Data Analysis Machine Learning and Applications Episode 2 Part 1 pot

... watermark database.Table 1. Averaged precision and recall at N /2 for the watermark database.Classes 1 2 3 4 5 6 7 8 9 10 11 12 13 14 N 322 11 5 13 9 71 91 44 19 7 12 6 99 33 14 31 17 416 P(N /2) .4 92 .24 3 ... 416 P(N /2) .4 92 .24 3 . 21 4 .14 4 .10 9 .24 4 .17 3 .097 .4 42 .068 .19 0 .8 02 .556 .28 3R(N /2) . 528 .13 9 .3 02 .19 7 .088 .1 82 .1 52 .19 1 .26 3 .0 61 .14 3 . 710 .3 52 .3 61 29 6 Triebel et al. data point p whose ... Situations 27 5 -16 00 -14 00- 12 0 0 -10 00-800-600-400 -20 0 0 5 10 15 20 25 30log likelihoodtime (s)passingaborted passingfollow -20 0 -10 0 0 10 0 20 0 300 400 500 600 4 6 8 10 12 14 16 18 20 ...

Data Analysis Machine Learning and Applications Episode 2 Part 6 potx

... 3 61 108560 1 826 10 936 026 28 518 3 18 16 923 1 0 21 51 1948 12 3 47 02 1 924 30044079003 526 98 18 58 24 719 8 924 00 18 98 26 326 628 20000 514 5 17 75 22 800903 9 21 910 6 010 056 1 922 704600 19 6997 618 60 18 47 1 826 18 1 619 4 81 924 18 5 818 9 817 7 51 922 19 6 918 47 10 8560 1 826 10 936 026 28 518 3 18 16 923 1 0 21 51 1948 12 3 47 02 1 924 30044079003 526 98 18 58 24 719 8 924 00 18 98 26 326 628 20000 514 5 17 75 22 800903 9 21 910 6 010 056 1 922 704600 19 6997 618 60 18 47 1 826 18 1 619 4 81 924 18 5 818 9 817 7 51 922 19 6 918 47Fig. ... and while some cover a wide Analysis of Stock Markets 3 61 108560 1 826 10 936 026 28 518 3 18 16 923 1 0 21 51 1948 12 3 47 02 1 924 30044079003 526 98 18 58 24 719 8 924 00 18 98 26 326 628 20000 514 5 17 75 22 800903 9 21 910 6 010 056 1 922 704600 19 6997 618 60 18 47 1 826 18 1 619 4 81 924 18 5 818 9 817 7 51 922 19 6 918 47 10 8560 1 826 10 936 026 28 518 3 18 16 923 1 0 21 51 1948 12 3 47 02 1 924 30044079003 526 98 18 58 24 719 8 924 00 18 98 26 326 628 20000 514 5 17 75 22 800903 9 21 910 6 010 056 1 922 704600 19 6997 618 60 18 47 1 826 18 1 619 4 81 924 18 5 818 9 817 7 51 922 19 6 918 47Fig. ... 3 61 108560 1 826 10 936 026 28 518 3 18 16 923 1 0 21 51 1948 12 3 47 02 1 924 30044079003 526 98 18 58 24 719 8 924 00 18 98 26 326 628 20000 514 5 17 75 22 800903 9 21 910 6 010 056 1 922 704600 19 6997 618 60 18 47 1 826 18 1 619 4 81 924 18 5 818 9 817 7 51 922 19 6 918 47 10 8560 1 826 10 936 026 28 518 3 18 16 923 1 0 21 51 1948 12 3 47 02 1 924 30044079003 526 98 18 58 24 719 8 924 00 18 98 26 326 628 20000 514 5 17 75 22 800903 9 21 910 6 010 056 1 922 704600 19 6997 618 60 18 47 1 826 18 1 619 4 81 924 18 5 818 9 817 7 51 922 19 6 918 47Fig. 2. Reduced adjacency matrix entries for the...

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Data Analysis Machine Learning and Applications Episode 2 Part 2 ppsx

... byAnselin (19 88b):LMAO=[ûW 2 û/ˆV 2 ] 2 T 22 −(T 21 A) 2 ˆvar(Û), (6)LMAU=[ûBBW 1 y] 2 Hrho−HTUˆvar(ˆT)HTU, (7)where T 21 A= tr[W 2 W 1 A 1 +W 2 W 1 A 1 ], ... 20 01) , FSG (Kuramochi and Karypis 20 01) ,MoSS/MoFa (Borgelt and Berthold 20 02) , gSpan (Yan and Han 20 02) , Closegraph(Yan and Han 20 03), FFSM (Huan et al. 20 03), and Gaston (Nijssen and Kok 20 04).A ... I −ÛW 1 , T=(EOV 12 ) , B = I −ÔW 2 ,HU= trW 2 + tr(BW B 1 )(BW B 1 )+ 1 V 2 (BW X E)(BW X E) and HTU=⎛⎝ 1 V 2 (BX)BW X Etr(WB 1 )BW B 1 +trWW B 1 0⎞⎠,...

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Data Analysis Machine Learning and Applications Episode 2 Part 3 pps

... O=0.5W 2 W 2 , U=0.50 0.05 0 .1 0 .15 0 .2 00 .1 0 .2 0.30.40.50.60.70.80.9 1 U,Opowerc) SARAR (1, 1): LMAW 1 W 1 , O=0.5W 1 W 1 , U=0.5W 2 W 2 , O=0.5W 2 W 2 , U=0.50 0.05 0 .1 0 .15 0 .2 00 .1 0 .2 0.30.40.50.60.70.80.9 1 U,Ocorrect ... O=0.5W 1 W 1 , U=0.5W 2 W 2 , O=0.5W 2 W 2 , U=0.50 0.05 0 .1 0 .15 0 .2 00 .1 0 .2 0.30.40.50.60.70.80.9 1 U,Opowerb) SARAR (1, 1): GMM opt.inst. WaldW 1 W 1 , O=0.5W 1 W 1 , U=0.5W 2 W 2 , ... lname asoundex.lname 11 525 6 WESTERHEIDE W236 20 00 01 BESTEWEIDE B233 20 00 02 WESTERWELLE W2363.3 Candidate selectioncandidates (data1 , data2 , method, selvars1, selvars2, key1, key2, )provides an...

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Data Analysis Machine Learning and Applications Episode 2 Part 4 doc

... RPart1 QUEST CTreeJ4.8 010 020 3J4.8(cv) 17 0 0 0 5 3 25 RPart0 18 18 0 0 13 15 64RPart1 18 18 16 0 14 15 81 QUEST 15 13 5 4 0 10 47CTree 18 14 3 2 8 0 4586 64 24 6 42 43Table 3. Median linear ... J4.8(cv) RPart0 RPart1 QUEST CTreeJ4.8 029 911 839J4.8(cv)408 911 9 41 RPart0560 710 735RPart16 410 8 625 QUEST 422 50 720 CTree76789037 26 20 27 38 49 37Complexity J4.8 J4.8(cv) RPart0 RPart1 QUEST ... other use is made of such data than statistical analysis. 1 2 3 4 5 6 7 8 9 10 11 12 00.5 1 1.5 2 2.533.544.55x 10 12 Applications Volume (in byte)UnknownWebP2PFTPNewsMailDBOthersControlGamesStreamingChatupstream...

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Data Analysis Machine Learning and Applications Episode 2 Part 5 pps

... (in byte) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 20 406080typical day cluster numberCluster 6 (5%)Global 1 2 345 678 9 1 0 11 12 1 3 14 0 20 406080Typical customer 6Typical day 12 Typical ... 20 30 40 50 60 70 8000 .2 0.40.60.8 1 Typical day 12 0 5 10 15 20 25 0 1 2 345x 10 70 5 1 0 1 5 2 0 2 50 2 468 10 x 10 6cluster 6, application: p2p down ( 12 % ) volume (in byte)global ... 9 1 0 11 12 1 3 14 0 20 406080Global meanTypical customer 10 typical day cluster number Cluster 10 (3%)0 10 20 30 40 50 60 70 8000 .2 0.40.60.8 1 0 1 0 2 0 3040 50 6070 8000 .2 0.40.60.8 1 Typical...

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Data Analysis Machine Learning and Applications Episode 2 Part 7 docx

... 16 Cluster 1 1.88 0 0 0 1 1 0 0 0 0 010 0 010 Universal -0.09 1 1 1 1 1 1 1 1 1 111 111 1In Table 2, the second row represents whether each of the 16 actors belongs tocluster 1 (when the element is 1) ... 0. 12 3 7 Guadagni 0 .28 3 0 .16 68 Lamberteschi 0.086 0.0769 Medici 0.383 0.434 10 Pazzi 0.039 0 .11 7 11 Peruzzi 0.339 -0.385 12 Pucci 0.000 0.000 13 Ridolfi 0.3 01 0. 12 4 14 Salviati 0 .13 7 0 .23 6 15 ... each consists of;Subgroup 1: actors 1, 2, 3, 6, 7, 8, 9, 10 , ( 12 ) , 13 , 14 , and, 16 Subgroup 2: actors 4, 5, 11 , and, 15 The two subgroups are graphically shown in Figure 1, where the horizontal...

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Data Analysis Machine Learning and Applications Episode 2 Part 8 docx

... 13 (2) .39 (1) 37 (3) 26 (3)Beach 9.83 10 (2) .35 (1) 5.56 09 (2) .26 (1) 25 (3) 17 (3)Hotel servicesLeisure activities 11 . 72 20 (6) 02 (2) 7. 52 04 (6) . 02 (2) .04 (2) .20 (1) 01 (4) . 01 ... (3). 01 (3) 03 (5) .03 (1) 01 (5)Furnishing 10 .15 .10 (1) 13 (4) 12 . 49 .08 (1) 09 (4) 03 (3) .06 (2) 01 (3) .03 (2) Catering 12 . 17 19 (5) .03 (3) 13 .29 07 (5) 01 (3). 12 (1) 07 (4) . 02 (2) 04 ... (4) .10 (2) .10 (1) Hotel facilitiesLocation 7.78 24 (2) .24 (1) 12 . 84 32 (2) . 32 (1) Type of building 9.09 .08 (2) 22 (3) 8.36 03 (2) 12 (3) .14 (1) .15 (1) Outside facilities 7.40 .25 (1) .00...

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Data Analysis Machine Learning and Applications Episode 2 Part 9 pdf

... PWAttribute 1 Level 1 0.0 32 0.044 0. 020 Level 2 18 .2 % 0 .11 7 14 .4 % 0.080 21 .8 % 0 .15 4Level 3 0 .14 0 0.095 0 .18 4Attribute 2 Level 1 0 .10 6 0 .17 7 0.036Level 2 0 .15 7 0 .18 3 0 .1 32 Level 3 27 .5 %0 .23 8 28 .5 ... %0 .23 8 28 .5 %0 .23 6 26 .5 %0 .24 0Level 4 0.0 81 0.036 0. 12 4 Attribute 3 Level 1 0 .25 6 0. 317 0 .19 6Level 2 0 .14 5 0 .13 1 0 .15 9Level 3 29 .2 %0 .16 333 .1 %0 .16 5 25 .4 %0 .16 1Level 4 0. 019 0. 017 0. 0 21 Attribute ... pseudo R 2 1 -13 1.49 28 0.97 28 9.97 .00 .23 2 -11 7.04 27 6.09 29 7.09 .09 . 81 3 -10 0.96 26 7. 92 300. 92 .08 . 92 4 -89.76 26 9. 52 314 . 52 .11 . 92 5 - 82. 62 279 .24 336 .24 .11 .95Classifying Contemporary Marketing...

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Data Analysis Machine Learning and Applications Episode 2 Part 10 docx

... 337.35 10 Q( .10 ) 38 37 35 30 22 12 5Q(.50) 23 8 23 7 23 6 23 0 22 2 21 2 20 1 Q(.90) 786 785 784 779 7 71 759 749ARL 19 9.77 19 6.56 18 3.44 15 1.96 10 5 .25 59 .10 28 .04SDRL 19 3.98 19 3. 91 193 .13 18 7.73 16 9.44 ... 13 3.97 93.43 28 Q( .10 ) 25 219 5333Q(.50) 14 0 13 7 12 4 86 12 5 4Q(.90) 456 4 52 438 399 325 20 5 44ARL 18 5 .15 17 0.90 11 8 . 21 43 .15 9 .17 4. 32 3.47SDRL 17 6 .11 17 5.05 1 62. 56 10 4.09 31. 47 4.67 0. 91 100 ... Z-StatisticD 1 Constant 1. 573 1 .27 2 1 .23 7Length 1. 1 31 0.4 32 2. 620 ∗Owner -1. 995 1. 130 -1. 765Brand 1 1.455 0.635 2. 2 92 ∗Consideration 0 .14 1 0.445 0. 316 Dealer 0.9 12 0.746 1 .22 3Class 2: Concomitant...

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