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POReplacing all whitespace between brackets using regex in Python
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19276731
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2013-10-09T15:49:09.693
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2013-10-09T16:15:24.133
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I need to replace all unnecessary whitespace between brackets in lisp-like code like this: <pre><code>(print (number->string ( (func? '+ '-) 12 13 14 15) ) ) </code></pre> So that it becomes: <pre><code>(display (number->string ((func? '+ '-) 12 13 14 15))) </code></pre> I know it looks really simple at first sight, but it's really tricky(at least to me). Here's what I've tried: <pre><code>In [14]: re.sub(r"\)[ \t]*\)","))",")) ) ",0) Out[14]: ')) ) ' In [15]: re.sub(r"\)[ \t]*\)","))",") ) ",0) Out[15]: ')) ' In [16]: re.sub(r"\)[ \t]+\)","))",")) ) ",0) Out[16]: '))) ' In [17]: re.sub(r"\)[ \t]+\)","))",")) ) ",0) Out[17]: '))) ' In [18]: re.sub(r"\)[ \t]+\)","))",") ) ) ",0) Out[18]: ')) ) ' In [19]: re.sub(r"\)[ \t]+\)","))",") ) ) ",1) Out[19]: ')) ) ' In [20]: re.sub(r"\)[ \t]+\)","))",") ) ) ",0) Out[20]: ')) ) ' In [21]: re.sub(r"\)[ \t]+\)","))",") ) ) ",0,re.M) Out[21]: ')) ) ' In [22]: re.sub(r"\).+\)","))",") ) ) ",0,re.M) Out[22]: ')) ' In [23]: re.sub(r"\)\s+\)","))",") ) ) ",0,re.M) Out[23]: ')) ) ' In [24]: re.sub(r"\)([ \t]*)\)","))",") ) ",0) Out[24]: ')) ' In [25]: re.sub(r"\)([ \t]*)\)","",") ) ",0) Out[25]: ' ' In [26]: re.sub("\)([ \t]*)\)","))",") ) ",0) Out[26]: ')) ' In [27]: re.sub("\)([ \t]*)\)","))",") ) )",0) Out[27]: ')) )' In [28]: re.sub("\)([ \t]*)\)","))",") ) )",0) Out[28]: ')) )' In [29]: re.sub("\)([ \t]*?)\)","))",") ) )",0) Out[29]: ')) )' In [30]: re.sub("\)(.+?)\)","))",") ) )",0) Out[30]: ')) )' </code></pre> In every case above, <code>re.sub</code> replaces the whitespace between the first brackets only and leaves the rest of the string the same way it was before. I need a regexp to replace all the whitespace.
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<python><regex>
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Replacing all whitespace between brackets using regex in Python
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