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    <p>Since you are dealing with nested parenthesis, regex can't help you much here. It cannot easily convert the string to the list, as you wanted at the end.</p> <p>You would better go with parsing the string yourself. You can try this code, just to meet your requirement at the end:</p> <h2>Parsing string into list without loosing parenthesis:</h2> <pre><code>def parse(s): li = [] open = 0 closed = False start_index = -1 for index, c in enumerate(s): if c == '(': if open == 0: start_index = index open += 1 elif c == ')': open -= 1 if open == 0: closed = True elif closed: li.append(s[start_index: index + 1]) closed = False elif open == 0 and c.isdigit(): li.append(c) return li </code></pre> <p>This will give you for the string <code>'1/(2/(3)2/4)s/5'</code> the following list:</p> <pre><code>['1', '(2/(3)2/4)s', '5'] </code></pre> <p>and for the string <code>'1/(15/-23)s/4'</code>, as per your changed requirement, this gives:</p> <pre><code>['1', '(15/-23)s', '4'] </code></pre> <p>Now, you need to take care of the breaking the parenthesis further up to get different list elements.</p> <hr /> <h2>Expanding the strings with parenthesis to individual list elements:</h2> <p>Here you can make use of a regex, by just dealing with inner-most parenthesis at once:</p> <pre><code>import re def expand(s): ''' Group 1 contains the string inside the parenthesis Group 2 contains the digit or character `s` after the closing parenthesis ''' match = re.search(r'\(([^()]*)\)(\d|s)', s) if match: group0 = match.group() group1 = match.group(1) group2 = match.group(2) if group2.isdigit(): # A digit after the closing parenthesis. Repeat the string inside s = s.replace(group0, ((group1 + '/') * int(group2))[:-1]) else: s = s.replace(group0, '/'.join(group1.split('/') + group1.split('/')[::-1])) if '(' in s: return expand(s) return s li = parse('1/(15/-23)2/4') for index, s in enumerate(li): if '(' in s: s = expand(s) li[index] = s.split('/') import itertools print list(itertools.chain(*li)) </code></pre> <p>This will give you the required result:</p> <pre><code>['1', '15', '-23', '-23', '15', '4'] </code></pre> <p>The above code iterates over the list generated from <code>parse(s)</code> method, and then for each element, recursively expands the inner most parenthesis.</p>
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    1. POPython/Regex - Expansion of Parentheses and Slashes
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    1. USRohit Jain
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    1. POPython/Regex - Expansion of Parentheses and Slashes
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    1. COSee my edit. It doesn't seem to work with numbers >10 or negatives. I didn't say that in my post, but one of the other solutions seems to handle them ok.
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      1. USScott B
    2. CO@ScottB. Well, you never said, you need to handle number > 10, or negative numbers, as you said. But still, good that you got your issue resolved.
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      1. USRohit Jain
    3. CORight, I didn't mention it, but I'll probably use a combination of Joowani's post and your second half to expand the parentheses. Thanks.
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