reverse a regexp substitution

Given a string transformation expressed as a regexp substitution, is there a convenient way to reverse that transformation, preferably in python?

For example, given the transformation

def f(x): return re.sub('foo((:?bar)?)', r'\1', x)

which transforms 'foobar' to 'bar' and 'foo' to the empty string, I would like to obtain

def g(x): return re.sub('((:?bar)?)', r'foo\1', x)

which does the reverse, in the sense that

f(g(x)) == x

Obviously not all regexp substitutions are 1:1 mappings, but my wishful thinking is that they can all be reversed in the sense of obtaining one possible input value g(x) that would yield a given output x from the original substitution.

Wh开发者_C百科y would I want to do this? I'd like to generate URLs for arbitrary filesystem paths based on parsing AliasMatch directives in an Apache config file.

Your example doesn't work; ('(bar)?', 'foo$1') is not the reverse of ('foo(bar)?', '$1').

If you try it (dropping the group substitution for the moment):

import re
re.sub(r'(bar)?', 'foo', 'xyz')

you get fooxfooyfoozfoo.

That's because (bar)? matches the null string, and the null string exists at every possible location in the string.

I'd suggest that with problems like this in such a simple example, you might want to seek a different approach to whatever you're actually trying to do.

(I dropped the group above because that doesn't actually work in Python. If you do re.sub(r'(bar)?', r'foo\1', 'xyz'), it'll fail with an exception if (bar)? was skipped. You can fix this with ((bar)?), so the outer grouping is never omitted. This is incidental to your problem, though.)

Well, practically what happens in this example is:

xy? -> $1
y?  -> x$1

I'm not entirely sure what you're after with this - more examples wouldn't hurt - but I think that general formula could be expanded to cover more at least simple cases.