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PORegex: How do I capture a group after an optional capturing group using regular expressions?
 

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  1. PORegex: How do I capture a group after an optional capturing group using regular expressions?
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    <p>Suppose I have the following strings:</p> <pre><code>s1=u'--FE(-)---' s2=u'--FEM(-)---' s3=u'--FEE(--)-' </code></pre> <p>and I want to match F,E,E,M and the content of the parentheses in different groups.</p> <p>I have tried the following regular expression:</p> <pre><code>u'^.-([F])([EF]*)([E]+)[^FEM]?(M*)?(\\(.*\\))?.*$' </code></pre> <p>This expressions give the following groups and spans for the different strings:</p> <pre><code>s1 -&gt; 'F',(2,3) , '',(3,3) , 'E',(3,4) , '',(5,5) , None,(-1,-1) s2 -&gt; 'F',(2,3) , '',(3,3) , 'E',(3,4) , 'M',(4,5) , (-),(5,8) s3 -&gt; 'F',(2,3) , 'E',(3,4) , 'E',(4,5) , '',(6,6) , None,(-1,-1) </code></pre> <p>For s2, I get the wanted behaviour, a matching of the contents of the parentheses, but for s1 and s3 I don't. </p> <p>How do I create a regular expression that will match the content of the parentheses even if I don't have a proper match for the group containing 'M's?</p> <p>EDIT:</p> <p>The answer by DWilches resolved the initial issue using the regular expression</p> <pre><code>'^.-(F)([EF]*)(E+)[^FEM]??(M*)(\(.*\)).*?$' </code></pre> <p>However, the parentheses group is also optional. The following short python script clarifies the problem:</p> <pre><code>s1=u'--FE(-)---' s2=u'--FEM(-)--' s3=u'--FEE(--)-' s4=u'--FEE-M(---)--' s5=u'--FE-M-(-)-' s6=u'--FEM--' s7=u'--FE-M--' ll=[s1,s2,s3,s4,s5,s6,s7] import re rr1=re.compile(u'^.-(F)([EF]*)(E+)[^FEM]??(M*)[^FEM]??(\(.*\)).*?$') rr2=re.compile(u'^.-(F)([EF]*)(E+)[^FEM]??(M*)[^FEM]??(\(.*\))?.*?$') for s in ll: b=rr1.search(s) print s if b: print " '%s' '%s' '%s' '%s' '%s' " % (b.group(1), b.group(2), b.group(3), b.group(4), b.group(5)) else: print 'No match' print '######' </code></pre> <p>For <code>rr1</code>, the output is:</p> <pre><code>--FE(-)--- 'F' '' 'E' '' '(-)' ###### --FEM(-)-- 'F' '' 'E' 'M' '(-)' ###### --FEE(--)- 'F' 'E' 'E' '' '(--)' ###### --FEE-M(---)-- 'F' 'E' 'E' 'M' '(---)' ###### --FE-M-(-)- 'F' '' 'E' 'M' '(-)' ###### --FEM-- No match ###### --FE-M-- No match ###### </code></pre> <p>It is OK for the first 5 strings, but not for the two last, since it requires the parentheses.</p> <p>The <code>rr2</code>, however, adding <code>?</code> to <code>(\(.*\))</code>, yields the following output:</p> <pre><code>--FE(-)--- 'F' '' 'E' '' '(-)' ###### --FEM(-)-- 'F' '' 'E' 'M' '(-)' ###### --FEE(--)- 'F' 'E' 'E' '' '(--)' ###### --FEE-M(---)-- 'F' 'E' 'E' '' 'None' ###### --FE-M-(-)- 'F' '' 'E' '' 'None' ###### --FEM-- 'F' '' 'E' 'M' 'None' ###### --FE-M-- 'F' '' 'E' '' 'None' ###### </code></pre> <p>This is ok for <code>s1,s2,s3</code> and <code>s6</code>.</p> <p>Some modification is needed to yield the desired output: getting the <code>M</code> if it exists and the content of the parentheses if the parentheses exist.</p>
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