python - 将嵌套括号树转换为嵌套列表

Question

我有一个树结构文件，其中括号用来表示树。这是将其转换为 Python 嵌套列表的代码

def foo(s):
    def foo_helper(level=0):
        try:
            token = next(tokens)
        except StopIteration:
            if level != 0:
                raise Exception('missing closing paren')
            else:
                return []
        if token == ')':
            if level == 0:
                raise Exception('missing opening paren')
            else:
                return []
        elif token == '(':
            return [foo_helper(level+1)] + foo_helper(level)
        else:
            return [token] + foo_helper(level)
    tokens = iter(s)
    return foo_helper()    

由 how to parse a strign and return nested array .

这里它工作正常，当字符长度为 1 时。对于单词或句子，同样不能正常工作。我的树样本是:

( Satellite (span 69 74) (rel2par Elaboration)
        ( Nucleus (span 69 72) (rel2par span)
          ( Nucleus (span 69 70) (rel2par span)
            ( Nucleus (leaf 69) (rel2par span) (text _!MERRILL LYNCH READY ASSETS TRUST :_!) )
            ( Satellite (leaf 70) (rel2par Elaboration) (text _!8.65 % ._!) )
          )
          ( Satellite (span 71 72) (rel2par Elaboration)
            ( Nucleus (leaf 71) (rel2par span) (text _!Annualized average rate of return_!) )
            ( Satellite (leaf 72) (rel2par Temporal) (text _!after expenses for the past 30 days ;_!) )
          )
        )
        ( Satellite (span 73 74) (rel2par Elaboration)
          ( Nucleus (leaf 73) (rel2par span) (text _!not a forecast_!) )
          ( Satellite (leaf 74) (rel2par Elaboration) (text _!of future returns ._!) )
        )
      )

这里，输出需要是['satellite',['span','69','74'].........] 但是对于给定的函数，我得到的是 ['s ','a',''......['s','p','a','n','7','3']。 ...............]

如何修改？

Answer 1

我假设您想使用 _! 来表示带空格的字符串。然后我使用正则表达式拆分表达式:

from re import compile
resexp = compile(r'([()]|_!)')
…
  tokens = iter(resexp.split(s))
…

我的结果是(使用深度为 4 的 pprint)

$ python lispparse.py  | head
['\n',
 [' Satellite ',
  ['span 69 74'],
  ' ',
  ['rel2par Elaboration'],
  '\n        ',
  [' Nucleus ',
   ['span 69 72'],
   ' ',
   ['rel2par span'],

我进一步完善了它:

tokens = iter(filter(None, (i.strip() for i in resexp.split(s))))

得到:

$ python lispparse.py  
[['Satellite',
  ['span 69 74'],
  ['rel2par Elaboration'],
  ['Nucleus',
   ['span 69 72'],
   ['rel2par span'],
   ['Nucleus', [...], [...], [...], [...]],
   ['Satellite', [...], [...], [...], [...]]],
  ['Satellite',
   ['span 73 74'],
   ['rel2par Elaboration'],
   ['Nucleus', [...], [...], [...]],
   ['Satellite', [...], [...], [...]]]]]