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...
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...
Data Analysis Machine Learning and Applications Episode 1 Part 2 potx
... Cal
P
ANN ,1 v–rest,Dirichlet
0.973 0.963 0. 815 0.809 0.9 71 0.952
P
LDA
0.980 0.972 0. 713 0.737 0.862 0.835
P
QDA
0.980 0.969 0.7 71 0.7 61 0. 9 14 0.866
P
Logistic Regression
0.973 0.9 64 0.5 61 0.633 0. 843 ... I
z
j
,6
j
0.000 0 .4 41 0.8 21 0.8 71
CC MK G
CC
= M ·Kf
T
j
= I
z
j
,6
j
0.000 0.0 54 0. 217 0.8 01
LLDA J = 5, n = 500 -
0.000 0.0 31 0.207 0.869
MDA S
k
= M - 0.00...
Data Analysis Machine Learning and Applications Episode 1 Part 3 docx
... 20 40 60 80 10 0 12 0 14 0 16 0 18 0 200
0
20
40
60
80
10 0
12 0
14 0
16 0
18 0
200
0 20 40 60 80 10 0 12 0 14 0 16 0 18 0 200
0
20
40
60
80
10 0
12 0
14 0
16 0
18 0
200
Fig. 2. Problem fourclass (Schoelkopf and ... 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 .17
2 28....
Data Analysis Machine Learning and Applications Episode 1 Part 5 pdf
... dendrograms
Q
2
0 1 3 4 12 20 32 64
0 f 0022256
1
0 f 10 0000
3
01f 00 000
4
200f 342 2
12
2003f 322
20
20 04 3 f 21
32
5002 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 ... 0.382
Indonesia
0. 616 0 . 14 4 0. 240
Italy 0. 044
0.950 0.006
Japan 0 .13 4
0. 846 0.020
Norway 0.08...
Data Analysis Machine Learning and Applications Episode 1 Part 6 docx
... data.
grandfather 0.000 0.0 24 0. 012
0.965 0.000
grandmother 0.005 0 .13 4 0. 016
0. 840 0.005
granddaughter 0 .11 3 0. 242 0.0 54
0 .46 6 0 .12 5
grandson 0 .13 4 0 .11 1 0.052
0.5 81 0 .12 2
brother
0. 612 0.282 0.0 24 0.082 0.000
sister
0.579 ... 0.3 91 0.026 0.002 0.002
father 0.099
0. 546 0 .12 2 0 .15 8 0.075
mother 0.089
0.6 54 0 .13 6 0.0 54 0.066
daughter 0.000
1. 00...
Data Analysis Machine Learning and Applications Episode 1 Part 7 doc
... parameter S = 0.75,
mean vectors z
1
=(0,0, 0)
, z
2
=(3,5,8)
, and covariance matrices
6
1
=
⎛
⎝
3.02 .42 .4
2 .43 .02 .1
2 .42 .11 .3
⎞
⎠
, 6
2
=
⎛
⎝
4. 02 .42 .4
2 .43 .52 .1
2 .42 .13 .2
⎞
⎠
.
In the second setting, ... bias MSE S.Cov Length
NM 0.0 015 0. 04 31 93.8% 0 .16 04
M
2
0.0052 0. 043 7 94. 0% 0 .16 61
M
3
0.0 043 0. 043 5 94. 0% 0 .16 51
M
4
0.0059 0. 044 2...
Data Analysis Machine Learning and Applications Episode 1 Part 8 ppsx
... ∗umbh+ 0 .13 35∗tie+ 0.20 41 textiles+ 0. 211 4 bag
+0 .17 91 wat + 0 .12 92∗mous+ 0.08 81 scul+ 0.2322∗pens
(1)
13 8 Christian Hennig and Pietro Coretto
CAMPBELL, N. A. (19 84) : Mixture models and atypical ... %
Umbre =1
Umbre=2
Umbre=3
Umbre =4
Keyri =1
Keyri=2
Keyri=3
Keyri =4
Tie =1
Tie=2
Tie=3
Tie =4
Hat =1
Hat=2
Hat=3
Hat =4
Kerf1 =1
Kerf1=2
Kerf1=3
Kerf1 =4
Kerf2 =1...
Data Analysis Machine Learning and Applications Episode 1 Part 9 doc
... B
0
= B and Y
0
= Y the design and response data matrices, respec-
tively. Define t
1
= B
0
w
1
and u
1
= Y
0
c
1
as the first MAPLSS components, where
the weighting unit vectors w
1
and c
1
are ... La Revue de Modulad, 31, 1 31.
D’ AMBRA, L. and LAURO, N. (19 89): Non symetrical analysis of three-way contingency
tables. Multiway Data Analysis, 3 01 315 .
ESCOFIER...
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 ... viol
c4–c4 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 v...
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