如何在Python中将nltk树（Stanford）转换为newick格式？

Question

我有这个斯坦福树，我想把它转换成newick格式。

(ROOT 
    (S 
     (NP (DT A) (NN friend)) 
     (VP 
     (VBZ comes) 
     (NP 
      (NP (JJ early)) 
      (, ,) 
      (NP 
      (NP (NNS others)) 
      (SBAR 
       (WHADVP (WRB when)) 
       (S (NP (PRP they)) (VP (VBP have) (NP (NN time)))))))))) 

Answer 1

可能有方法只是使用字符串处理来做到这一点，但我会解析它们并递归地以newick格式打印它们。一个有点最小的实现：

import re 

class Tree(object): 
    def __init__(self, label): 
     self.label = label 
     self.children = [] 

    @staticmethod 
    def _tokenize(string): 
     return list(reversed(re.findall(r'\(|\)|[^ \n\t()]+', string))) 

    @classmethod 
    def from_string(cls, string): 
     tokens = cls._tokenize(string) 
     return cls._tree(tokens) 

    @classmethod 
    def _tree(cls, tokens): 
     t = tokens.pop() 
     if t == '(': 
      tree = cls(tokens.pop()) 
      for subtree in cls._trees(tokens): 
       tree.children.append(subtree) 
      return tree 
     else: 
      return cls(t) 

    @classmethod 
    def _trees(cls, tokens): 
     while True: 
      if not tokens: 
       raise StopIteration 
      if tokens[-1] == ')': 
       tokens.pop() 
       raise StopIteration 
      yield cls._tree(tokens) 

    def to_newick(self): 
     if self.children and len(self.children) == 1: 
      return ','.join(child.to_newick() for child in self.children) 
     elif self.chilren: 
      return '(' + ','.join(child.to_newick() for child in self.children) + ')' 
     else: 
      return self.label 

注意，当然，信息会在转换过程中丢失，因为只有终端节点被保留。用法：

>>> s = """(ROOT (...""" 
>>> Tree.from_string(s).to_newick() 
...