Python regex \number adding \0x

Question

I am tring to go a simple regex replace on a string in python. This is my code:

>>> s = "num1 1 num2 5"
>>> re.sub("num1 (.*?) num2 (.*?)","1 \1 2 \2",s)

I would expect an output like this, with the \numbers being replaced with their corresponding groups.

'1 1 2 5'

However, this is the output I am getting:

'1 \x01 2 \x025'

And I'm kinda stumped as to why the \x0s are their, and not what I would like to be there. Many thanks for any help

Answer 1

You need to start using raw strings (prefix the string with r):

>>> import re
>>> s = "num1 1 num2 5"
>>> re.sub(r"num1 (.*?) num2 (.*?)", r"1 \1 2 \2", s)
'1 1 2 5'

Otherwise you would need to escape your backslashes both for python and for the regex, like this:

>>> re.sub("num1 (.*?) num2 (.*?)", "1 \\1 2 \\2", s)
'1 1 2 5'

(this gets really old really fast, check out the opening paragraphs of the python regex docs

Answer 2

\1 and \2 are getting interpreted as octal character code escapes, rather than just getting passed to the regex engine. Using raw strings r"\1" instead of "\1" prevents this interpretation.

>>> "\17"
'\x0f'
>>> r"\17"
'\\17'