Sébastien Ollier , Sandrine Pavoine and Pierre...

1

Comparing and classifyingmulti-scale spatial patterns

Sébastien Ollier*, Sandrine Pavoine and Pierre Couteron

Laboratoire de Biométrie et Biologie EvolutiveCNRS, UMR 5558,

Université Claude Bernard, Lyon1FRANCE

90th Annual Meeting (ESA), August 7-12 2005, Montreal, Canada

2

Introduction• Remotely sensed data to understand vegetation/land surface interface

• One sampling unit

PATTERN ANALYSIS

2-D

AutocorelogramFractal dimension

VariogramPeriodogramScalogram...

1-D

VARI

AN

CE

SCALES

kV

k• Many sampling units

1-D

2-D

How to compare many sampling unitson the basis of their spatial patterns?

How to assess the relative importanceof scales in a set of sampling units ?

DALE et al. (2002) Ecography

3

I. Pattern quantification1. Two Term Local Variance2. Mean Block Size Variance3. Spectral Analysis

II. Pattern ordination1. Scale standardization2. Ordination by a centred PCA

III.Application to laser profiles1. Data set2. Results

HILL (1973) J. Ecol.

RIPLEY (1978) J. Ecol.

GREIG-SMITH (1952)Ann. bot.

OLLIER et al. (2003)Rem. Sens. Env.

4

I. Pattern quantification

5


( ) ( ) ( )

( )

2 2 21 1 2 2 3 3 4

1

22 1 2 3 4

2

1

1

ttlv x x x x x xS

ttlv x x x xS

= − + − + −

= + − −

PATTERN ANALYSIS( )1 2 3 4, , ,t x x x x=x

1 1b =

2 2b =

1 4n =

2 2n =

Scalogram

1. Two Term Local Variance

6

( )

( )

1 1 2 3 4

2 1 2 2 3 3 4

, , ,

, ,

t

t

x x x x

x x x x x x

=

= + + +

x

x

k k=x H x

1 4

2

1 1 0 00 1 1 00 0 1 1

=

=

H Id

H1



( ) ( ) ( )

( )

2 2 21 1 2 2 3 3 4

1

22 1 2 3 4

2

1

1

ttlv x x x x x xS

ttlv x x x xS

= − + − + −

= + − −


1 1b =

2 2b =

1 4n =

2 2n =

Scalogram


7

1 1b =

1

0 1 0 01 0 1 00 1 0 10 0 1 0

M

=

2 2b =

2

0 0 10 0 01 0 0

M =

2

1

1 0 0 00 2 0 00 0 2 00 0 0 1

N

=

2

1 0 00 0 00 0 1

N =

1 2x x+ 2 3x x+ 3 4x x+

1 2 3 4x x x x


( )

( )

1 1 2 3 4

2 1 2 2 3 3 4

, , ,

, ,

t

t

x x x x

x x x x x x

=

= + + +

x

x

k k=x H x

1 4

2

1 1 0 00 1 1 00 0 1 1

=

=

H Id

H1


( ) ( ) ( )

( )

2 2 21 1 2 2 3 3 4

1

22 1 2 3 4

2

1

1

ttlv x x x x x xS

ttlv x x x xS

= − + − + −

= + − −


1 1b =

2 2b =

1 4n =

2 2n =

Scalogram


8

3( )t

k k k kk

k

ttlvS−

=x N M x


( )

( )

1 1 2 3 4

2 1 2 2 3 3 4

, , ,

, ,

t

t

x x x x

x x x x x x

=

= + + +

x

x

k k=x H x

1 4

2

1 1 0 00 1 1 00 0 1 1

=

=

H Id

H1


( ) ( ) ( )

( )

2 2 21 1 2 2 3 3 4

1

22 1 2 3 4

2

1

1

ttlv x x x x x xS

ttlv x x x xS

= − + − + −

= + − −


1 1b =

2 2b =

1 4n =

2 2n =

Scalogram


1 1b =

1

0 1 0 01 0 1 00 1 0 10 0 1 0

M

=

2 2b =

2

0 0 10 0 01 0 0

M =

2

1

1 0 0 00 2 0 00 0 2 00 0 0 1

N

=

2

1 0 00 0 00 0 1

N =

1 2x x+ 2 3x x+ 3 4x x+

1 2 3 4x x x x

9

1 1b =

2 2b = 1 2x x+ 3 4x x+

1 2 3 4x x x x

( )

ttlv witht

kk

kt

k k k k k

S=

= −

x A x

A H N M H



( ) ( ) ( )

( )

2 2 21 1 2 2 3 3 4

1

22 1 2 3 4

2

1

1

ttlv x x x x x xS

ttlv x x x xS

= − + − + −

= + − −


1 1b =

2 2b =

1 4n =

2 2n =

Scalogram


3( )t

k k k kk

k

ttlvS−

=x N M x

( )

( )

1 1 2 3 4

2 1 2 2 3 3 4

, , ,

, ,

t

t

x x x x

x x x x x x

=

= + + +

x

x

k k=x H x

1 4

2

1 1 0 00 1 1 00 0 1 1

=

=

H Id

H1

2

10

2. Mean Block Size Variance


( )1 2 3 4, , ,t x x x x=x

1 1b =

2 2b =

3 4b =

1 4n =

2 2n =

3 1n =

( ) ( )

( )

2 21_ 2 1 2 3 4

1

22_ 4 1 2 3 4

2

1 1 12 2

1 14

msbs x x x xS

msbs x x x xS

= − + −

= + − −

PATTERN ANALYSIS

11

( )

( )

( )

1 1 2 3 4

2 1 2 1 2 3 4 3 4

3 1 2 3 4 1 2 3 4

, , ,

, , ,

, ... ,

t

t

t

x x x x

x x x x x x x x

x x x x x x x x

=

= + + + +

= + + + + + +

x

x

x

tk k=x H x1

_ 1 1 11

1 1k k k k k k

k kb b+ + ++

= −y H x H x

2

1 2

2 1

3 4

4 3

12

x xx xx xx x

− − − −

1 2 3 4

1 2 3 4

3 4 1 2

3 4 1 2

14

x x x xx x x xx x x xx x x x

+ − − + − − + − − + − −

1 4

2

3 4 4

1 1 0 01 1 0 00 0 1 10 0 1 1t

=

=

=

H Id

H

H 1 1

1_ 2y 2_ 4y

3_ 1 _ 1 _ 1t

k k k k k kmsbs + + += y y



( )1 2 3 4, , ,t x x x x=x

1 1b =

2 2b =

3 4b =

1 4n =

2 2n =

3 1n =

( ) ( )

( )

2 21_ 2 1 2 3 4

1

22_ 4 1 2 3 4

2

1 1 12 2

1 14

msbs x x x xS

msbs x x x xS

= − + −

= + − −

PATTERN ANALYSIS

12

_ 1_ 1

1 1 1 11 1

msbs with

1 1 1 1

tk k

k kk

t t t tk k k k k k k k k

k k k k

S

b b b b

++

+ + + ++ +

=

= − −

x A x

A H H H H H H H H

( )

( )

( )

1 1 2 3 4

2 1 2 1 2 3 4 3 4

3 1 2 3 4 1 2 3 4

, , ,

, , ,

, ... ,

t

t

t

x x x x

x x x x x x x x

x x x x x x x x

=

= + + + +

= + + + + + +

x

x

x

tk k=x H x1

_ 1 1 11

1 1k k k k k k

k kb b+ + ++

= −y H x H x

2

1 2

2 1

3 4

4 3

12

x xx xx xx x

− − − −

1 2 3 4

1 2 3 4

3 4 1 2

3 4 1 2

14

x x x xx x x xx x x xx x x x

+ − − + − − + − − + − −

1_ 2y 2_ 4y

3_ 1 _ 1 _ 1t

k k k k k kmsbs + + += y y



( )1 2 3 4, , ,t x x x x=x

1 1b =

2 2b =

3 4b =

1 4n =

2 2n =

3 1n =

( ) ( )

( )

2 21_ 2 1 2 3 4

1

22_ 4 1 2 3 4

2

1 1 12 2

1 14

msbs x x x xS

msbs x x x xS

= − + −

= + − −

PATTERN ANALYSIS

13

3.Spectral analysis


( )1 2 3 4, , ,t x x x x=x

( ) ( ) ( )

( ) ( )

2 2

1 1 11 11

2

2 212

1 1 cos sin

1 1 cos

n n

i ii i

n

ii

I x i x iS n

I x iS n

ω ω ω

ω ω

= =

=

= +

=

∑ ∑

∑

1k =

1k =1

2nπω =

2k =2ω π=

Periodogramcos

cos

sin

1c

1s

2c

PATTERN ANALYSIS

14

( )

( )

I with

1

tk

kk

t tk k k kk

S

n

ω =

= +

x A x

A c c s s

3.Spectral analysis


( )1 2 3 4, , ,t x x x x=x

( ) ( ) ( )

( ) ( )

2 2

1 1 11 11

2

2 212

1 1 cos sin

1 1 cos

n n

i ii i

n

ii

I x i x iS n

I x iS n

ω ω ω

ω ω

= =

=

= +

=

∑ ∑

∑

1k =

1k =1

2nπω =

2k =2ω π=

Periodogramcos

cos

sin

1c

1s

2c

PATTERN ANALYSIS

15

Pattern quantificationMean square block size analaysis

Two term local varance

Spectral analysis

1-D

VARI

AN

CE

SCALES

kV

k

1-D

...

• Many sampling units

• One sampling unit Vt

kk

kS=

x A x

TABLE Z

16

How to choose the scaling coeficient ?

Pattern quantificationMean square block size analaysis

Two term local varance

Spectral analysis

1-D

VARI

AN

CE

SCALES

kV

k

1-D

...

• Many sampling units

• One sampling unit Vt

kk

kS=

x A x

TABLE Z

kS

17

II. Pattern ordination

18

1. Scale standardization

EUCLIDEAN STANDARDIZATION

( )k kS trace A=

• Two term local variance

161, 2, 4,8

nk==

kS

1k = 2 4 8

30 52 72 16

19


EUCLIDEAN STANDARDIZATION SPECTRAL STANDARDIZATION

( )k kS trace A= ( )1k kS Aλ=


161, 2, 4,8

nk==

kS

1k = 2 4 8

30 52 72 16 kS

1k = 2 4 84 9 26 16

firsteigenvectors

20


EUCLIDEAN STANDARDIZATION SPECTRAL STANDARDIZATION

( )k kS trace A= ( )1k kS Aλ=


161, 2, 4,8

nk==

kS

1k = 2 4 8

30 52 72 16 kS

1k = 2 4 84 9 26 16

• Mean block size variance

kS 8 4 2 1

• Spectral analysis

kS 2 2 2 1

kS 1 1 1 1

kS 1 1 1 1

21

2.Ordination by a centred PCA

SCALES

1-D

...

TABLE Z

Vk values

TRA

NSE

CTS Centered

PrincipalComponentAnalysis

1

p

q...

...• Eigenvalues

• Coordinates onPrincipal Components

• Coordinates onPrincipal Axes

Package ade4 : http://pbil.univ-lyon1.fr/ADE-4Software R : http://cran.univ-lyon1.fr/R

22

III. Application

23

1. Data set

OLLIER et al. (2003) Rem. Sens. Env.

N

E

S

O

N

E

S

O

The study site

2.5 km

24

1. Data set

N

E

S

O

N

E

S

O2.5 km

The study site

• ALLUVIAL PLAINSvery simple

and flat relief

• COMPLEX RELIEF FORMS< 60 m

gently slopes

• COMPLEX RELIEF FORMS> 60 m

steep slopesC

A

B

25

1. Data set

N

E

S

O

N

E

S

O2.5 km

The study site


and flat relief


gently slopes


steep slopes

A

26

1. Data set

N

E

S

O

N

E

S

O2.5 km

The study site


and flat relief


gently slopes


steep slopes

A

B

27

1. Data set

N

E

S

O

N

E

S

O2.5 km

The study site


and flat relief


gently slopes


steep slopes

B

A

C

28

1. Data set

N

E

S

O

N

E

S

O2.5 km

Laser data

29

1. Data set

264 laser transects

N

E

S

O

N

E

S

O2.5 km

264 laser transects (n=64)

30

Pattern analysis and ordination

264 scalograms (block size = 1,2,4,8,16,32)


TTLV

31

Pattern analysis and ordination

264 scalograms (block size = 1,2,4,8,16,32)


TTLVPCA

32

2.Results

• Eigenvalues

33

d = 0.2

ttg_1 ttg_2

ttg_4

ttg_8 ttg_16

ttg_32

2.Results

• Eigenvalues

• Principal axis

34

d = 0.2

ttg_1 ttg_2

ttg_4

ttg_8 ttg_16

ttg_32

2.Results

• Coordinates on principal components

• Eigenvalues

• Principal axis

35

d = 0.2

1 2 3

4

5 ttg_1 ttg_2

ttg_4

ttg_8 ttg_16

ttg_32

2.Results

• FIVE CLUSTERS• Coordinates on principal components

• Eigenvalues

• Principal axis

36

d = 0.2

1 2 3

4

5 ttg_1 ttg_2

ttg_4

ttg_8 ttg_16

ttg_32

2.Results

• FIVE CLUSTERS• Coordinates on principal components

• Eigenvalues

• Principal axis

ttg_1 ttg_2 ttg_4 ttg_8 ttg_16 ttg_32

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

1 23

4

5

37

d = 0.2

1 2 3

4

5 ttg_1 ttg_2

ttg_4

ttg_8 ttg_16

ttg_32


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

1 23

4

5

2.Results

9191248C

234223248B

186752A54321

38

Conclusion

• An efficient method to summarizeand classify 1-D spatial patterns

• A very general approach

• Generalization to 2-D sampling unitsCOUTERON et al. (2005) J. Appl. Ecol.

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