Métodos cuantitativos para la diversidad lingüística

Transcription

Harald Hammarstrom
March 2015, Mexico
Hammarstrom
Métodos Diversidad Lingüística
March 2015, Mexico
1 / 55
Para la diversidad lingüística = Métodos que no están baseados
en los hechos de una(s) lengua(s) especíca(s)
Se considerarán métodos
Estos métodos
humanos
computacionales
no reemplacen a los humanos, sino ayudan a los
Ejemplos de lo que se puede hacer a partir de
Clase #1 datos textuales: Morfología automática
Clase #2 datos textuales anotados: Morfología, clases de palabras,
sintaxis
Clase #3 un base de datos de características: Unas técnicas para
visualización
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Estructura de clases
Se van a presentar técnicas
I Lo importante es obtener una familiaridad con las ideas
I No hay que entender todas detalles
No habrán prácticas
No hace falta tener una computadora en clase
Si quieres probar algo con tus propios datos
I Si sabes programación: Puedes implementar las técnicas tu mismo
I Si tu amigo sabe programación: Tu y tu amigo podéis implementar las
técnias
I En otro caso: Yo te puedo ayudar
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De datos textuales
Jimbokaksïni nósesi járhati jurhiatekuecha janocheaka, engaksïni
chéti ikiatspiricha ópandaaka, ka uirhipkurhaatiksïni ka iápuru
ísïksïni k'uaniraati, ka xerentskantaatiksïni. Iámendu jimini
anapuechanksïni uándikucheati, ka nóksï ma tsakapu tsïntsikata
jurajkuati, jimbokari no kurhanguska imani jurhiatekuani engari
juka p'urhembenhani.
Que se puede hacer con solo datos textuales?
Sin otra información de la lengua, solo se puede utilizar
...
frecuencias
Por ejemplo, frecuentemente las palabras terminan en -i, se repite
mucho la palabra ka etc
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Ejemplo de diferencias signicativas de frecuencias (Inglés)
# different words in corpus having suffix
4000
3500
3000
2500
2000
1500
1000
500
0
playing
laying
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aying
ying
ing
ng
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g
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Ejemplo de diferencias signicativas de frecuencias (Español)
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Aprendizaje Morfológico sin Supervisión
Aprendizaje Morfológico sin Supervisión = Unsupervised Learning
of Morphology (ULM)
Input: Datos textuales crudos (sin anotación) de una lengua natural
Output: Una descripción de la estructura de las palabras en los
material textuales dados
Con: Lo menos supervisión, i.e. parámetros, datos auxiliares
anotados, selección de modeles durante el desarrollo etc,
posible
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Porqué estudiar el problema ULM?
Teoría Lingüística
I Harris (1955), testeado con ingl'es (Harris, 1967)
I Andreev (1959, 1963, 1965b, 1967) testeado con Albanés, Armenio,
Bulgaro, Checo, Inglés, Estonés, Francés, Alemán, Hausa, Hungaro,
Latviano, Ruso, Serbo-Croato, Swahili y Vietnamés
Eliminación del léxico
I En los 1980s tener un lexicón entero sobrepasaba la capacidad de la
memoria de una computadora (Klenk, 1985a,b, Wothke, 1985)
I Ahora esta motivación ya no existe
Aprendizaje de lengua por niños
I P.e., Brent et al. 1995, Batchelder 1997, Brent 1999, Clark 2001,
Goldwater 2007
Arrancar una procesadora morfológica computacional
I P.e., Goldsmith 2001 y muchos más (véanse mas tarde)
Ayudar a la descripción y documentación lingüística
I P.e., Hammarström et al. 2008, Monson et al. 2008c Kirschenbaum
et al. 2012, Palmer et al. 2010
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ULM: Pasamos revista
Morfología: Para esta revista lo denimos como incluyendo lo siguimente
Derivacional y ineccional
En número de ajos que se puede añadir es nito
El orden de ajos es jo (no se puede permutar)
Algunos enfoques: También clíticos, composición,
morfología no-concatenativa, incorporación y ajos
lexicales
Palabras ortográcas: Solo se consideran palabras ortográcas (i.e. no se
trata el problema de buscar la separación de palabras de
lenguas cuya ortgrafía no marca limites de palabras)
Dependencia de lengua: Angunos enfoques si tienen un sesgo explicito o
implicito a ciertas lenguas
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ULM: Variedades de Output
Lista de ajos
↑
Misma-raiz decisión
↑
Analisis
↑
Lista de paradigmas
↑
Lexicon+Paradigmas
Hammarstrom
Una lista de ajos.
Dos palabras dadas, decidir si son inecciones de la misma raiz.
Dada una palabra, analizarla en raiz y
ajo(s).
Una lista de paradigmas.
Una lista de los paradigmas y una lista de
las raíces con indicaciones a qué paradigma
pertenece.
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Trabajos hasta hoy #1
Harris 1955, 1968, 1970
Andreev 1965a, Andreev 1967, Chapter
2, Andreev 1965a, Eliseeva 1965, Fedulova 1965, Fihman 1965a,b, Fitialova 1965,
Hol'm 1965, Jakubajtis 1965, Jaku²eva
1965,
Kordi
1965,
Malahovskij
1965,
Melkumjan 1965, Oºigova 1965, Panina
1965, Per²ikov 1965
Gammon 1969
Lehmann 1973, 7193
de Kock and Bossaert 1969, 1974, 1978
Faulk and Gustavson 1990
Hafer and Weiss 1974
Klenk and Langer 1989
Langer 1991
Redlich 1993
Klenk 1991, 1992
Flenner 1992, 1994, 1995
Janÿen 1992
Juola et al. 1994
Brent 1993, 1999, Brent et al. 1995, Snover
2002, Snover and Brent 2001, 2003, Snover
et al. 2002
Deligne 1996, Deligne and Bimbot 1997
Yvon 1996
Kazakov 1997, Kazakov and Manandhar
1998, 2001
Jacquemin 1997
Cromm 1997
Gaussier 1999
Hammarstrom
Model
Superv.
Experimentación
Qué aprende?
C
C
T
T
English
Vietnamese
Hungarian (I)
Segmentation
Segmentation
C
C
C
T
T
T
English
French (I)
French/Spanish
C
C
C
C
C
C
C
C
C
C
T
T
T+SP
T+SP
T
T+SP
T+SP
T+SP
T
T
C
C
C
T
T
T
English (I)
English (IR)
German
German
English (I)
Spanish
Spanish
French
English
English/ChildEnglish/Polish/
French
English/French (I)
French (I)
French/English
C
C
C
T
T
T
English
German
French/
English (I)
to
Segmentation
Segmentation
Lexicon+
Paradigms
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Lexicon+
Paradigms
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Model
Superv.
Experimentación
Qué aprende?
Déjean 1998a,b
C
T
Ax Lists
Medina Urrea 2000, 2003, 2006
Schone 2001, Schone and Jurafsky 2000,
2001
Belkin and Goldsmith 2002, Goldsmith
2000, 2001, Goldsmith et al. 2001, Goldsmith 2006, Hu et al. 2005b, Xanthos et al.
2006
Baroni 2000, 2003
C
C
T
T
Turkish/English/
Korean/French/
Swahili/
Vietnamese (I)
Spanish
English
C
T
English (I)
Lexicon+
Paradigms
C
T
C
C
T
T
Child-English/
English
Korean
Assamese
Ax List
Cho and Han 2002
Sharma and Das 2002, Sharma et al. 2002,
2003
Baroni et al. 2002
Bati 2002
C/NC
C/NC
T
T
English/German (I)
Amharic
Creutz 2003, 2006, Creutz and Lagus
2002, 2004, 2005a,b,c, 2007, Creutz et al.
2005a,b, Hirsimäki et al. 2003
Kontorovich et al. 2003
Medina-Urrea 2006, 2008, Medina Urrea
and Díaz 2003
Mayeld and McNamee 2003, McNamee
and Mayeld 2007
Hadouche 2002, Zweigenbaum et al. 2003
Calderone 2008, Pirrelli et al. 2004, Pirrelli
and Herreros 2007
Johnson and Martin 2003
Katrenko 2004
C
T
Finnish/Turkish/
English
C
C
T
T
English
Segmentation
Chuj/Rarámuri/Czech Ax List
-
-
C
C
T
T
8 West European
Same-stem
languages (IR)
Medical French
Segmentation
Italian/English/Arabic Unclear
C
C
T
T
Inuktitut
Ukrainian
Ax List
Segmentation
Segmentation
Lexicon+
Paradigms
Related word pairs
Lexicon+
Paradigms
Segmentation
Unclear
Lexicon+
Paradigms
avar et al. 2004a,b, avar et al. 2006a,b
C
T
Child-English
Unclear
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Model
Superv.
Experimentación
Qué aprende?
C
T
Segmentation
C/NC
T
C
C
C/NC
C
T
T
T
English/Spanish/
Mapudungun (I)
30-ish mostly European type languages
English
English
Unclear
Italian/English
Segmentation
Rewrite Rules
Segmentation
Segmentation
Unclear
Segmentation
C
C
C
T
T
-
Catalan
English/German
Maori to Warlpiri
Paradigms
Segmentation
Same-stem
C
T
Finnish/Turkish/English
Keshava and Pitler 2005
Johnsen 2005
Atwell and Roberts 2005
Dang and Choudri 2005
ur Rehman and Hussain 2005
Jordan et al. 2005, 2006
Goldwater 2007, Goldwater et al. 2005,
Naradowsky and Goldwater 2009
Freitag 2005
Golcher 2006
C
C
C
C
C
C
C
T
T
T
T
T
T
T
English/Child-English
Segmentation+
Related
sets
words
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
Segmentation
C
C
T
-
English
English/German
Arabsorkhi and Shamsfard 2006
Chan 2006, Chan 2008, 101-139
Demberg 2007
C
C
C/NC
T
T
T
Dasgupta 2007, Dasgupta and Ng
2006, 2007, Dasgupta and Ng. 2007
C
T
Monson 2004, 2009, Monson et al.
2007a,b, 2008a, 2004, 2008b,c
Wicentowski 2002, 2004, Yarowsky and
Wicentowski 2000
Gelbukh et al. 2004
Argamon et al. 2004
Goldsmith et al. 2005, Hu et al. 2005a
Bacchin et al. 2002a,b, 2005, Nunzio
et al. 2004
Oliver 2004, Chapter 45
Bordag 2005a,b, 2007a,b,c
Hammarström 2005, 2006a,b, 2007,
2009a,b
Bernhard 2005a,b, 2006, 2007a,b
Hammarstrom
+
of
Segmentation
Lexicon+
Paradigms
Persian
Segmentation
English
Paradigms
English/German/Finnish/ Segmentation
Turkish
Bengali
Segmentation
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Model
Superv.
Experimentación
Qué aprende?
De Pauw and Wagacha 2007
Tepper 2007, Tepper and Xia 2008
Xanthos 2007
C/NC
C/NC
NC
T
T+RR
T
Gikuyu
English/Turkish
Arabic
Majumder et al. 2008, 2007
C
T
Zeman 2007, 2008a,b
C
-
Kohonen et al. 2008
Goodman 2008
Golénia 2008
Pandey and Siddiqui 2008
C
C
C
C
T
T
T
T
French/Bengali/French/
Bulgarian/Hungarian
Czech/English/German/
Finnish
Turkish/Russian
Hindi
Segmentation
Analysis
Lexicon+
Paradigms
Analysis
Johnson 2008
Snyder and Barzilay 2008
C
C/NC
T
T
Spiegler et al. 2008
Moon et al. 2009
Poon et al. 2009
C
C
C
T
T
T
Hammarstrom
Segmentation+
Paradigms
Segmentation
Segmentation
Segmentation
Segmentation+
Paradigms
Sesotho
Segmentation
Hebrew/Arabic/Aramaic/ Segmentation
English
Zulu
Segmentation
English/Uspanteko
Segmentation
Arabic/Hebrew
Segmentation
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Estado del Arte?
Una comparación justa de precisión no es posible ya que
I Gran variedad en ambiciones y presuposiciones
I Specicaciones no completas
I Ad-hoc umbrales de valores (thresholds)
Hubo una competición
MorphoChallenge
http: // www. cis. hut. fi/ morphochallenge2009/
I Segmentación de Finés, Inglés, Alemán, Arabe y Turco
I La mayoría de sistemas no tomaron parte, incluso varios que se
consideran informalmente los mejores
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Análisis morfológico sin-supervisión (ULM) vs sistemas
basados en reglas
Citación del Preface de State of the Art
ed. Mahlow and Piotrowski 2009 p vi:
in Computational Morphology
En la tarea de análisis de morfemas de MorphoChallenge, los
análises hechos por los algoritmos de los participantes
fueron comparados con un gold standard lingüístico.
En Morpho Challenge 2008 [3], el mejor sistema por Alemán
logró un F-measure de 54.06%. La mejor cifra de recall fue
59.51% (este sistema logró 49.53% precision), el mejor
Estas cifras son
demasiadamente bajas para considerarse útiles en la
mayoría de las aplicaciones, y en particular en aplicaciones
resultado de precision ...
interactivas. Si comparamos los resultados de Morpho
Challenge con las cifras presentadas en el primer
la decisión de concentrarse en
sistemas basados en reglas se hace evidente.
Morpholympics [4],
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Técnicas en ULM
Buscar segmentos frecuentes/sobrerepresentados
Buscar el confín de la segmentación
Buscar paradigmas (ajos que ocurren sistematicamente en la(s)
misma(s) raíc(es)
Vamos a presentar unas de las más simples + una combinación
de las tres
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Notaciones Formuláicos
Letras minúsculas w , s, b, x, y , . . . : signican una cadena de letras
Por ejemplo, x = hotel, x = cama, o sea cualquier palabra
Letras mayúsculas W , X , S, . . . : signican un conjunto de palabras
Por ejemplo, W
= {hotel,
cama, . . . }
|x| signica la longitud de x si x es una palabra
Por ejemplo,
|cama| = 4
|W | signica el número de miembros de W si W es un conjunto
Por ejemplo,
{hotel, cama, mono} = 3
w = xs signica la segmentacion de w en x seguido de s
Por ejemplo, cama = xs son las posibilidades x = c ,
s = ama o x = ca, s = ma o x = cam, s = a o x = cama,
s =00
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Buscar Segmentos Sobrerepresentados
Sobrerepresentación como más-frecuente-que-su-longitud: Un segmento x
de longitud |x| letras está sobrerepresentado cuando es más
frecuente que lo esperado por su longitud
f (x)
|Σ||x|
Sobrerepresentación como más-frecuente-que-sus-partes: x = c1 c2 . . . cn de
n letras está sobrerepresentado cuando es más frecuente de lo
esperado de la combinación de las frecuencias de sus partes
f (c1 c2 . . . cn )
f (c1 )f (c2 ) . . . (cn )
Sobrerepresentación como más-frecuente-cuando-sujo: La probabilidad de
ocurrencia como sujo divisada por la probabilidad en otras
posiciones (no nales)
Pf (x)
Pnf (x)
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Sobrerepresentación como más-frecuente-que-su-longitud:
Ejemplo
En la Biblia española hay 25 169 palabras diferentes
Entre ellas 556 terminan en
-ar:
f (−ar ) = 556
Hay 37 letras distintas en la Biblia española
Entonces la frencuencia esperada de -ar que es un segmento con
longitud 2 sería
1
25168 · ( )2 = 18.38
37
Entonces la sobrerepresentación de -ar será
556
≈ 30.25
18.38
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Sobrerepresentación como más-frecuente-que-sus-partes:
Ejemplo
Hay 556 palabras que terminan en
-ar:
f (−ar ) = 556
Hay 37 letras distintas en la Biblia española
a
ocurre con la probabilidad 13.2%
a
ocurre con la probabilidad 9.1%
Entonces la frencuencia esperada de
-ar
que está compuesto de a y r
25169 · 0.091 · 0.132 ≈ 301.5
Entonces la sobrerepresentación de
-ar
será
556
≈ 1.84
301.5
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Sobrerepresentación como más-frecuente-cuando-sujo:
Ejemplo
Hay 556 palabras que terminan en -ar: f (−ar ) = 556
Entonces la probabilidad nal de -ar es
556
= 0.022
25169
Hay 462 palabras donde -ar- ocurre no-nal (dentro de la palabras)
En las 25 169 palabras hay 162 036 posiciones no nales
Entonces la probabilidad no-nal de -ar es
462
= 0.00285
162036
Entonces la sobrerepresentación de -ar será
0.022
≈ 7.72
0.00285
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Considera la distribución de letras en la posición inmediatamente
antes de un un sujo posiblemente segmentable, p.e., Inglés
ing
3258
t 640
l 329
r 317
d 258
n 249
k 216
s 170
h 149
y 131
p 130
g 117
...
...
Hammarstrom
i
o
a
u
e
h
l
n
ng
3352
3258
35
26
26
4
1
1
1
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Considera la distribución de letras en la posición inmediatamente
antes de un un sujo posiblemente segmentable, p.e., Español
er
5956
t 1068
d
677
n
662
s
486
l
374
k
337
r
332
g
305
i
298
m
242
p
184
...
...
Hammarstrom
r
12258
e 5956
a 4059
o 1249
i 1101
u
237
r
46
h
1
g
1
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Heuristicas sobre el confín de la segmentación
Número de letras distintas (Harris 1951 original) LPH:
LPH = |{c|xcs ∈ W }|
Entropía de ocurrencias de letras LPE:
LPE = −
X fW (cs)
fW (s)
log2
fW (cs)
fW (s)
El sesgo de la letra más frecuentemente ocurriente LPM:
LPM =
1
− maxc (fW (cs)/fW (s))
1
1−
|Σ|
LPM(ing ) ≈ 0.833, LPM(ng ) ≈ 0.029
LPM(t) ≈ 0.712.
pero también
Las tres medidas son altamente correlacionadas!
LPH&LPE
LPE&LPM
LPM&LPH
r
r -rank
Hammarstrom
0.872
0.957
0.729
0.999
0.998
0.996
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Extracción de ajos
Combinamos sobrerepresentación y heuristicas del confín de la
segmentación:
I Sobrerepresentación como más-frecuente-cuando-sujo: (RA)
I Heuristica del confín: LPM
Da un tanteo para cada sujo s : Z (s) = RA(s) · LPM(s)
El tanteo da una lista
arriba y basura abajo
ordenada
con sujos verdaderos probablemente
No queda claro la mejor manera de delimitar los sujos verdaderos
desde la basura
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Resultados Experimentales 1: Biblia en Inglés
-ed
-eth
-ted
-iah
-ly
-ings
-ing
-ity
-edst
15448.4
12797.1
11899.4
11587.5
10571.2
8038.9
7292.8
6917.6
6844.7
Hammarstrom
-ites
-seth
-ned
-s'
-nded
-ts
-ah
-ness
-s
5370.2
5081.6
4826.7
4305.2
3833.8
3783.1
3766.9
3679.3
3407.3
-ions
-est
-sed
-y
-leth
-nts
-ied
-ened
-ers
2684.5
2452.6
2313.7
2239.2
2166.3
2122.6
1941.7
1834.9
1819.5
-ered
-ded
-neth
...
-ig
-io
-ti
1796.7
1582.2
1540.0
...
0.0
0.0
0.0
March 2015, Mexico
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Resultados Experimentales 2: Biblia en Español
s
n
as
a
en
aren
eras
e
nas
na
39908.25
28210.99
16352.31
12545.77
9180.55
8796.45
8445.28
6401.39
6391.66
4342.84
Hammarstrom
an
ares
ernas
nen
os
es
r
are
ar
et
3902.40
3546.75
3263.43
2980.62
2903.22
2849.25
2741.61
2671.10
2534.42
2203.46
ren
sen
das
is
aron
te
les
ras
er
idas
2117.22
1996.00
1949.44
1649.77
1639.59
1630.10
1559.92
1475.85
1474.87
1450.58
gen
t
tas
rna
aros
den
adas
da
onen
la
1438.76
1420.67
1415.96
1334.99
1311.10
1242.33
1211.48
1012.60
1008.23
952.42
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Resultados Experimentales 3: Biblia en Turco
-larna
-larndan
-lerinin
-lerden
-inden
-iyorlard
-arak
-iyorsunuz
-inin
-dlar
-lere
Hammarstrom
71645.4
47941.9
43917.3
36294.0
35258.2
28716.2
27774.1
25403.1
25045.5
20718.7
20718.2
-ip
-dan
-siniz
-ndan
-ndan
-yorlard
-acaksnz
-adlar
-lerinden
-nden
-sinin
20431.2
19468.4
10391.1
18556.3
18226.3
18097.1
16751.1
16587.9
15201.1
14082.2
13493.9
-nin
-yorsunuz
-larla
-en
-ten
-siniz
-madlar
-lardan
-. . .
-iy
-yo
12340.9
12135.0
12069.7
11513.5
11424.0
11043.0
10958.9
10428.1
...
0.0
0.0
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Resultados Experimentales 4: Biblia en Tarasco
pka
apka
ni
empka
ani
antani
skia
ejka
ka
apiringa
245707.20
169135.60
98790.17
67553.61
64718.98
60578.56
52050.93
49687.70
43963.20
43006.52
Hammarstrom
ntani
i
piringa
aani
ini
akia
ti
aaka
sïni
ajka
38342.84
37914.21
35125.87
33818.15
33591.51
32199.03
24665.53
24647.32
22963.65
22615.23
ksï
jka
aka
tajka
ntajka
pti
rini
aspti
tani
jti
18514.77
17184.20
17164.30
17093.86
16941.85
16618.10
15787.82
15764.78
15364.25
14944.13
tia
chani
hani
stia
axapka
spka
nhani
nhajka
kia
eni
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14
14
14
14
14
14
14
14
13
13
30 / 55
Segmentación: Prejos vs. Sujos
Sueco
0.097
-en
0.086
-na
0.036
-ade
0.035
-a
0.034
-ar
0.033
-er
0.033
-as
0.032
-s
0.031
-de
0.031
...
...
för-
Hammarstrom
Inglés
0.132
-eth
0.109
-iah
0.099
-ly
0.090
-ings
0.068
-ing
0.062
-ity
0.059
-edst
0.058
-ites
0.046
-s'
0.036
...
...
-ed
Swahili
0.100
wa0.095
ali0.065
nita0.059
aka0.049
ni0.046
ku0.044
ata0.042
ha0.032
a0.031
...
...
-a
March 2015, Mexico
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Buscar Paradigmas
Paradigma = Un conjunto de ajos que sistematicamente en la(s)
misma(s) raic(es):
Buscar paradigmas es un problema dicilísimo:
I El número de paradigmas posibles en teoría es exponencial (en el
número de sujos)
I Paradigmas no necesitan ser sin silapamiento (tipicamente en lenguas
naturales sí silapan)
=>
Paradigmas tipicamente tienen miembros muy frecuentes y tambíen
miembros poco frecuentes
Casi cada palabra ocurre solamente en un par de todas las formas de su
paradigma
I Los datos desde los que trabajamos es material crudo textual
F
F
Hammarstrom
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Sesgo de Paradigmas: Ejemplo
Los tres paradigmas suecos más frecuentes
Adjective 1st decl
P.e. gul
-a
2022
-
1821
-t
1572
-e
221
-are
208
-s
114
-aste
90
-ast
46
-as
39
-es
13
-ts
4
-ares
1
Hammarstrom
Noun 3rd decl
P.e. tid
-
1619
-en
1141
-er
1072
-erna
583
-s
310
-ens
259
-ernas
136
-ers
40
Verb 1st decl
P.e. lag-a
1001
-ade
948
-ar
883
-at
579
-as
482
-ande
423
-ad
387
-ades
273
-ats
207
-andes
5
-ads
3
March 2015, Mexico
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Número de formas a la palabra: Ejemplo
Por los tres paradigmas más frecuentes, cuántas palabraas ocurren en
cuántas formas de su paradigma?
# forms
1
2
3
4
5
6
7
8
9
10
11
Adjective 1st decl
# lemmas
1107
703
714
210
54
29
12
4
2
-
Hammarstrom
Noun 3rd decl
# lemmas
1142
538
341
200
105
41
15
5
4
-
Verb 1st decl
# lemmas
606
397
289
227
160
106
76
53
46
13
March 2015, Mexico
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Lista de Quotientes
Primeramente, denir listas de quotientes Hx (y ) : S W → [0, 1] as:
Hx (y ) =
y
ed
00
s
e
es
er
ers
ion
y
ings
ions
in
ation
0
s
ingly
or
able
ive
ors
ations
er 0 s
ment
ly
...
Hammarstrom
|s|sx ∈ W ∧ sy ∈ w |
|s|sx ∈ w |
Hing (y )
y
ing
0.59
00
0.41
0.25
0.24
0.19
0.12
0.10
0.07
0.05
0.05
0.03
0.03
0.03
e
s
es
er
ion
ers
y
ions
ation
able
ings
0
s
or
in
ly
ive
ingly
al
ment
ors
ations
0.03
0.03
0.02
0.02
0.02
0.02
0.01
0.01
0.01
0.01
...
...
Hed (y )
0.42
0.33
0.21
0.20
0.17
0.08
0.07
0.05
0.04
0.03
0.03
0.02
0.02
0.02
0.02
0.01
0.01
0.01
0.01
0.01
0.01
0.01
0.01
...
March 2015, Mexico
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Calidad de Paradigmas
{ing,ed,,
ed
ing
s
e
es
ers
er
y
0
s
ion
er 0 s
d
ly
ings
in
ered
ation
ions
...
s,er}
{ing,ation,'s,xt}
00
1.51
0.94
0.78
0.73
0.61
0.48
0.47
0.24
0.18
ed
st
e
s
nd
sted
sts
nder
nding
nds
nded
xts
ar
ll
aring
nt
sting
nts
0.17
0.15
0.13
0.12
0.12
0.12
0.10
0.10
0.09
0.08
...
...
[2, 1, 0, 3, 7]
0.60
0.60
0.60
0.60
0.42
0.40
0.40
0.40
0.40
0.40
...
[32, 662, 661, 87352]
Medida de calidad de paradigmas
VI (P):
Suma de rangos de los miembros de
relativa a la suma optimal (que depende de
VI (P) =
Hammarstrom
1.83
1.10
0.80
0.64
0.64
0.60
0.60
0.60
0.60
2
|P|
y es 0
P,
+ . . . + |P| − 1):
|P|(|P| − 1)
P
x∈P rankP (x)
March 2015, Mexico
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Crecer Paradigmas
G (P) = argmaxp∈{P∪{s}|s∈S W } VI (p)
P
if G (P) = P
∗
G (P) =
G ∗ (G (P)) if G (P) 6= P
Otra alternativa más osada es permitir también expulsiones:
G (P) = argmaxp∈{P}∪{P
Hammarstrom
xor s|s∈S W } VI (p)
March 2015, Mexico
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Crecer Paradigmas: Ejemplos
P
{-ation}
{-ated, -ation}
{-ate, -ated, -ation}
{-ate, -ated, -ating, -ation}
{-ate, -ated, -ating, -ation, -ations}
P
{-xt}
{-xt, -n}
{-xt, -n, -ns}
{-n, -ns}
...
VI(P)
0.00
0.14
0.40
0.75
1.00
VI(P)
0.00
0.04
0.12
0.55
...
Nota: el estrechez de un paradigma es independiente de la calidad de
segmentación: E.g. VI ({0 xcellent 0 ,0 xcellently 0 }) = 1.0.
Hammarstrom
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Ejemplo de Resultados: Poláco
-a
-
-a¢
-a
-acie
-ach
-aj¡
-ami
-aª
-e
-ali
-em
-aªa
-om
-aªo
-owe
-asz
-owej
-am
-owi
-ane
-owych
-anie
-u
-aniem
-y
-aniu
-ania
-ano
-asz
Hammarstrom
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Una segmentadora sin supervisión para lenguas de
morfología con un espacio
Podemos combinar los tres componentes
I Segmentación
I Sobrerepresantación
I Paradigmas
Se hace una segmentadora completamente sin supervisión
Para una lengua concatenativa con una morfología de un espacio, p.e.,
Indonesio
Hammarstrom
March 2015, Mexico
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Segmentación para el hombre pobre
Input: Un conjunto de palabtas W de un corpus de texto
Un conjunto S de ajos salientes
1
2
de
Extracción de ajos de W
Purga
Por cada s ∈ S , its paradigm P(s)
3
from
Alternación en W
Dado w = xy ∈ W , es que x selecciona P(y ) o no?
4
Decisión de segmentación
Hammarstrom
March 2015, Mexico
41 / 55
Cuándo debe una palabra w segmentarse?
w = xy debe segmentarse si (y solo si):
-y
es un ajo saliente Y
x- selecciona P(y ) (tiende a ocurrir con los sujos que pertenecen al
paradigma de -y)
Intuición:
w = xy
-y saliente?
x- selecciona P(y )?
Segmentación?
play-ing
yes
yes
yes
s-ing
yes
no
no
P(ing ) = {-ing,
Hammarstrom
drea-m
no
yes
no
-ed, -s, -, . . . }
March 2015, Mexico
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Extracción: Sujos en Indonesio
-anmu
-nya
-anku
-lah
-kanlah
-hnya
-inya
-atnya
-kan
-tnya
87195.4
73694.0
51923.8
39535.9
27933.2
20915.8
19677.7
18361.2
18318.4
14237.1
-iel
-snya
-rlah
-annya
-ilah
-anya
-nmu
-mu
-kannya
-an
10000.7
9635.2
9351.8
7802.0
7534.3
7005.4
6531.7
6350.9
5981.8
5931.3
-arlah
-atlah
-anlah
-nglah
-anglah
-akanlah
-hlah
-ah
-nku
-ya
...
-aadil
-aadai
-aaan
Hammarstrom
5702.8
5148.6
4522.9
4121.3
3990.4
3882.9
3658.6
3596.1
3359.6
3203.6
0.0
0.0
0.0
0.0
March 2015, Mexico
43 / 55
Purga de ajos
Quedarse solo con los sujos que son la mejor disección de por lo
menos una palabra
0
UW
= {s|s = argmaxs 0 /w ZW (s 0 ) for some w ∈ W }
0 = {anmu,
UW
nya, anku, lah, kan, nmu, mu, nku, i, ezer,
zabad, inadab, ihud, nadab, arif, obab, ezib, ilene, laf, ilo,
ore, e}
Siguen existir algunos sujos espurio, como
muy escasos
-ilene,
da igual ya que son
Un error grave es la falta de -an (purgado ya que -mu y -ku
frecuentemente aparacen junto a el Indonesio no es puremente una
lengua de un espacio!)
Hammarstrom
March 2015, Mexico
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Listas de quotientes: Indonesio
y
nya
00
mu
ku
kan
lah
an
i
kanlah
kah
ilah
annya
kannya
t
k
anmu
n
m
h
s
ng
l
inya
ya
...
Hammarstrom
Hnya (y )
1.000
0.913
0.261
0.153
0.071
0.057
0.043
0.039
0.034
0.016
0.015
0.014
0.011
0.008
0.008
0.008
0.007
0.007
0.007
0.005
0.005
0.005
0.004
0.004
...
y
mu
00
nya
ku
kan
lah
an
kanlah
i
kah
annya
ilah
anmu
n
m
k
ng
mulah
t
anku
wan
ya
ta
s
...
Hmu (y )
1.000
0.943
0.749
0.393
0.063
0.059
0.056
0.045
0.040
0.022
0.022
0.019
0.015
0.014
0.014
0.012
0.011
0.011
0.009
0.009
0.008
0.007
0.007
0.007
...
March 2015, Mexico
45 / 55
Calidad de Paradigmas: Indonesio
{nya,
mu, , ku}
2.789
nya
1.652
mu
1.004
ku
0.572
lah
0.243
kan
0.231
an
0.197
i
0.157
kanlah
0.137
annya
0.075
ilah
0.068
kah
0.065
n
0.049
anmu
0.047
m
0.043
t
0.037
k
0.036
anku
0.033
ng
0.032
h
0.031
mulah
0.031
ya
0.031
ta
0.029
s
0.027
...
...
[0, 1, 2, 3]
Hammarstrom
{nya,
s, a, ya}
mu
ku
snya
i
kan
lah
an
n
anya
slah
skan
k
ng
san
t
kanlah
m
si
r
h
l
amu
nya
...
[24, 32, 50, 79]
1.025
0.273
0.166
0.098
0.092
0.089
0.076
0.069
0.064
0.058
0.057
0.053
0.053
0.052
0.048
0.045
0.044
0.043
0.042
0.042
0.042
0.030
0.030
0.027
...
March 2015, Mexico
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Crecer Paradigmas: Indonesio
P
VI (P)
('nya',)
0.000
(, 'nya')
0.333
(, 'mu', 'nya')
0.750
(, 'ku', 'mu', 'nya')
1.000
P
VI (P)
('s',)
0.0
('s', 'snya')
0.077
('s', 'smu', 'snya')
0.273
('s', 'sku', 'smu', 'snya')
0.667
('s', 'san', 'sku', 'smu', 'snya')
0.833
('s', 'san', 'skan', 'sku', 'smu', 'snya')
0.882
Hammarstrom
March 2015, Mexico
47 / 55
Decisión de segmentación: Intuición
Considera:
'gadisnya' el
-nya
es una ocurrencia verdadera del sujo
'hanya' termina con la secuencia de letras
-nya
-nya
por azar
No tenemos lexicon de raíces para decirnos que hay una raiz
gadis- pero no hay una raiz ha-!
Hay 247 palabras en W que empiezan con
ha-:
Ninguna de las otras 246 continuaciones son - , -mu o -ku,
i.e., los ajos del paradigma
G ∗ (nya)
sistematicamente
alternatando con -nya!
Hay 3 palabras en W que empiezan con
gadis-
Las otras dos son 'gadis' y 'gadismu', i.e., con - y -mu!
Hammarstrom
March 2015, Mexico
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Decisión de segmentación: Formalmente
0
Dada una palabra w = xs donde s ∈ UW
Denir CW (x) = {y |xy ∈ W } como el cunjunto de continuationes
de una raiz x
Cuánto del contenido de CW (x) aparece dentro versus fuera del
paradigma de s ?
El conjunto de continuaciones dentro I = CW (x) ∩ G ∗ (s)
El conjunto de continuaciones fuera O = CW (x) \ G ∗ (s)
αW (G ∗ (s)) es la proporción de palabras en W que terminan en (algún
miembro de) G ∗ (s)
w = xs debe segmentarse si esta proporción sobrepasa ≥ 1:
(1 − α)|O|
α|I |−1
Hammarstrom
March 2015, Mexico
49 / 55
Decisión de segmentación: Ejemplo
w
x
s
G ∗ (s)
αW G ∗ (s)
CW (x)
I
O
|O| · log(1 − α)
(|I | − 1) · log α
Segmentatación justicada?
Hammarstrom
'gadisnya'
'hanya'
gadis-
ha-
-nya
-nya
{00 , ku, mu, nya}
{00 , ku, mu, nya}
{00 , nya, mu}
{00 , nya, mu}
{}
{nya, sratmu, ncurkanlah, rta, rod, . . . }
{nya}
{sratmu, ncurkanlah, rta, rod, . . . }
0.205
0.0
-3.2
0.0 > −3.2 yes
0.205
-56.9
0.0
−56.9 < 0.0 no
March 2015, Mexico
50 / 55
Evaluación
Datos gold standard para evaluación:
100 palabras seleccionadas por azar de W
Segmentadas a mano con respecto a prejos y sujos
P.e., di-rencana-kan-nya y meng-erut-kan
Un número total de 64 segmentaciones se fueron encontradas en las
100 palabras
Procedimiento:
El algoritmo fue aplicado a las 100 palabras, una vez para sujos y una
vez para prejos.
Hammarstrom
March 2015, Mexico
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Resultados
58 de las segmentationes fueron encontradas correctamente
No se encontraron segmentaciones spurias
6 segmentaciones faltan
Precision: 58/58=100% y Recall: 58/64=90.6%
Todos los 6 de las segmentaciones faltantes fueron palabras con sujos
apilados como -kan-lah, o con -an nal.
Hammarstrom
March 2015, Mexico
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Segmentadora de un espacio sin supervisión: Discussión
Un conjunto bastante largo de heurísticas
No hay prueba matemática con rigor, pero por lo menos todas
heurísticas tienen una intuición clara
No hay umbrales de valores
Pero:
La morfología del Indonesio es muy fácil
El procedimiento no puede aplicarse facilmente a lenguas con más que
un espacio en su morfología
I La extracción de ajos encuentra varios ajos apilados, pero no todos
I El crecimiento de paradigmas encuentra paradigmas demasiadamente
pequeños cuando se aplica a lenguas con más que un espacio
Hammarstrom
March 2015, Mexico
53 / 55
Conclusiones
No hay una teoría madura por ULM
La técnicas posiblemente pueden
I Solucionar el problema con lenguas de morfología de un espacio
I Acercar el problema con lenguas de morfología de más que un espacio
Precisión de aplicaciones con más ambición es questionable
El uso de ULM para facilitar o ayudar transcripción no está explorado
Hammarstrom
March 2015, Mexico
54 / 55
Thank you
Hammarstrom
March 2015, Mexico
55 / 55
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