Python parsing bracketed blocks

Question

What would be the best way in Python to parse out chunks of text contained in matching brackets?

\"{ { a } { b } { { { c } } } }\"

should i

Answer 1

You could also parse them all at once, though I find the {a} to mean "a" rather than ["a"] slightly weird. If I've understood the format correctly:

import re
import sys


_mbrack_rb = re.compile("([^{}]*)}") # re.match doesn't have a pos parameter
def mbrack(s):
  """Parse matching brackets.

  >>> mbrack("{a}")
  'a'
  >>> mbrack("{{a}{b}}")
  ['a', 'b']
  >>> mbrack("{{a}{b}{{{c}}}}")
  ['a', 'b', [['c']]]

  >>> mbrack("a")
  Traceback (most recent call last):
  ValueError: expected left bracket
  >>> mbrack("{a}{b}")
  Traceback (most recent call last):
  ValueError: more than one root
  >>> mbrack("{a")
  Traceback (most recent call last):
  ValueError: expected value then right bracket
  >>> mbrack("{a{}}")
  Traceback (most recent call last):
  ValueError: expected value then right bracket
  >>> mbrack("{a}}")
  Traceback (most recent call last):
  ValueError: unbalanced brackets (found right bracket)
  >>> mbrack("{{a}")
  Traceback (most recent call last):
  ValueError: unbalanced brackets (not enough right brackets)
  """
  stack = [[]]
  i, end = 0, len(s)
  while i < end:
    if s[i] != "{":
      raise ValueError("expected left bracket")
    elif i != 0 and len(stack) == 1:
      raise ValueError("more than one root")
    while i < end and s[i] == "{":
      L = []
      stack[-1].append(L)
      stack.append(L)
      i += 1
    stack.pop()
    stack[-1].pop()
    m = _mbrack_rb.match(s, i)
    if m is None:
      raise ValueError("expected value then right bracket")
    stack[-1].append(m.group(1))
    i = m.end(0)
    while i < end and s[i] == "}":
      if len(stack) == 1:
        raise ValueError("unbalanced brackets (found right bracket)")
      stack.pop()
      i += 1
  if len(stack) != 1:
    raise ValueError("unbalanced brackets (not enough right brackets)")
  return stack[0][0]


def main(args):
  if args:
    print >>sys.stderr, "unexpected arguments: %r" % args
  import doctest
  r = doctest.testmod()
  print r
  return r[0]

if __name__ == "__main__":
  sys.exit(main(sys.argv[1:]))

Answer 2

If you want to use a parser (lepl in this case), but still want the intermediate results rather than a final parsed list, then I think this is the kind of thing you were looking for:

>>> nested = Delayed()
>>> nested += "{" + (nested[1:,...]|Any()) + "}"
>>> split = (Drop("{") & (nested[:,...]|Any()) & Drop("}"))[:].parse
>>> split("{{a}{b}{{{c}}}}")
['{a}{b}{{{c}}}']
>>> split("{a}{b}{{{c}}}")
['a', 'b', '{{c}}']
>>> split("{{c}}")
['{c}']
>>> split("{c}")
['c']

That might look opaque at first, but it's fairly simple really :o)

nested is a recursive definition of a matcher for nested brackets (the "+" and [...] in the definition keep everything as a single string after it has been matched). Then split says match as many as possible ("[:]") of something that is surrounded by "{" ... "}" (which we discard with "Drop") and contains either a nested expression or any letter.

Finally, here's a lepl version of the "all in one" parser that gives a result in the same format as the pyparsing example above, but which (I believe) is more flexible about how spaces appear in the input:

>>> with Separator(~Space()[:]):
...     nested = Delayed()
...     nested += Drop("{") & (nested[1:] | Any()) & Drop("}") > list
...
>>> nested.parse("{{ a }{ b}{{{c}}}}")
[[['a'], ['b'], [[['c']]]]]

Answer 3

Here is a solution I came up with for a similar use case. This was loosely based on the accepted psuedo code answer. I didn't want to add any dependencies for external libraries:

def parse_segments(source, recurse=False):
    """
    extract any substring enclosed in parenthesis
    source should be a string
    """
    unmatched_count = 0
    start_pos = 0
    opened = False
    open_pos = 0
    cur_pos = 0

    finished = []
    segments = []

    for character in source:
        #scan for mismatched parenthesis:
        if character == '(':
            unmatched_count += 1
            if not opened:
                open_pos = cur_pos
            opened = True

        if character == ')':
            unmatched_count -= 1

        if opened and unmatched_count == 0:
            segment = source[open_pos:cur_pos+1]
            segments.append(segment)
            clean = source[start_pos:open_pos]
            if clean:
                finished.append(clean)
            opened = False
            start_pos = cur_pos+1

        cur_pos += 1

    assert unmatched_count == 0

    if start_pos != cur_pos:
        #get anything that was left over here
        finished.append(source[start_pos:cur_pos])

    #now check on recursion:
    for item in segments:
        #get rid of bounding parentheses:
        pruned = item[1:-1]
        if recurse:
            results = parse_tags(pruned, recurse)
            finished.expand(results)
        else:
            finished.append(pruned)

    return finished