Regex: Remove all match plus one char before all

[Ofnuts]

Warning, aspirin required.

This is quite tricky because you can have anything before |BS|, including another |BS|. And covering your rear with something such as [\|]|BS| isn't general enough because it prevents backspacing over a |. and in a regexp you can't express something like "not this string"...

So, you have to attack at the other end: use a regexp that will match any character followed by whole sequence of consecutive |BS|. Due to the greedy way things are matched, this will always include the whole sequence of consecutive |BS|, so you initial character cannot be itself part of a |BS|.

Then look at the fine print in the specs of re.sub(), it looks for non-overlapping occurences of the pattern, so the search for the next match starts after the end of the current match... which is after the end of the sequence of |BS|, so in a sequence of |BS|you will only process one per call to sub().

So in practice, we look for a character followed by a |BS| followed by zero or more other |BS| (captured in a group) and replace that by just that captured group:

import re

pattern=re.compile(r'.\|BS\|((\|BS\|)*)')

def noBS(s):
    print '------------'
    previous=''
    while s!=previous:
        previous=s
        s=re.sub(pattern,r'\1',s)
        print s # this shows that the two sequences of |BS| are processed in parallel 
    return s

print noBS("it |BS||BS||BS|this is one|BS||BS||BS|an example")
print noBS("it |BS||BS||BS| |BS|this is one|BS||BS||BS|an example")
print noBS("it |BS||BS||BS| |BS|this is o n  e|BS||BS||BS||BS||BS||BS|an example")
# The first 'BS|' gets backspaced over due to missing leading '|'... 
print noBS("it BS||BS||BS||BS||BS||BS||BS|this is o n  e|BS||BS||BS||BS||BS||BS|an example")

Output for he last one:

Output:
it BS|BS||BS||BS||BS||BS|this is o n  |BS||BS||BS||BS||BS|an example
it B|BS||BS||BS||BS|this is o n |BS||BS||BS||BS|an example
it |BS||BS||BS|this is o n|BS||BS||BS|an example
it|BS||BS|this is o |BS||BS|an example
i|BS|this is o|BS|an example
this is an example

Unfortunately, I don't think you can avoid n explicit iteration.