以编号格式转换斯坦福依赖项

Question

我使用的是Stanford dependency parser，我得到以下句子的输出

我在睡梦中射杀了一只大象

>>>python dep_parsing.py 
[((u'shot', u'VBD'), u'nsubj', (u'I', u'PRP')), ((u'shot', u'VBD'), u'dobj', (u'elephant', u'NN')), ((u'elephant', u'NN'), u'det', (u'an', u'DT')), ((u'shot', u'VBD'), u'nmod', (u'sleep', u'NN')), ((u'sleep', u'NN'), u'case', (u'in', u'IN')), ((u'sleep', u'NN'), u'nmod:poss', (u'my', u'PRP$'))]

不过，需要编号的令牌作为输出，就像 它就在这里一样

nsubj(shot-2, I-1)
  root(ROOT-0, shot-2)
  det(elephant-4, an-3)
  dobj(shot-2, elephant-4)
  case(sleep-7, in-5)
  nmod:poss(sleep-7, my-6)
  nmod(shot-2, sleep-7)

到现在为止这是我的密码。

  from nltk.parse.stanford import StanfordDependencyParser
  stanford_parser_dir = 'stanford-parser/'
  eng_model_path = stanford_parser_dir + "stanford-parser-models/edu/stanford/nlp/models/lexparser/englishRNN.ser.gz"
  my_path_to_models_jar = stanford_parser_dir + "stanford-parser-3.5.2-models.jar"
  my_path_to_jar = stanford_parser_dir + "stanford-parser.jar"

  dependency_parser = StanfordDependencyParser(path_to_jar=my_path_to_jar, path_to_models_jar=my_path_to_models_jar)

  result = dependency_parser.raw_parse('I shot an elephant in my sleep')
  dep = result.next()
  a = list(dep.triples())
  print a

我怎么能有这样的输出？

Answer 1

编写遍历树的递归函数。第一次通过时，只需给单词分配数字即可。