使用pythonfinditer，如何替换每个匹配的字符串？

Question

我使用Python（实际上是pl/Python）在一个非常大的文本对象中连续地查找一系列regex匹配项。这很好用！每次匹配都是不同的结果，每次替换都将是不同的结果，最终基于循环内的查询。在

目前，我很乐意用任何文本替换rx中的每一个匹配，这样我就能理解它是如何工作的。有人能给我一个明确的例子来替换匹配的文本吗？在

匹配组（1） 似乎正确地指示了匹配的文本；这是这样做的吗？在

plan3 = plpy.prepare("SELECT field1,field2 FROM sometable WHERE indexfield = $1", 
  [ "text" ])

rx = re.finditer('LEFT[A-Z,a-z,:]+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

你想要^{}。在

import re

def repl(var):
  return var.group().encode('rot13')

print re.sub('[aeiou]', repl, 'yesterday')

yrstrrdny