Using find in python, how can I replace every line?

Question

I use Python (pl / python, actually) to find successively a series of regular expression matches in a very large text object. It works great! Each match represents a different result, and each replacement will differ from the result, ultimately, based on a query inside the loop.

At the moment, I would be happy to replace every match in rx with any text, so that I understand how it works. Can someone give me a clear example of replacing matched text?

match.group (1) seems to correctly indicate consistent text; is this a way to do something?

plan3 = plpy.prepare("SELECT field1,field2 FROM sometable WHERE indexfield = $1", [ "text" ]) rx = re.finditer('LEFT[AZ,az,:]+RIGHT)', data) # above does find my n matches... # ------------------- THE LOOP ---------------------------------- for match in rx: # below does find the 6 match objects - good! # match.group does return the text plpy.notice("-- MATCH: ", match.group(1)) # must pull out a substring as the 'key' to an SQL find (a separate problem) # (not sure how to split based on the colon:) keyfield = (match.group(1).split, ':') plpy.notice("---------: ",kefield) try: rv = plpy.execute(plan3, [ keyfield ], 1 ) # --- REPLACE match.group(1) with results of query # at this point, would be happy to replace with ANY STRING to test... except: plpy.error(traceback.format_exc()) # ------------------- ( END LOOP ) ------------------------------ 

Answer 1

Instead, you want re.sub() .

 import re def repl(var): return var.group().encode('rot13') print re.sub('[aeiou]', repl, 'yesterday') 

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