hi~u qua d~ tlm t~p ph6 bien va lu~t

ke't hQp trong CSDL biing cach ghlm dQphuc t~p cila nnh toaD va giam so lftn

truy c~p CSDL. Co hai lo~i thu~t gi.H du'Qcphat tri~n la thu~t giai khong tang

cu'ongva thU?tgiai tang cu'ong.

Trong thu~tgiai khong tang cu'ong, mo hlnh vector bi€u di~n t~p m~t hang

va baa dong d:i du\1Cd€ xu!t nhiim bi~u di€n CSDL thanh ngfi'canh nhi phan

niim trong bQnho may nnh va giam solu'c1ngt~p ung VieDdn tinh dQph6 bien

d~ DangcaDhi~u stIlt thu~t giai. ,

Trong thu~t giai tang cu'ong, thu~t giai (~O daD khai ni~m cila R. Godin

d:i du'Qcdi bien d€ fun t~p ph6 bie'n (it cac khai ni~m hlnh £huc £rong daD khai

ni~m. Thu~t ghHtren daDkhai ni~m ngoai kha Dang tang cu'ongcon co tnI di~m
"f,

la chi dn truy c~p CSDL mQ(Iftn'atiy nh!t la co th€ t~o daDkhai ni~m.

Ke' de'n la cac nghien CUumd rQng lu~t ke't hQptruy€n thong sang d~ng

lu~t ke't hQpphil d!nhva lu~tket hc;ipmo.

Cuoi clIngchttc1ngnay trlnh bay cac nghien cUudung lu~t ke't hc;iPlam lu~t

, phan lOpdfi'li~u va xay dl,l'ngh~ so ph1,1£huQcthuQctinh rod fQngtrong ly thuyet

t~p tho nhiim Dangcao khii Dangkhao sat mli'cdQph1,1thuQcgifi'acac ~p thuQc

tinh trong cae bai toaDphan lop dii' li~u g§n dung.