Im trying to just capture the following string: u00
because i need to replace it to \u00
.
Sometimes this characters appear with a \
before, in that case, i don't want to capture it.
At other times, the simbol is "
, i want to capture it, but just the u00
, not "u00
Im trying this:
file_modified = re.sub(r'[^\\|^\s](u00)', r'\\u00', original_file)
Im capturing the "
and i don't know how to skip it, i just want to capture u00
CodePudding user response:
Use a negative lookbehind assertion r'(?<!\\)(u00)'
. The will match u00
provided it is not preceded by \
.
CodePudding user response:
Just match it optionally:
file_modified = re.sub(r'\\?u00', r'\\u00', original_file)
Here,
\\?u00
- matches an optional\
andu00
\\u00
- is a replacement pattern that replaces with\u00
Thus, even if there was a \
before u00
, it won't disappear and won't get doubled, but if it was missing, it will be added.
See the Python demo:
import re
original_file = r"u00 because i need to replace it to \u00"
print(re.sub(r'\\?u00', r'\\u00', original_file))
# => \u00 because i need to replace it to \u00