使用python finditer,我该如何替换每个匹配的字符串？

Question

我正在使用Python(实际上是pl/python)在一个非常大的文本对象中连续查找一系列正则表达式匹配.这工作正常!每个匹配都是不同的结果,每个替换将是不同的结果,最终基于循环内的查询.

目前,我很乐意用任何文本替换rx中的每个匹配,这样我就能理解它是如何工作的.有人能给我一个替换匹配文本的明确例子吗？

match.group(1)似乎正确表示匹配的文本; 这是做事的方式吗？

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

你想要的re.sub().

import re

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

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

yrstrrdny