Data Analysis Machine Learning and Applications Episode 1 Part 2 potx

Data Analysis Machine Learning and Applications Episode 3 Part 9 docx

Data Analysis Machine Learning and Applications Episode 3 Part 9 docx

... R., 31 9 Bessler, Wolfgang, 499 Biemann, Chris, 577 Borgelt, Christian, 2 29 Bradley, Patrick E., 95 Brunner, Gerd, 237 Brusch, Michael, 431 Burgard, Wolfram, 2 69, 2 93 Burkhardt, Hans, 11, 37 , 237 Calò, ... Wendelin, 2 69 Fernández-Aguirre, K., 1 83 Fessant, F., 34 3 Fiedler, Mathias, 2 29 Flodman, Pamela, 1 19 Franke, Markus, 35 5 Fried, Roland, 277 Gabriel, Thomas R.,...

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Data Analysis Machine Learning and Applications Episode 1 Part 1 doc

Data Analysis Machine Learning and Applications Episode 1 Part 1 doc

... able2 14 4 42 32 8 44 46 2 24 38 9 11 20 16 15 6 21 50 13 30 27 49 1 5 29 28 34 7 35 22 3 31 37 48 12 26 39 10 45 17 23 25 75 98 18 43 36 33 19 47 90 70 82 71 41 40 57 78 94 84 58 88 79 59 55 51 91 73 85 64 61 65 62 80 96 89 83 95 10 0 63 54 74 92 53 72 87 76 93 97 66 81 69 67 99 11 3 68 86 56 60 77 52 13 9 13 4 13 0 10 3 13 8 10 9 14 0 14 3 11 4...

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Data Analysis Machine Learning and Applications Episode 1 Part 2 potx

Data Analysis Machine Learning and Applications Episode 1 Part 2 potx

... Proc. of 26 th DAGM-Symposium. Springer, 22 0 22 7. HAASDONK, B. and BURKHARDT, H. (20 07): Invariant kernels for pattern analysis and machine learning. Machine Learning, 68, 35– 61. SCHÖLKOPF, B. and ... Networks, 12 (5), 987–997. TITSIAS, M.K. and LIKAS, A. (20 02) : Mixtures of Experts Classification Using a Hierarchi- cal Mixture Model. Neural Computation, 14 , 22 21...

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Data Analysis Machine Learning and Applications Episode 1 Part 3 docx

Data Analysis Machine Learning and Applications Episode 1 Part 3 docx

... 14 :55. 23 10 :55.70 14 : 21. 99 1. 37 1. 04 Classification Time 03 : 13 .60 00 :14 . 73 00 :14 . 63 13 .14 13 . 23 Classif. Accuracy % 95.78 % 91. 01 % 91. 01 % 1. 05 1. 05 USPS RBF H1-SVM H1-SVM RBF/H1 RBF/H1 (Min-Max) Kernel ... 2.62 3. 87 77 .30 46.67 2 28. 83 88. 41 18.06 2.50 1 68.54 7.44 2.54 0.00 SRNG 1 2 3 4 4 0.00 0.56 2.08 53. 33 3 0.67 3. 60 81. 12 44 .1...

Ngày tải lên: 05/08/2014, 21:21

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Data Analysis Machine Learning and Applications Episode 1 Part 4 pptx

Data Analysis Machine Learning and Applications Episode 1 Part 4 pptx

... 0.89 642 0.763 84 0. 712 12 0.85838 ¯r 0.5 313 0 0 .4 41 1 9 0.56066 0 .44 540 0 .45 403 0.39900 0. 618 83 0. 747 30 ccr 98.22% 98.00% 94. 44% 90.67% 97 .11 % 89.56% 98.89% 98 .44 % 11 a 0. 043 35 0. 043 94 0.00 012 0. 043 88 ... 0. 547 46 0.6 013 9 0.27 610 0 .46 735 0.58050 0 .49 842 0.33303 0.5 017 8 b 0. 910 71 0. 848 88 0 .48 550 0.73720 0. 813 17 0.79 644 0.72899 0. 744 62 6 a 0. 6...

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Data Analysis Machine Learning and Applications Episode 1 Part 5 pdf

Data Analysis Machine Learning and Applications Episode 1 Part 5 pdf

... dendrograms Q 2 0 1 3 4 12 20 32 64 0 f 0022 256 1 0 f 10 0000 3 01f 00 000 4 200f 3422 12 2003f 322 20 2004 3 f 21 32 50 02 2 2 f 5 64 6002 2 1 5 f Fig. 1. 2-adic valuations for D. 0 1 0 1 0 1 2 0 1 3 0 1 4 0 1 5 0 1 6 0 1 0 64 32 4 20 12 ... random initialization data set COPK-Means ssALife with U*C Atom 71 100 Chainlink 65. 7 10 0 Hepta 10 0 10 0 L...

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Data Analysis Machine Learning and Applications Episode 1 Part 6 docx

Data Analysis Machine Learning and Applications Episode 1 Part 6 docx

... data. grandfather 0.000 0.024 0. 012 0. 965 0.000 grandmother 0.005 0 .13 4 0. 0 16 0.840 0.005 granddaughter 0 .11 3 0.242 0.054 0. 466 0 .12 5 grandson 0 .13 4 0 .11 1 0.052 0.5 81 0 .12 2 brother 0. 61 2 0.282 0.024 0.082 ... 0.000 sister 0.579 0.3 91 0.0 26 0.002 0.002 father 0.099 0.5 46 0 .12 2 0 .15 8 0.075 mother 0.089 0 .65 4 0 .13 6 0.054 0. 066 daughter 0.000 1. 000 0.0...

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Data Analysis Machine Learning and Applications Episode 1 Part 7 doc

Data Analysis Machine Learning and Applications Episode 1 Part 7 doc

... follows: E (t +1) |···∼W  2G +2gD,(2h+ 2 K  k =1 6 (t) 1 k ) 1  , S (t +1) |···∼D(J+ n 1 , ,J +n K ), z (t +1) k |···∼N  (n k 6 (t) 1 k + <) 1 (n k 6 (t) 1 k y k + <[),(n k 6 (t) 1 k + <) 1  , 6 1( t +1) k |···∼W ⎛ ⎝ 2D ... + (1+ m 1 ) ˆ B, and a 95% confidence interval for Q is given by ˆ Q±t Q,0. 975 ˆ T 1/ 2 , where the degrees of freedom are Q =(m...

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Data Analysis Machine Learning and Applications Episode 1 Part 8 ppsx

Data Analysis Machine Learning and Applications Episode 1 Part 8 ppsx

... case 0 .84 0. 68 0 .82 0 .84 0 .83 0. 68 0.72 0 .88 0. 91 0 .85 0 .85 0. 68 0. 78 0.77 0.72 0 .89 0.66 0.96 0.66 0.93 0.90 0.73 0 .87 0 .88 0 .83 0. 78 0.64 0 .86 0. 78 [ 0.79 0 .89 0. 91 0. 48 0.57 0.49 0.60 0.62 0. 71 0. 71 0.64 0.69 0. 58 0.67 0.66 0.65 0. 61 0.74 0.75 0.72 0.69 0. 58 0.53 0.65 0.45 0.70 0.76 0.75 0.73 umbh tie textiles bag wat mous scul pens 0 ....

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Data Analysis Machine Learning and Applications Episode 1 Part 9 doc

Data Analysis Machine Learning and Applications Episode 1 Part 9 doc

... Ncube, 19 85; Lowry et al., 19 92 ; Jackson, 19 91 ; Liu, 19 95 ; Kourti and MacGregor, 19 96 , Mac- Gregor, 19 97 ). In particular, we focus on the approach based on PLS components proposed by Kourti and ... Gervini and Rousson (2004). At the end, the optimal 202 Rosaria Lombardo, Amalia Vanacore and Jean-Francỗois Durand limits (Wu and Wang, 19 97 ; Jones and Woodal...

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Data Analysis Machine Learning and Applications Episode 1 Part 10 ppt

Data Analysis Machine Learning and Applications Episode 1 Part 10 ppt

... 1 1 1 1 ∗ c4–g4 1 1 1 1 1 c4–a4 1 1 1 1 1 c4–c5 1 1 0 1 ∗ 1 instrument notes flu guit pian trum viol c4–c4 0 0 1 ∗ 1 ∗ 1 c4–e4 1 1 1 1 1 ∗ c4–g4 1 1 1 1 1 c4–a4 1 1 1 1 1 c4–c5 1 1 1 ∗ 1 ∗ 1 4.3 ... 1 1 1 1 1 c4–e4 0 1 0 0 1 c4–g4 0 0 0 0 0 c4–a4 1 1 1 0 0 c4–c5 1 1 1 1 1 instrument notes flu guit pian trum viol c4–c4 1...

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Data Analysis Machine Learning and Applications Episode 3 Part 1 pdf

Data Analysis Machine Learning and Applications Episode 3 Part 1 pdf

... 1 035 11 4 15 9 13 17 158 0 . 31 4 Polynomial (3rd degree) 12 42 633 516 14 767 17 158 0.6 43 RBF 860 6 51 498 15 149 17 158 0.640 Coulomb 0 11 24 25 16 009 17 158 0 .14 8 M1 287 850 299 15 722 17 158 0.505 M2 * 19 1 ... Me NonV B T3 Ratios VB T2 NonV B T2 VB T3 NonV B T3 Test1 Test2 ROS 9.52 3. 93 5 .39 2 .16 1 03 11 6 ROE 6.7 3. 83 3 .3 2. 01 111 11 0 ROI 6.85...

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Data Analysis Machine Learning and Applications Episode 3 Part 2 pdf

Data Analysis Machine Learning and Applications Episode 3 Part 2 pdf

... Artificial Intelligence, pp. 43 52, July 1998. BURKE, R. (20 02) : Hybrid Recommender Systems: Survey and Experiments. User Modeling and User-Adapted Interaction. vol. 12( 4), pp. 33 1 37 0. HERLOCKER, J.L., ... EachMovie, containing 2, 558,871 votes from 61, 1 32 users on 1,6 23 movies, and the MovieLens100k dataset, contain- ing 100,000 ratings from 9 43 users on 1,6 82 movies. Th...

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Data Analysis Machine Learning and Applications Episode 3 Part 3 pps

Data Analysis Machine Learning and Applications Episode 3 Part 3 pps

... Weibull K=2 K =3 K=4 K=5 separate 233 39.27 232 02. 23 230 40.01 229 43. 11 main.g 233 55.66 230 58.25 22971.86 228 63. 43 main.p 235 03. 73 233 68.77 231 65.60 230 68.47 int.gp 235 72.21 234 22.51 233 05. 63 230 75.76 main.gp ... 128 5 4 4 4 120 120 126 3 2 15 10 13 105 108 107 32 322229497 93 436 3029747979 5414 233 676564 ±10 234 3 231 787075 35 9 535 4455142 48564672 435 19 590...

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Data Analysis Machine Learning and Applications Episode 3 Part 4 potx

Data Analysis Machine Learning and Applications Episode 3 Part 4 potx

... Equation 3 and Figure 4: P nìm = U nìc ÃS cìc ÃV cìm (3) 2.69 0.57 2.22 4. 25 0.78 3. 93 2.21 0. 04 3. 17 1 .38 2.92 4. 78 P nìi -0.61 0.28 -0.29 -0.95 -0. 74 0. 14 U nìc 8.87 0 0 4. 01 S cìc -0 .47 -0.28 ... efc(0.17) Portal 1 0.1126 0.20 54 0.2815 0 .35 18 0. 640 8 0.7685 0 .36 Portal 2 0. 142 5 0.2050 0.1 836 0.2079 0.1965 0. 233 8 0.18 Portal 3 0.0058 0. 245 5 0.02...

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