使用re.findall（）替换所有匹配项

Question

Using re.findall() I've managed to get return multiple matches of a regex in a string. However my object returned is a list of matches within the string. This is not what I want.

What I want is to replace all matches with something else. I've tried to use similar syntax as you would use in re.sub to do this as so:

import json
import re

regex = re.compile('([a-zA-Z]\"[a-zA-Z])', re.S)

filepath = "C:\\Python27\\Customer Stuff\\Austin Tweets.txt"

f = open(filepath, 'r')
myfile = re.findall(regex, '([a-zA-Z]\%[a-zA-Z])', f.read())
print myfile

However, this creates the following error:

Traceback (most recent call last):
  File "C:/Python27/Customer Stuff/Austin's Script.py", line 9, in <module>
    myfile = re.findall(regex, '([a-zA-Z]\%[a-zA-Z])', f.read())
  File "C:\Python27\lib\re.py", line 177, in findall
    return _compile(pattern, flags).findall(string)
  File "C:\Python27\lib\re.py", line 229, in _compile
    bypass_cache = flags & DEBUG
TypeError: unsupported operand type(s) for &: 'str' and 'int'

Can anyone assist me within the last bit of syntax I need to replace all matches with something else within the original Python object?

EDIT:

In line with comments and answers received, here is me trying to sub one regex with another:

import json
import re

regex = re.compile('([a-zA-Z]\"[a-zA-Z])', re.S)
regex2 = re.compile('([a-zA-Z]%[a-zA-Z])', re.S)

filepath = "C:\\Python27\\Customer Stuff\\Austin Tweets.txt"

f = open(filepath, 'r')
myfile = f.read()
myfile2 = re.sub(regex, regex2, myfile)
print myfile

This now produces the following error:

Traceback (most recent call last):
  File "C:/Python27/Customer Stuff/Austin's Script.py", line 11, in <module>
    myfile2 = re.sub(regex, regex2, myfile)
  File "C:\Python27\lib\re.py", line 151, in sub
    return _compile(pattern, flags).sub(repl, string, count)
  File "C:\Python27\lib\re.py", line 273, in _subx
    template = _compile_repl(template, pattern)
  File "C:\Python27\lib\re.py", line 258, in _compile_repl
    p = sre_parse.parse_template(repl, pattern)
  File "C:\Python27\lib\sre_parse.py", line 706, in parse_template
    s = Tokenizer(source)
  File "C:\Python27\lib\sre_parse.py", line 181, in __init__
    self.__next()
  File "C:\Python27\lib\sre_parse.py", line 183, in __next
    if self.index >= len(self.string):
TypeError: object of type '_sre.SRE_Pattern' has no len()

Answer 1

import re

regex = re.compile('([a-zA-Z]\"[a-zA-Z])', re.S)
myfile =  'foo"s bar'
myfile2 = regex.sub(lambda m: m.group().replace('"',"%",1), myfile)
print(myfile2)

Answer 2

If I understand your question correctly, you're trying to replace a quotation mark between two characters with an percent sign between those characters.

There are several ways to do this with re.sub ( re.findall doesn't do replacements at all, so your initial attemps were always doomed to fail).

An easy approach would be to change your pattern to group the letters separately, and then use a replacement string that includes backreferences:

pattern = re.compile('([a-zA-Z])\"([a-zA-Z])', re.S)
re.sub(pattern, r'\1%\2', text)

Another option would be to use a replacement function instead of a replacement string. The function will be called with a match object for each match found in the text, and its return value is the replacement:

pattern = re.compile('[a-zA-Z]\"[a-zA-Z]', re.S)
re.sub(pattern, lambda match: "{0}%{2}".format(*match.group()), text)

(There are probably lots of other ways of implementing the lambda function. I like string formatting.)

However, probably the best approach is to use a lookahead and a lookbehind in your pattern to make sure your quotation mark is between letters without actually matching those letters. This lets you use the trivial string '%' as the replacement:

pattern = re.compile('(?<=[a-zA-Z])\"(?=[a-zA-Z])', re.S)
re.sub(pattern, '%', text)

This does have very slightly different semantics than the other versions. A text like 'a"b"c' will have both quotation marks replaced, while the previous codes would only replace the first one. Hopefully this is an improvement!

Answer 3

As suggested in comment, use re.sub() :

myfile = re.sub(regex, replacement, f.read())

where, replacement is the string your matches will be substituted with.

Answer 4

I find it clearer to use a function to do this type of substitution rather than a lambda. It makes it easy to perform any number of transformations on the matched text prior to replacing the text:

import re

def replace_double_quote(match):
    text = match.group()
    return text.replace('"', '%')

regex = re.compile('([a-zA-Z]\"[a-zA-Z])')
myfile = 'foo"s bar and bar"s foo'
regex.sub(replace_double_quote, myfile)

This returns foo%s bar and bar%s foo . Note that it replaces all matches.