Replacing string regex python not working

Question

The program should get each of these punctuation characters replaced by that character surrounded by two spaces.

I tried this:

sent = re.sub('[!,\'!?.]',' \1 ', sent)

but it just printed some weird icon instead of those punctuation characters

This was done using python 3.

We need to see input, output and expected output to help you. — Olivier Melançon, Feb 28 '18 at 15:59
@ctwheels there is not much more to add, should be an answer! — Olivier Melançon, Feb 28 '18 at 16:03
The lesson here is to always use raw strings when working with regex — DeepSpace, Feb 28 '18 at 16:06

score 2 · Answer 1 · answered Feb 28 '18 at 16:03

2

The string "\1" was interpreted as ascii codepoint 1 \x01. To prevent this from happening, use the raw string r' \1'. Also, to use a back-reference, you should use parentheses. This is the result:

>>> sent = "!,\'!?."
>>> sent = re.sub(r'([!,\'!?.])',r' \1 ', sent)
>>> sent
" !  ,  '  !  ?  . "

answered Feb 28 '18 at 16:03

Zweedeend

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Thanks that did the trick. – shittyflute Feb 28 '18 at 16:10

score 0 · Answer 2 · answered Feb 28 '18 at 16:04

As per my original comment, your replacement is \1 but you never created a capture group. Surround your regex in () as the following suggests. Also, you need to escape \ or make it a raw string.

re.sub(r"([,'!?.])", r' \1 ', sent)

See code in use here

Replacing string regex python not working

2 Answers2