Data Analysis Machine Learning and Applications Episode 1 Part 8 ppsx
Data Analysis Machine Learning and Applications Episode 1 Part 8 ppsx
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Data Analysis Machine Learning and Applications Episode 1 Part 1 doc
... able1
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4
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Data Analysis Machine Learning and Applications Episode 1 Part 2 potx
... 0.7 61 0.694 0.677 0 .88 6 0 .83 2
P
all–pairs,Dirichlet
0 .89 3 0.755 0.720 0. 688 0 .88 8 0.7 71
P
1 v–rest,no
0 .83 3 0.539 0. 688 0.570 0 .88 5 0.464
P
1 v–rest,map
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P
1 v–rest,assign
0 .86 7 ... 0. 619
P
1 v–rest,no
0.973 0. 6 18 0 .80 3 0.646 0.9 81 0. 588
P
1 v–rest,map
0.973 0.942 0 .80 3 0. 785 0.9 78 0.9 21
P
1 v–rest,assign
0.9...
Data Analysis Machine Learning and Applications Episode 1 Part 3 docx
... 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 Smola ... 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 .17...
Data Analysis Machine Learning and Applications Episode 1 Part 4 pptx
... 0.3 511 3 0.4 788 5
b 0 .83 880 0 .8 7 18 3 0.56074 0.75 584 0 .86 282 0. 81 3 95 0. 710 85 0.79 0 18
7
a 0 .10 84 8 0 .11 946 0.00 517 0.09267 0 .10 945 0 .11 88 3 0.00 389 0.00659
b 0 .80 072 0 .87 399 0.27965 0 .87 892 0.9 488 2 ... centroid
2
a 0. 380 47 0.53576 0.00022 0 .11 912 0.42 288 0.2 511 4 0.00527 0.00032
b 0 .84 2 18 0.90705 0.72206 0 .12 010 0.99 680 0. 417 96 0.304 51 0...
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
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20
2004 3 f 21
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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 ... 0.0 08
Bolivia
0.666 0.056 0.2 78
Canada 0. 0 18
0. 980 0.002
Chile
0.632 0.356 0. 012
Egypt
0.750 0.070 0 . 18 0
France...
Data Analysis Machine Learning and Applications Episode 1 Part 6 docx
... 0.000 0.024 0. 012
0.965 0.000
grandmother 0.005 0 .13 4 0. 016
0 .84 0 0.005
granddaughter 0 .11 3 0.242 0.054
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grandson 0 .13 4 0 .11 1 0.052
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brother
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0.579 ... Name) (Product Name) (Price)
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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 1)...
Data Analysis Machine Learning and Applications Episode 1 Part 9 doc
... 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, B. (19 83 ): Généralisation de ... 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...
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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