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copied!<p>Focusing only on the example you're giving in this Q, and the Q's title about converting vertical trees to S-expressions, something like...:</p> <pre><code>import re import sys samp='''S =NP ==(DT +def) the == (N +ani) man =VP ==V walks'''.splitlines() relinelev = re.compile(r'(=*)(.*)') reclean = re.compile(r'\s*\((\S+)[^)]*\)') def clean(line): return reclean.sub(r'\1', line) def reparse(tree=samp): stack = [-1] for line in tree: equals, rest = relinelev.match(line).groups() linelev = len(equals) while linelev < stack[-1]: sys.stdout.softspace = False print ')', curlev = stack.pop() if linelev == stack[-1]: sys.stdout.softspace = False print ')', else: stack.append(linelev) print '(%s' % clean(rest), while stack[-1] >= 0: sys.stdout.softspace = False print ')', stack.pop() print reparse() </code></pre> <p>seems to work, and outputs</p> <pre><code>(S (NP (DT the) (N man)) (VP (V walks))) </code></pre> <p>I realize you're trying to do much more "cleaning" than I'm doing here, but that can be concentrated in the <code>clean</code> function, leaving <code>reparse</code> to deal with the Q's title. If you don't want to print as you go, but rather return the result as a string, the changes are of course quite minor:</p> <pre><code>def reparse(tree=samp): stack = [-1] result = [] for line in tree: equals, rest = relinelev.match(line).groups() linelev = len(equals) while linelev < stack[-1]: result[-1] += ')' curlev = stack.pop() if linelev == stack[-1]: result[-1] += ')' else: stack.append(linelev) result.append('(%s' % clean(rest)) while stack[-1] >= 0: result[-1] += ')' stack.pop() return ' '.join(result) </code></pre>

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