Python doubling some backslashes while leaving others

Asked Oct 07 '20 at 13:28

Active Oct 07 '20 at 13:36

Viewed 21 times

I am defining the following string in python:

s = '(?<=\n\d{8})(.*?)(?=\n\d{8})'

When I print it, python returns:

'(?<=\n\\d{8})(.*?)(?=\n\\d{8})'

Why it is doubling the backslash before the d?

I have also tried to make the string raw:

s = r'(?<=\n\d{8})(.*?)(?=\n\d{8})'

and in this case, python print output is:

'(?<=\\n\\d{8})(.*?)(?=\\n\\d{8})'

What is the reason behind both of these behaviours?

edited Oct 07 '20 at 13:36

CEO1111

asked Oct 07 '20 at 13:28

1

If you print it with `print(s)` that's not the output you'll get – khelwood Oct 07 '20 at 13:29
Yeah but why just the second backslash? and not both n and d – Tommaso Bendinelli Oct 07 '20 at 13:33
3

Because if you put unescaped `\n` in a string literal, you don't get a ```\``` followed by `n`; you get the newline character `\n`. Whereas in the version where you use a raw-string literal, you actually get ```\``` followed by `n`, so the backslash is doubled in the repr. – khelwood Oct 07 '20 at 13:35
you can write it as response and I will accept – Tommaso Bendinelli Oct 07 '20 at 13:44

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