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69 lines
2.4 KiB
Plaintext
69 lines
2.4 KiB
Plaintext
.. Copyright (C) 2001-2020 NLTK Project
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.. For license information, see LICENSE.TXT
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.. see also: gluesemantics.doctest
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==============================================================================
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Glue Semantics
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==============================================================================
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>>> from nltk.sem.glue import *
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>>> nltk.sem.logic._counter._value = 0
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--------------------------------
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Initialize the Dependency Parser
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--------------------------------
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>>> from nltk.parse.malt import MaltParser
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>>> tagger = RegexpTagger(
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... [('^(John|Mary)$', 'NNP'),
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... ('^(sees|chases)$', 'VB'),
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... ('^(a)$', 'ex_quant'),
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... ('^(every)$', 'univ_quant'),
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... ('^(girl|dog)$', 'NN')
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... ])
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>>> depparser = MaltParser(tagger=tagger)
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--------------------
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Automated Derivation
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--------------------
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>>> glue = Glue(depparser=depparser)
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>>> readings = glue.parse_to_meaning('every girl chases a dog'.split())
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>>> for reading in sorted([r.simplify().normalize() for r in readings], key=str):
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... print(reading.normalize())
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all z1.(girl(z1) -> exists z2.(dog(z2) & chases(z1,z2)))
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exists z1.(dog(z1) & all z2.(girl(z2) -> chases(z2,z1)))
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>>> drtglue = DrtGlue(depparser=depparser)
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>>> readings = drtglue.parse_to_meaning('every girl chases a dog'.split())
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>>> for reading in sorted([r.simplify().normalize() for r in readings], key=str):
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... print(reading)
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([],[(([z1],[girl(z1)]) -> ([z2],[dog(z2), chases(z1,z2)]))])
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([z1],[dog(z1), (([z2],[girl(z2)]) -> ([],[chases(z2,z1)]))])
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--------------
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With inference
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--------------
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Checking for equality of two DRSs is very useful when generating readings of a sentence.
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For example, the ``glue`` module generates two readings for the sentence
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*John sees Mary*:
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>>> from nltk.sem.glue import DrtGlue
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>>> readings = drtglue.parse_to_meaning('John sees Mary'.split())
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>>> for drs in sorted([r.simplify().normalize() for r in readings], key=str):
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... print(drs)
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([z1,z2],[John(z1), Mary(z2), sees(z1,z2)])
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([z1,z2],[Mary(z1), John(z2), sees(z2,z1)])
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However, it is easy to tell that these two readings are logically the
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same, and therefore one of them is superfluous. We can use the theorem prover
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to determine this equivalence, and then delete one of them. A particular
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theorem prover may be specified, or the argument may be left off to use the
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default.
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>>> readings[0].equiv(readings[1])
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True
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