Regex to find variants of “Google”












4












$begingroup$


From the HackerRank question definition:




The word google can be spelled in many different ways.



E.g. google, g00gle, g0oGle, g<>0gl3, googl3, GooGIe etc...



Because



g = G



o = O = 0 = () = = <>



l = L = I



e = E = 3



That's the problem here to solve.



And the match has to be only this single word, nothing more, nothing
less.



E.g.



"g00gle" = True, "g00gle " = False, "g google" = False, "GGOOGLE" =
False, "hey google" = False




What I'm looking in review: How can I make my code more pythonic? 



#!/usr/bin/env python3

# -*- coding: utf-8 -*-



import re



g = ["g", "G"]

o = ["o", "O", "0", "()", "", "<>"]

l = ["l", "L", "I"]

e = ["e", "E", "3"]



regex = (g, o, o, g, l, e)

regex = ((re.escape(y) for y in x) for x in regex)

regex = ("(?:{})".format("|".join(x)) for x in regex)

regex = "^{}$".format("".join(regex))



print(bool(re.match(regex, input())))





python python-3.x programming-challenge regex







share|improve this question











$endgroup$

















    4












    $begingroup$


    From the HackerRank question definition:




    The word google can be spelled in many different ways.



    E.g. google, g00gle, g0oGle, g<>0gl3, googl3, GooGIe etc...



    Because



    g = G



    o = O = 0 = () = = <>



    l = L = I



    e = E = 3



    That's the problem here to solve.



    And the match has to be only this single word, nothing more, nothing
    less.



    E.g.



    "g00gle" = True, "g00gle " = False, "g google" = False, "GGOOGLE" =
    False, "hey google" = False




    What I'm looking in review: How can I make my code more pythonic? 



    #!/usr/bin/env python3
    
# -*- coding: utf-8 -*-
    

    
import re
    

    
g = ["g", "G"]
    
o = ["o", "O", "0", "()", "", "<>"]
    
l = ["l", "L", "I"]
    
e = ["e", "E", "3"]
    

    
regex = (g, o, o, g, l, e)
    
regex = ((re.escape(y) for y in x) for x in regex)
    
regex = ("(?:{})".format("|".join(x)) for x in regex)
    
regex = "^{}$".format("".join(regex))
    

    
print(bool(re.match(regex, input())))





    python python-3.x programming-challenge regex







    share|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      From the HackerRank question definition:




      The word google can be spelled in many different ways.



      E.g. google, g00gle, g0oGle, g<>0gl3, googl3, GooGIe etc...



      Because



      g = G



      o = O = 0 = () = = <>



      l = L = I



      e = E = 3



      That's the problem here to solve.



      And the match has to be only this single word, nothing more, nothing
      less.



      E.g.



      "g00gle" = True, "g00gle " = False, "g google" = False, "GGOOGLE" =
      False, "hey google" = False




      What I'm looking in review: How can I make my code more pythonic? 



      #!/usr/bin/env python3
      
# -*- coding: utf-8 -*-
      

      
import re
      

      
g = ["g", "G"]
      
o = ["o", "O", "0", "()", "", "<>"]
      
l = ["l", "L", "I"]
      
e = ["e", "E", "3"]
      

      
regex = (g, o, o, g, l, e)
      
regex = ((re.escape(y) for y in x) for x in regex)
      
regex = ("(?:{})".format("|".join(x)) for x in regex)
      
regex = "^{}$".format("".join(regex))
      

      
print(bool(re.match(regex, input())))





      python python-3.x programming-challenge regex







      share|improve this question











      $endgroup$




      From the HackerRank question definition:




      The word google can be spelled in many different ways.



      E.g. google, g00gle, g0oGle, g<>0gl3, googl3, GooGIe etc...



      Because



      g = G



      o = O = 0 = () = = <>



      l = L = I



      e = E = 3



      That's the problem here to solve.



      And the match has to be only this single word, nothing more, nothing
      less.



      E.g.



      "g00gle" = True, "g00gle " = False, "g google" = False, "GGOOGLE" =
      False, "hey google" = False




      What I'm looking in review: How can I make my code more pythonic? 



      #!/usr/bin/env python3
      
# -*- coding: utf-8 -*-
      

      
import re
      

      
g = ["g", "G"]
      
o = ["o", "O", "0", "()", "", "<>"]
      
l = ["l", "L", "I"]
      
e = ["e", "E", "3"]
      

      
regex = (g, o, o, g, l, e)
      
regex = ((re.escape(y) for y in x) for x in regex)
      
regex = ("(?:{})".format("|".join(x)) for x in regex)
      
regex = "^{}$".format("".join(regex))
      

      
print(bool(re.match(regex, input())))





      python python-3.x programming-challenge regex




      python python-3.x programming-challenge regex






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 8 hours ago









      200_success

      129k15153416




      129k15153416










      asked 8 hours ago









      422_unprocessable_entity422_unprocessable_entity

      1,88231750




      1,88231750






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          Honestly there's not much code to comment on, it's also a programming challenge, so there's not much incentive to make it any better than it is. I too would have done it your way.



          The only objective criticism I have is I'd just add a function regex_options to build the non-capturing group.



          Other than that I'd apply this to the creation of g, o, l and e. As I think it's a little cleaner. But you may want to perform it before "".join, and after regex = (g, o, o, g, l, e).



          import re
          

          

          
def regex_options(options):
          
    options = (re.escape(o) for o in options)
          
    return "(?:{})".format("|".join(options))
          

          

          
g = regex_options(["g", "G"])
          
o = regex_options(["o", "O", "0", "()", "", "<>"])
          
l = regex_options(["l", "L", "I"])
          
e = regex_options(["e", "E", "3"])
          

          
regex = "^{}$".format("".join((g, o, o, g, l, e)))
          

          
print(bool(re.match(regex, input())))


          If this were professional code, I'd suggest using an if __name__ == '__main__': block, and possibly a main function.






          share|improve this answer









          $endgroup$









          • 2




            $begingroup$
            I like this solution better, because it doesn't repeatedly redefine regex and change its type.
            $endgroup$
            – 200_success
            7 hours ago











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Honestly there's not much code to comment on, it's also a programming challenge, so there's not much incentive to make it any better than it is. I too would have done it your way.



          The only objective criticism I have is I'd just add a function regex_options to build the non-capturing group.



          Other than that I'd apply this to the creation of g, o, l and e. As I think it's a little cleaner. But you may want to perform it before "".join, and after regex = (g, o, o, g, l, e).



          import re
          

          

          
def regex_options(options):
          
    options = (re.escape(o) for o in options)
          
    return "(?:{})".format("|".join(options))
          

          

          
g = regex_options(["g", "G"])
          
o = regex_options(["o", "O", "0", "()", "", "<>"])
          
l = regex_options(["l", "L", "I"])
          
e = regex_options(["e", "E", "3"])
          

          
regex = "^{}$".format("".join((g, o, o, g, l, e)))
          

          
print(bool(re.match(regex, input())))


          If this were professional code, I'd suggest using an if __name__ == '__main__': block, and possibly a main function.






          share|improve this answer









          $endgroup$









          • 2




            $begingroup$
            I like this solution better, because it doesn't repeatedly redefine regex and change its type.
            $endgroup$
            – 200_success
            7 hours ago
















          4












          $begingroup$

          Honestly there's not much code to comment on, it's also a programming challenge, so there's not much incentive to make it any better than it is. I too would have done it your way.



          The only objective criticism I have is I'd just add a function regex_options to build the non-capturing group.



          Other than that I'd apply this to the creation of g, o, l and e. As I think it's a little cleaner. But you may want to perform it before "".join, and after regex = (g, o, o, g, l, e).



          import re
          

          

          
def regex_options(options):
          
    options = (re.escape(o) for o in options)
          
    return "(?:{})".format("|".join(options))
          

          

          
g = regex_options(["g", "G"])
          
o = regex_options(["o", "O", "0", "()", "", "<>"])
          
l = regex_options(["l", "L", "I"])
          
e = regex_options(["e", "E", "3"])
          

          
regex = "^{}$".format("".join((g, o, o, g, l, e)))
          

          
print(bool(re.match(regex, input())))


          If this were professional code, I'd suggest using an if __name__ == '__main__': block, and possibly a main function.






          share|improve this answer









          $endgroup$









          • 2




            $begingroup$
            I like this solution better, because it doesn't repeatedly redefine regex and change its type.
            $endgroup$
            – 200_success
            7 hours ago














          4












          4








          4





          $begingroup$

          Honestly there's not much code to comment on, it's also a programming challenge, so there's not much incentive to make it any better than it is. I too would have done it your way.



          The only objective criticism I have is I'd just add a function regex_options to build the non-capturing group.



          Other than that I'd apply this to the creation of g, o, l and e. As I think it's a little cleaner. But you may want to perform it before "".join, and after regex = (g, o, o, g, l, e).



          import re
          

          

          
def regex_options(options):
          
    options = (re.escape(o) for o in options)
          
    return "(?:{})".format("|".join(options))
          

          

          
g = regex_options(["g", "G"])
          
o = regex_options(["o", "O", "0", "()", "", "<>"])
          
l = regex_options(["l", "L", "I"])
          
e = regex_options(["e", "E", "3"])
          

          
regex = "^{}$".format("".join((g, o, o, g, l, e)))
          

          
print(bool(re.match(regex, input())))


          If this were professional code, I'd suggest using an if __name__ == '__main__': block, and possibly a main function.






          share|improve this answer









          $endgroup$



          Honestly there's not much code to comment on, it's also a programming challenge, so there's not much incentive to make it any better than it is. I too would have done it your way.



          The only objective criticism I have is I'd just add a function regex_options to build the non-capturing group.



          Other than that I'd apply this to the creation of g, o, l and e. As I think it's a little cleaner. But you may want to perform it before "".join, and after regex = (g, o, o, g, l, e).



          import re
          

          

          
def regex_options(options):
          
    options = (re.escape(o) for o in options)
          
    return "(?:{})".format("|".join(options))
          

          

          
g = regex_options(["g", "G"])
          
o = regex_options(["o", "O", "0", "()", "", "<>"])
          
l = regex_options(["l", "L", "I"])
          
e = regex_options(["e", "E", "3"])
          

          
regex = "^{}$".format("".join((g, o, o, g, l, e)))
          

          
print(bool(re.match(regex, input())))


          If this were professional code, I'd suggest using an if __name__ == '__main__': block, and possibly a main function.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 7 hours ago









          PeilonrayzPeilonrayz

          25.5k337107




          25.5k337107








          • 2




            $begingroup$
            I like this solution better, because it doesn't repeatedly redefine regex and change its type.
            $endgroup$
            – 200_success
            7 hours ago














          • 2




            $begingroup$
            I like this solution better, because it doesn't repeatedly redefine regex and change its type.
            $endgroup$
            – 200_success
            7 hours ago








          2




          2




          $begingroup$
          I like this solution better, because it doesn't repeatedly redefine regex and change its type.
          $endgroup$
          – 200_success
          7 hours ago




          $begingroup$
          I like this solution better, because it doesn't repeatedly redefine regex and change its type.
          $endgroup$
          – 200_success
          7 hours ago


















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