re.sub在python中：详细模式不适用于替换模式吗？

Question

Is there any way to get around this limitation of re.sub? It is not fully functional for verbose mode (with back reference here) in the replace pattern; it does not eliminate whitespace or comments (yet it does interpret backreferences properly).

import remport re

ft1=r"""(?P<test>[0-9]+)"""
ft2=r"""\g<test>and then: \g<test> #this remains"""

print re.sub(ft1,ft2,"front 1234 back",flags=re.VERBOSE) #Does not work 
#result: front 1234and then: 1234 #this remains back

re.VERBOSE does not apply to the replacement pattern... Is there a work-around? (Simpler than working with groups after an re.match.)

Answer 1

Here is the only way I have found to "compile" an re replace expression for sub. There are a few extra constraints: both spaces and newlines have to be written like spaces are written for the re match expression (in square brackets: [ ] and [\\n\\n\\n]) and the whole replace expression should have a verbose newline at the beginning.

An example: this searches a string and detects a word repeated after /ins/ and /del/, then replaces those occurrences with a single occurrence of the word in front of .

Both the match and the replace expressions are complex, which is why I want a verbose version of the replace expression.

===========================

import re

test = "<p>Le petit <ins>homme à</ins> <del>homme en</del> ressorts</p>"


find=r"""
    <ins>
    (?P<front>[^<]+)          #there is something added that matches 
    (?P<delim1>[ .!,;:]+)     #get delimiter
    (?P<back1>[^<]*?)
    </ins>
    [ ]
    <del>
    (?P=front)
    (?P<delim2>[ .!,;:]+)
    (?P<back2>[^<]*?)
    </del>
"""
replace = r"""
    <<<<<\g<front>>>>>         #Pop out in front matching thing
    <ins>
    \g<delim1>
    \g<back1>
    </ins>
    [ ]     
    <del>    
    \g<delim2>             #put delimiters and backend back
    \g<back2>
    </del>
"""

flatReplace = r"""<<<<<\g<front>>>>><ins>\g<delim1>\g<back1></ins> <del>\g<delim2>\g<back2></del>"""


def compileRepl(inString):

    outString=inString
    #get space at front of line
    outString=re.sub(r"\n\s+","\n",outString)
    #get space at end of line
    outString=re.sub(r"\s+\n","",outString) 
    #get rid of comments
    outString=re.sub(r"\s*#[^\n]*\n","\n",outString)
    #preserve space in brackets, and eliminate brackets
    outString=re.sub(r"(?<!\[)\[(\s+)\](?!\[)",r"\1",outString)
    # get rid of newlines not in brackets
    outString=re.sub(r"(?<!\[)(\n)+(?!\])","",outString)
    #get rid of brackets around newlines
    outString=re.sub(r"\[((\\n)+)\]",r"\1",outString)
    #trim brackets    
    outString=re.sub(r"\[\[(.*?)\]\]","[\\1]",outString)
    return outString


assert(flatReplace == compileRepl(replace))


print test
print compileRepl(replace)
print re.sub(find,compileRepl(replace),test, flags=re.VERBOSE)

#<p>Le petit <ins>homme à</ins> <del>homme en</del> ressorts</p>
#<<<<<\g<front>>>>><ins>\g<delim1>\g<back1></ins> <del>\g<delim2>\g<back2></del>
#<p>Le petit <<<<<homme>>>><ins> à</ins> <del> en</del> ressorts</p>

Answer 2

You can first use re.compile to compile regular expressions. Here, you can make use of re.VERBOSE flag. Later, you can pass these compiled expressions as argument to re.sub()