S-exression parsing

[George Sakkis]

The S-expression parser below works, but I wonder if it can be simplified; it's not as short and
straightforward as I would expect given the simplicity of S-expressions. Can you see a simpler
non-recursive solution ?

George

# usage
>>> parseSexpression("(a (b c) (d))")
['a', ['b', 'c'], ['d']]
>>> parseSexpression("(a (b c)) (d))")
ValueError: Unbalanced right parenthesis: (a (b c)) (d))
>>> parseSexpression("(a ((b c) (d))")
ValueError: Unbalanced left parenthesis: (a ((b c) (d))
>>> parseSexpression("(a (b c)) (d)")
ValueError: Single s-expression expected (2 given)
>>> parseSexpression("(a (b c)) e")
ValueError: Unenclosed subexpression (near e)


def parseSexpression(expression):
    subexpressions,stack = [],[]
    for token in re.split(r'([()])|\s+', expression):
        if token == '(':
            new = []
            if stack:
                stack[-1].append(new)
            else:
                subexpressions.append(new)
            stack.append(new)
        elif token == ')':
            try: stack.pop()
            except IndexError:
                raise ValueError("Unbalanced right parenthesis: %s" % expression)
        elif token:
            try: stack[-1].append(token)
            except IndexError:
                raise ValueError("Unenclosed subexpression (near %s)" % token)
    if stack:
        raise ValueError("Unbalanced left parenthesis: %s" % expression)
    if len(subexpressions) > 1:
        raise ValueError("Single s-expression expected (%d given)" % len(subexpressions))
    return subexpressions[0]


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