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10 X X sup(x) (X Y ) sup(x Y ) conf (X Y ) sup(x Y )/sup(x ) IBM R.Agrawal [11] 3.2 [10] k k + 1 [12] k k C k k L k k 1. C 1 C 1 2. C 1 L 1 3. L 1 C C k % 4
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12 t/yes t/no m/yes m/no : [11] 3.1 (t/yes) (t/no) (m/yes) (m/no) {m/yes} {t/yes} 50/75 = 67% ({t/yes} ) 80/100 = 80% X Y X Y X Y S X S Y S XY 6
13 N X Y 2 T dep T dep = N (S XY S X ) 2 S X S Y (1 S Y )(1 S X ) (3.1) T dep 0 X Y α T dep < x 2 1 (α) X Y X Y 3.4 [13] [14] IF=THEN IF-THEN A IF-THEN Y1 1(Y1) A(X) 7
14 1 A 20 Y 0 10 Y 1 30 N 1 20 N 1 10 Y 0 30 N 1 20 N 0 20 Y 1 3.2: 1 1 A A : Y1 Y2 (Y2) A(X) A 1 A Y1 A Y2 A X k H(S) = H(X) = p i log k p i i=1 p i X k a i (1 i k) Y1 A 3.3 H(X) = 5/8 log 5/8 3/8 log 3/8 = 0.95 Y1 8
15 A A : Y2 H(X Y 1 = yes) = 2/4 log 2/4 2/4 log 2/4 = 1 Y1 H(X Y 1 = no) = 3/4 log 3/4 1/4 log 1/4 = H(X Y 1) = 4/8 H(X Y 1 = yes) + 4/8 H(X Y 1 = no) = H(X) H(X Y 1) = = Y1 A Y2 A 3.4 H(X) = 5/8 log 5/8 3/8 log 3/8 = 0.95 Y2 H(X Y 2 = yes) = 4/4 log 4/4 0/4 log 0/4 = 0 Y2 H(X Y 2 = no) = 1/4 log 1/4 3/4 log 3/4 =
16 H(X Y 2) = 4/8 H(X Y 2 = yes) + 4/8 H(X Y 2 = no) = H(X) H(X Y 2) = = Y2 A Y2 A Y2 A NP [15] (greedy algorithm) [16][17] (main) 1. D 2. SPLIT(D) (SPLIT( D)) 1. IF D THEN (2),(3) D D 1 D 2 5. SPLIT(D 1 ) 10
17 6. SPLIT(D 2 ) D D D (cross validation) 2 N 1. N 2. N N (2) 4. (2),(3) 1 N = (overfitting) 11
18 2 SQL [18] 3.5 k-means N 1 N x 1 x 2 D(x 1, x 2 ) D(C 1, C 2 ) C 1 C 2 D(C 1, C 2 ) D(C 1, C 2 ) = D(C 1, C 2 ) = D(C 1, C 2 ) = 1 n 1 n 2 min D(x 1, x 2 ) x 1 C 1,x 2 C 2 max D(x 1, x 2 ) x 1 C 1,x 2 C 2 x 1 C 1 x 2 C 2 D(x 1, x 2 ) 12
19 D(C 1, C 2 ) = E(C 1 C 2 ) E(C 1 ) E(C 2 ) E(C i ) = x C i (D(x, c i )) 2 2 [19] k-means k- c i k i=1 x C i (D(x, C i )) 2 k-means O(NK) O(N 2 ) k-means [20] 13
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36 [17] Quinlan, J. R. C4.5:Programs for Machine Learning. Morgan Kaufmann, 1993 [18] [19] (1) vol.18, no.1, pp.59-65, 2003 [20] 30
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