将嵌套括号树转为嵌套列表

Question

我有一个树形结构的文件，里面用括号来表示这个树。下面是把它转换成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()    

这个方法是参考了 如何解析字符串并返回嵌套数组。

在这里，当字符长度为1时，这个方法运行得很好。但是对于单词或句子，它就不太好用了。 我的树的示例是：

( 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 ._!) )
        )
      )

在这里，我希望输出应该是 ['satellite',['span','69','74'].........]，但是用这个给定的函数，我得到的却是 ['s','a','t'...............['s','p','a','n','7','3']..............]

我该如何修改这个呢？

Answer 1

你应该不是直接在字符串上调用这个函数，而是要在一个词语列表上调用，也就是把字符串用split分开：

def parse(s):
    def parse_helper(level=0):
        try:
            token = next(tokens)
        except StopIteration:
            if level:
                raise Exception('Missing close paren')
            else:
                return []
        if token == ')':
            if not level:
                raise Exception('Missing open paren')
            else:
                return []
        elif token == '(':
            return [parse_helper(level+1)] + parse_helper(level)
        else:
            return [token] + parse_helper(level)
    tokens = iter(s)
    return parse_helper()

if __name__ == '__main__':
    with open('tree.thing', 'r') as treefile:
        tree = treefile.read()

    print(parse(tree.split()))

在这里，treefile包含了你发的那个数据结构，我得到了这个输出：

[['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', '._!)']]]]

Answer 2

我以为你想用 _! 来表示带空格的字符串。于是我用正则表达式把这个表达式拆分开了：

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

我得到的结果是（使用 pprint，深度设置为4）

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

我又稍微改进了一下，得到了：

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

最后得到了：

$ 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', [...], [...], [...]]]]]