Python: Faster regex replace -

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I have a large set of large files and a set of "phrases" that need to be replaced in each file .
"Commercial logic ..." bans many restrictions: \ n \ n

  • Matching should be insensitive
  • White pages, tabs and new rows can not be ignored in regedx

    My solution (see below) is a little bit slow and how can it be optimized in terms of string and replacement? 

      data = open ("INPUT__FILE"). Phrase in phrases for read () o = open ("OUTPUT_FILE", "w"): # These are the set of words I'm talking about B1, B2 = string. Strip ( ). Partition ("") regex = re.compile (r "% s \ * \ t * \ n *% S"% (b1, b2), re.IGNORECASE) data = regex.sub (b1 + "_" + b2 , Data) o. Write (Data)   
      Update : By changing all the text to less case and  re.IGNORECASE    
     
     
     You can reconstruct your regexp for each file: For the phrase in phrases < B>, b2 = str (phrase):. Partition ("") for precompiled.append (b1 + "_" + b2, re.compile (r "% s \ * \ t * \ n *% s"% (b1, b2, redirector)) In (Input, Output): Open (Output, "W") as O: Input for I: Data = i.read () (Pattern, Regex) in precompiled: data = Regex.sub (pattern, data) o.write (data)   
     This is the same f or a file, but if you are repeating several files then you can reuse the regexes are doing.  
     Disclaimer: Can be in unused, typo.  
     [ update ], you can simplify regexp by changing different codes from  \ s * . I suspect that you have a bug in which you want to match  "\ t"  and not currently.

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