in reply to Help with regex - find captured pattern twice

A bit broader than you specified - this replaces any single letter and optional parenthetical value. 

$_= "c*theta*h^(1+psi)=(1-alpha)*y;\n1 = beta*(c/c(+1)*alpha*y(+1)/k+(
+1-delta));\ny = a*(k(-1)^alpha)*(h^(1-alpha));\nk = y-c+(1-delta)*k(-
+1);\nln(a) = rho*ln(a(-1))+ e;\n";

s/(?<!\pL)(\pL)(\([\d+-]+\))?(?!\pL)/(${1}SS*exp($1$2))/g;

print;

#prints
(cSS*exp(c))*theta*(hSS*exp(h))^(1+psi)=(1-alpha)*(ySS*exp(y));
1 = beta*((cSS*exp(c))/(cSS*exp(c(+1)))*alpha*(ySS*exp(y(+1)))/(kSS*ex
+p(k))+(1-delta));
(ySS*exp(y)) = (aSS*exp(a))*((kSS*exp(k(-1)))^alpha)*((hSS*exp(h))^(1-
+alpha));
(kSS*exp(k)) = (ySS*exp(y))-(cSS*exp(c))+(1-delta)*(kSS*exp(k(-1)));
ln((aSS*exp(a))) = rho*ln((aSS*exp(a(-1))))+ (eSS*exp(e));

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Re^2: Help with regex - find captured pattern twice
by pachydermic (Beadle) on Apr 11, 2013 at 13:52 UTC

    Wow. What the heck is that monstrosity?! Well done.

    Would you mind explaining that regex? I think that \pL references a unicode character class (from this page: http://perldoc.perl.org/perlrecharclass.html) but I'm still kind of baffled by what's going on...
[reply]

      You might read it as this: 

      s/
  (?<!        # 2. NOT preceded by
     \pL      #    a letter
  )

  (           # 1. Capture
     \pL      #    a letter
  )

  (           # 3. Capture
     \(       #    open paren;
     [\d+-]+  #    one or more digits, plus, minus;
     \)       #    close paren
  )?          # (Zero or one time)

  (?!         # 4. NOT followed by
     \pL      #    a letter
  )

/(${1}SS*exp($1$2))/gx;

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