我在python3中使用stanford依赖解析器来解析一个句子,它返回一个依赖图。
import pickle
from nltk.parse.stanford import StanfordDependencyParser
parser = StanfordDependencyParser('stanford-parser-full-2015-12-09/stanford-parser.jar', 'stanford-parser-full-2015-12-09/stanford-parser-3.6.0-models.jar')
sentences = ["I am going there","I am asking a question"]
with open("save.p","wb") as f:
pickle.dump(parser.raw_parse_sents(sentences),f)
它出错了:
AttributeError: Can't pickle local object 'DependencyGraph.__init__.<locals>.<lambda>'
我想知道是否可以使用或不使用pickle保存依赖图。
答案 0 :(得分:3)
关注instructions to get a parsed output。
(请参阅What is CoNLL data format?和What does the dependency-parse output of TurboParser mean?)
$ export STANFORDTOOLSDIR=$HOME
$ export CLASSPATH=$STANFORDTOOLSDIR/stanford-parser-full-2015-12-09/stanford-parser.jar:$STANFORDTOOLSDIR/stanford-parser-full-2015-12-09/stanford-parser-3.6.0-models.jar
$ python
>>> from nltk.parse.stanford import StanfordDependencyParser
>>> dep_parser=StanfordDependencyParser(model_path="edu/stanford/nlp/models/lexparser/englishPCFG.ser.gz")
>>> sent = "The quick brown fox jumps over the lazy dog."
>>> output = next(dep_parser.raw_parse("The quick brown fox jumps over the lazy dog."))
>>> type(output)
<class 'nltk.parse.dependencygraph.DependencyGraph'>
>>> output.to_conll(style=4) # The *style* parameter just means that we want 4 columns in the CONLL format
u'The\tDT\t4\tdet\nquick\tJJ\t4\tamod\nbrown\tJJ\t4\tamod\nfox\tNN\t5\tnsubj\njumps\tVBZ\t0\troot\nover\tIN\t9\tcase\nthe\tDT\t9\tdet\nlazy\tJJ\t9\tamod\ndog\tNN\t5\tnmod\n'
>>> with open('sent.conll', 'w') as fout:
... fout.write(output.to_conll(4))
...
>>> exit()
$ cat sent.conll
The DT 4 det
quick JJ 4 amod
brown JJ 4 amod
fox NN 5 nsubj
jumps VBZ 0 root
over IN 9 case
the DT 9 det
lazy JJ 9 amod
dog NN 5 nmod
>>> from nltk.parse.dependencygraph import DependencyGraph
>>> output = DependencyGraph.load('sent.conll') # Loads any CONLL file, that might contain 1 or more sentences.
>>> output # list of DependencyGraphs
[<DependencyGraph with 10 nodes>]
>>> output[0] # the first DependencyGraph, the one we have saved
<DependencyGraph with 10 nodes>
>>> print output[0]
defaultdict(<function <lambda> at 0x10e83c758>, {0: {u'ctag': u'TOP', u'head': None, u'word': None, u'deps': defaultdict(<type 'list'>, {u'ROOT': [], u'root': [5]}), u'lemma': None, u'tag': u'TOP', u'rel': None, u'address': 0, u'feats': None}, 1: {u'ctag': u'DT', u'head': 4, u'deps': defaultdict(<type 'list'>, {}), u'tag': u'DT', u'address': 1, u'word': u'The', u'lemma': u'The', u'rel': u'det', u'feats': u''}, 2: {u'ctag': u'JJ', u'head': 4, u'deps': defaultdict(<type 'list'>, {}), u'tag': u'JJ', u'address': 2, u'word': u'quick', u'lemma': u'quick', u'rel': u'amod', u'feats': u''}, 3: {u'ctag': u'JJ', u'head': 4, u'deps': defaultdict(<type 'list'>, {}), u'tag': u'JJ', u'address': 3, u'word': u'brown', u'lemma': u'brown', u'rel': u'amod', u'feats': u''}, 4: {u'ctag': u'NN', u'head': 5, u'deps': defaultdict(<type 'list'>, {u'det': [1], u'amod': [2, 3]}), u'tag': u'NN', u'address': 4, u'word': u'fox', u'lemma': u'fox', u'rel': u'nsubj', u'feats': u''}, 5: {u'ctag': u'VBZ', u'head': 0, u'deps': defaultdict(<type 'list'>, {u'nmod': [9], u'nsubj': [4]}), u'tag': u'VBZ', u'address': 5, u'word': u'jumps', u'lemma': u'jumps', u'rel': u'root', u'feats': u''}, 6: {u'ctag': u'IN', u'head': 9, u'deps': defaultdict(<type 'list'>, {}), u'tag': u'IN', u'address': 6, u'word': u'over', u'lemma': u'over', u'rel': u'case', u'feats': u''}, 7: {u'ctag': u'DT', u'head': 9, u'deps': defaultdict(<type 'list'>, {}), u'tag': u'DT', u'address': 7, u'word': u'the', u'lemma': u'the', u'rel': u'det', u'feats': u''}, 8: {u'ctag': u'JJ', u'head': 9, u'deps': defaultdict(<type 'list'>, {}), u'tag': u'JJ', u'address': 8, u'word': u'lazy', u'lemma': u'lazy', u'rel': u'amod', u'feats': u''}, 9: {u'ctag': u'NN', u'head': 5, u'deps': defaultdict(<type 'list'>, {u'case': [6], u'det': [7], u'amod': [8]}), u'tag': u'NN', u'address': 9, u'word': u'dog', u'lemma': u'dog', u'rel': u'nmod', u'feats': u''}})
请注意,StanfordParser
的输出是nltk.tree.Tree
而不是DependencyGraph
(这只是有人在树上发布类似问题。
$ export STANFORDTOOLSDIR=$HOME
$ export CLASSPATH=$STANFORDTOOLSDIR/stanford-parser-full-2015-12-09/stanford-parser.jar:$STANFORDTOOLSDIR/stanford-parser-full-2015-12-09/stanford-parser-3.6.0-models.jar
$ python
>>> from nltk.parse.stanford import StanfordParser
>>> parser=StanfordParser(model_path="edu/stanford/nlp/models/lexparser/englishPCFG.ser.gz")
>>> list(parser.raw_parse("the quick brown fox jumps over the lazy dog"))
[Tree('ROOT', [Tree('NP', [Tree('NP', [Tree('DT', ['the']), Tree('JJ', ['quick']), Tree('JJ', ['brown']), Tree('NN', ['fox'])]), Tree('NP', [Tree('NP', [Tree('NNS', ['jumps'])]), Tree('PP', [Tree('IN', ['over']), Tree('NP', [Tree('DT', ['the']), Tree('JJ', ['lazy']), Tree('NN', ['dog'])])])])])])]
>>> output = list(parser.raw_parse("the quick brown fox jumps over the lazy dog"))
>>> type(output[0])
<class 'nltk.tree.Tree'>
对于nltk.tree.Tree
,您可以将其输出为括号中的解析字符串,并将字符串读入Tree对象:
>>> from nltk import Tree
>>> output[0]
Tree('ROOT', [Tree('NP', [Tree('NP', [Tree('DT', ['the']), Tree('JJ', ['quick']), Tree('JJ', ['brown']), Tree('NN', ['fox'])]), Tree('NP', [Tree('NP', [Tree('NNS', ['jumps'])]), Tree('PP', [Tree('IN', ['over']), Tree('NP', [Tree('DT', ['the']), Tree('JJ', ['lazy']), Tree('NN', ['dog'])])])])])])
>>> str(output[0])
'(ROOT\n (NP\n (NP (DT the) (JJ quick) (JJ brown) (NN fox))\n (NP\n (NP (NNS jumps))\n (PP (IN over) (NP (DT the) (JJ lazy) (NN dog))))))'
>>> parsed_sent = str(output[0])
>>> type(parsed_sent)
<type 'str'>
>>> Tree.fromstring(parsed_sent)
Tree('ROOT', [Tree('NP', [Tree('NP', [Tree('DT', ['the']), Tree('JJ', ['quick']), Tree('JJ', ['brown']), Tree('NN', ['fox'])]), Tree('NP', [Tree('NP', [Tree('NNS', ['jumps'])]), Tree('PP', [Tree('IN', ['over']), Tree('NP', [Tree('DT', ['the']), Tree('JJ', ['lazy']), Tree('NN', ['dog'])])])])])])
>>> parsed_tree = Tree.fromstring(parsed_sent)
>>> type(parsed_tree)
<class 'nltk.tree.Tree'>