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PC1
PC2
PC3
PC4
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*** 57.28
*** 119.71
*** 48.27
品種効果
自由度174, 875
主成分
寄与率
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(15.0%)
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ő!Ʀ4ÛžƢ
ǠǠǠ
z
iq=
x
i Tu
q(
z
i1,...,
z
iq)
=
x
i T(
u
1
,...,
u
q)
°ƐƢƌ
z
i
T
=
x
iTU
⇔
z
i=
U
Tx
iz
i 1=
x
iT
u
1
UTU= u
1 Tu
1 u1 Tu
q
u
q
Tu
1 uq
Tu
q
⎡
⎣ ⎢ ⎢ ⎢ ⎢
⎤
⎦ ⎥ ⎥ ⎥ ⎥
=I ƌźƤƋŨ
ƎƑƋ
U
T
=
U
−1⇔
UU
T=
I
∴
x
i=
Uz
i=^8_H\ tW]% Se'Gu %
xi =U k λ1
0
0
⎛
⎝ ⎜ ⎜ ⎜ ⎜
⎞
⎠ ⎟ ⎟ ⎟ ⎟
?Y"% 18_H'jsCZ
kAxB'w
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TŠ
eigenfaces
Murphy KP (2012) "Machine Learning: A Probabilistic Perspective" The MIT Press.
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A
ƌ
B
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100
100
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A
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100/100 = 1
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0
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A
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A
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Principal co-ordinate analysis (PCO)
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B
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T• q¼'ðŐ,Ə0xžƢn"ƑƶǟǎǛƑňČƦūƶǟǎǛŐƑ
ǘǡƱǚƾǃĶŗŬŴƠĿƏÂƛƢũ
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b
ij=
x
ikx
jkk=1
q
∑
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d
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(
x
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x
jk)
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q
∑
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ik2k=1
q
∑
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x
ik2k=1
q
∑
−
2
x
ikx
jkk=1
q
∑
=
b
ii+
b
jj−
2b
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b
iji=1
n
∑
=
x
ikx
jkk=1 q
∑
=
i=1
n
∑
x
jkx
iki=1 k
∑
=
0
k=1
q
∑
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ǢĄŶǣ
• żƄŵƇƊŨŵơñƈũdij2
i=1
n
∑
= (bii+bjj−2bij)i=1
n
∑
= biii=1
n
∑
+nbjjdij2
j=1
n
∑
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n
∑
= bjjj=1
n
∑
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n
∑
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j=1
n
∑
i=1n
∑
= (bii+bjj−2bij)j=1
n
∑
i=1n
∑
=n biii=1
n
∑
+nn bjjj=1
n
∑
=2n biii=1
n
∑
• żƄŵƇƊŨŵơñƈũ
di. 2
=1
n dij
2
j=1
n
∑
=1n i=1bii
n
∑
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=1
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2
i=1
n
∑
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n
∑
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∑
i=1 n∑
=2n i=1bii
n
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2
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2
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I 7#&&
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B=XXT'qX
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A
(
u
1,...,
u
q
)
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(
u
1,...,
u
q)
λ
10
0
λ
q⎡
⎣
⎢
⎢
⎢
⎤
⎦
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1
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U
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u
i T
u
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u
i
=
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U
T
U
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T=
U
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UU
T=
I
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A
=
U
Λ
U
Tu
i*=
λ
iu
i ƌžƢƌŨA
=
U
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u
q *)
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T ƑXƒŨX
=
U
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diag(
λ
1 1/2
,
…
,
λ
q1/2
)
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PC2
PC4
主成分得点の差
0
1
2
3
0.2 0.4 0.6 0.8 1.0
0
1
2
3
4
5
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PC
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1
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0
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PC1 PC2 PC3 PC4 PC5 PC6 PC7 PC8 PC9 PC10
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http://konicaminolta.jp/instruments/knowledge/color/part2/06.html
II. ŏàĊ: ŕŻƣƄ»¹ƦĦƢǧŐƑãƏƒĠŲƎŰƨǍǙDŽƑĖƑ`¹
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102
1
10-2
10-4
10-6
10-8
10-10
10-12
10-14
FM
780
700
600
500
400 380
(nm)
780
700
600
520
440
360
(nm)
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Syafaruddin et al. (2006) Breed Sci 56:75-79
Ǣ?ďǣ
2*24,1hU9eo
f
(
x
)
Ʀg(
x
)
=
0
ƑƜƌƋ«aǢrǣ:žƢǢ5ø±ǣ
L(
x
,
λ
)
≡
f
(
x
)
+
λ
g(
x
)
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ǙƲǙǟƷǗő¡ (Lagrangian)
ǙƱǙǟƷǗ ¡ (Lagrange multiplier)
∇L
=
∇f
+
λ
∇g
=
0
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Ǣ?ďǣ
ǙƲǙǟƷǗƑ®j ¡ÃƑSĤ
http://www.isigas.com/LagrangeMP.html Ə0Ŵơ§ŰĬ¦ŵůơƘž
g
(
x
1,
x
2)
=
0
f
(
x
1,
x
2)
∇
f
∇
g
g
(x)=0
ƌ
f
(x)
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:a
ƊŰ
&
#
∇
f
−
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g
=
0
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ǙƲǙǟƷǗ®j ¡ÃƑ
g(x
1,
x
2)
=
x
1+
x
2−
1
=
0
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(
x
1,
x
2)
=
1
−
x
1 2−
x
22
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L
(
x
,
λ
)
=
1
−
x
1 2
−
x
2 2
+
λ
(
x
1
+
x
2−
1)
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∂
L
∂
x
1
=
−
2
x
1
+
λ
=
0
∂
L
∂
x
2=
−
2
x
2+
λ
=
0
∂
L
∂λ
=
x
1+
x
2−
1
=
0
'
x
1=
1
2
,
x
2=
1
2
,
λ
=
1
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3
¼'Ƒ\G
z
1i=
u
11x
1i+
+
u
1mx
mi=
u
1 Tx
i
V
(
z
1)
=
1
n
−
1
(
z
1i−
z
1)
2
i=1
n
∑
=
s
jku
1ju
1k=
u
1 TVu
1k=1
m
∑
j=1
m
∑
sjkƒxjƌxkƑ*0
Ʀ
u
Tu
=
u
112
+
+
u
1M
2
=
1
5ø± ƑƋ«a:žƢ
z1Ƒ0
ƌžƢũ
L
(
u
1
,
λ
)
=
u
1 TVu
1
−
λ
(
u
1 Tu
1
−
1)
Ƒ«a:∂
L
∂
u
1=
2
Vu
1−
2
λ
u
1=
0
(
V
−
λ
I
)
u
1
=
0
T#Oş
V(z2)=u 2
T Vu
2
L
(
u
2
,
λ
,
ν
)
=
u
2 TVu
2
−
λ
(
u
2 Tu
2
−
1)
−
ν
(
u
2 Tu
1
)
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3
¼'Ƒ\GƒŨŻƠƏ
...
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z
2i=
u
21x
1i+
+
u
2mx
mi=
u
2T
x
iu
2 Tu
2
=
u
212
+
+
u
2M 2=
1
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u
2 Tu
1
=
u
21u
11+
+
u
2Mu
1M=
0
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ƑƋ«a:žƢũ Ʀ
Ƒ«a:
∂
L
∂
u
2=
2
Vu
2−
2
λ
u
2−
ν
u
1=
0
u
2 T
u
1
=
u
1 Tu
2
=
0,
u
1 Tu
1
=
1
Ɵơν
=
0
(
V
−
λ
I
)
u
=
0
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1T'`%
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