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6-hop myrrh example (from Damian). Market agency targeting advertising to friends of customers: Entities: 1. advertisements 2. markets 3. merchants 4.

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Presentation on theme: "6-hop myrrh example (from Damian). Market agency targeting advertising to friends of customers: Entities: 1. advertisements 2. markets 3. merchants 4."— Presentation transcript:

1 6-hop myrrh example (from Damian). Market agency targeting advertising to friends of customers: Entities: 1. advertisements 2. markets 3. merchants 4. products 5. reviews 6. customers 7. friends The hop logic would be: 1. Advertisements target specific markets. 2. Markets have particular merchants 3. Merchants sell individual products. 4. Products have reviews. 5. Reviews are provided by customers. 6. Customers have friends. Sell=S(F,G) IsUsedBy= R(E,F) 0001 0010 0001 0100 1001 01111000 1100 1 2 3 4 E=Markets F=merchants 2345 1 2 3 4 G=products AA CC RatedAt(G,H) 0001 1010 0001 0101 H=rating 2345 AreGivenBy=U(H,I) 1001 01011000 1100 1 2 3 4 I=Customers IsAFriend?(I,J) 0001 1010 0001 0101 J=Person 2345 D=Ads 2345 targetsQ(D,E) 1101 00011101 1100

2 r r r v v r m r r v v v r r v m v v r v v r v FAUST Oblique. formula: P (X o d)<a X any set of vectors. To separate r s from v s : Using means_midpoints as cutpoints, calculate a as follows: a Viewing m r, m v as vectors ( e.g., m r ≡origin  pt_m r ), a = ( m r +(m v -m r )/2 ) o d = (m r +m v )/2 o d d D≡ mrmv.D≡ mrmv. Let d = D/|D|. What if d points away from the intersection,, of the Cut-hyperplane (Cut-line in this 2-D case) and the d-line (as it does for class=V, where d = (m v  m r )/|m v  m r | ? Then a is the negative of the distance shown (the angle is obtuse so its cosine is negative). But each v o d is a larger negative number than a=(m r +m v )/2 o d, so we still want v o d < ( 1*m v +1*m r ) o d 1 + 1 or v o d < ½(m v +m r ) o d Next we will take a std-based alternative linear combination of m v and m r

3 Oblique FAUST (means midpt level-0 case): R G ir1 ir2 means 62.83 95.29 108.12 89.50 1 48.84 39.91 113.89 118.31 2 87.48 105.50 110.60 87.46 3 77.41 90.94 95.61 75.35 4 59.59 62.27 83.02 69.95 5 69.01 77.42 81.59 64.13 7 R G irR1 ir2 stds 8 15 13 9 1 8 13 13 19 2 5 7 7 6 3 6 8 8 7 4 6 12 13 13 5 5 8 9 7 7 Oblique level-0 (Oblique without eliminating classes as they are predicted) 1's 2's 3's 4's 5's 7's True Positives: 322 199 344 145 174 353 False Positives: 28 3 80 171 107 74 NonOblique lev-0 1's 2's 3's 4's 5's 7's True Positives: 99 193 325 130 151 257 Class Totals-> 461 224 397 211 237 470 NonOblq lev-1 50% 1's 2's 3's 4's 5's 7's True Positives: 212 183 314 103 157 330 False Positives: 14 1 42 103 36 189

4 r r r v v r m r r v v v r r v m v v r v v r v P X dot d<a = P  d i X i <a FAUST Oblique coordinate-wise std ratio level-0: D≡ m r  m v, d=D/|D| To separate r from v: Using the coordinate-wise std ratios cutpoint, calculate a as follows: d Viewing m r, m v as vectors, a = ( m r + m v ) o d std r +std v std r std r +std v std v Just as there is no median for a set of vectors, there is no std either. What is meant by the purple expression above is, for each coordinate (dimension) one calculates the stds of those coordinate values of the r and v vector sets, the ratios with those coordinate values of m r and m v. Is that the same as projecting the r-set and v-set onto the d-line (using R o d and V o d) and then using the stds of those shadow lengths to adjust the cutpoint? This is different and is a better way to do it? The next slide shows this better approach.

5 r r r v v r m r r v v v r r v m v v r v v r v pm r | P X dot d<a = P  d i X i <a FAUST Oblique: X any set of vectors. D≡ m r  m v, d=D/|D| To separate r from v: Using the mean and std of the projections cutpoint, calculate a as follows: d r|r| r|r| |r|r |r|r |r|r pm v | v|v| v|v| |v|v |v|v |v|v a = pm r + (pm v -pm r ) = pstd r +pstd v pstd r pm r *pstd r + pm r *pstd v + pm v *pstd r - pm r *pstd r pstd r +pstd v By pm r, we mean this distance, m r o d, which is also mean{r o d|r  R} By pstd r, std{r o d|r  R} next? pm r + (pm v -pm r ) = pstd v +2pstd r 2pstd r pm r *2pstd r + pm r *pstd v + pm v *2pstd r - pm r *2pstd r 2pstd r +pstd v In this case the predicted classes will overlap (i.e., a given sample point may be assigned multiple classes) therefore we will have to order the class predictions.

6 Oblique FAUST (level-0 case): R G ir1 ir2 means 62.83 95.29 108.12 89.50 1 48.84 39.91 113.89 118.31 2 87.48 105.50 110.60 87.46 3 77.41 90.94 95.61 75.35 4 59.59 62.27 83.02 69.95 5 69.01 77.42 81.59 64.13 7 R G irR1 ir2 stds 8 15 13 9 1 8 13 13 19 2 5 7 7 6 3 6 8 8 7 4 6 12 13 13 5 5 8 9 7 7 Oblique level-0 using means of midpoints 1's 2's 3's 4's 5's 7's True Positives: 322 199 344 145 174 353 False Positives: 28 3 80 171 107 74 NonOblique lev-0 1's 2's 3's 4's 5's 7's True Positives: 99 193 325 130 151 257 Class Totals-> 461 224 397 211 237 470 NonOblq lev-1 50% 1's 2's 3's 4's 5's 7's True Positives: 212 183 314 103 157 330 False Positives: 14 1 42 103 36 189 Oblique level-0 using means and stds of projections (w/o class elimination) 1's 2's 3's 4's 5's 7's True Positives: 359 205 332 144 175 324 False Positives: 29 18 47 156 131 58 Oblique lev-0, means, stds of projs (with class elimination in 2,3,4,5,6,7,1 order) Note that no elimination occurs! 1's 2's 3's 4's 5's 7's True Positives: 359 205 332 144 175 324 False Positives: 29 18 47 156 131 58

7 P X dot d<a = P  d i X i <a FAUST Oblique: X any set of vectors. D≡ m r  m v, d=D/|D| To separate r from v: Using the mean and std of the projections cutpoint, and: r r r v v r m r r v v v r r v m v v r v v r v pm r | d r|r| r|r| |r|r |r|r |r|r pm v | v|v| v|v| |v|v |v|v |v|v a = pm r + (pm v -pm r ) = pstd v +2pstd r 2pstd r pm r *2pstd r + pm r *pstd v + pm v *2pstd r - pm r *2pstd r pstd r +2pstd v Oblique level-0 using means and stds of projections 1's 2's 3's 4's 5's 7's True Positives: 359 205 332 144 175 324 False Positives: 29 18 47 156 131 58 Oblique level-0 using means and stds of projections, doubling pstd r as above 1's 2's 3's 4's 5's 7's True Positives: 410 212 277 179 199 324 False Positives: 114 40 113 259 235 58 Class Totals-> 461 224 397 211 237 470 Oblique l-0, means, stds of projs, doubling pstd r, classify, eliminate in 2,3,4,5,7,1 order 1's 2's 3's 4's 5's 7's True Positives: 309 212 277 154 163 248 False Positives: 22 40 65 211 196 27 So the number of FPs is drastically reduced and TPs somewhat reduced. Is that better? If we parameterize 2 and adjust to maximize TPs and minimize FPs, what is the optimal multiplier parameter value? Code question: In a case stmt, if 1st case is true, does the 2nd case get evaluated? It doesn't if coded: If (case 1) then action1 Else if (case 2) then action2 Else if (case 3) then action3, etc.


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