python re.sub regex

Question

Python keeps returning a string with a broken character.

python

test = re.sub('handle(.*?)', '<verse osisID="lol">\1</verse>', 'handle a bunch of random text here.')
print test

what I want

<verse osisID="lol">a bunch of random text here.</verse>

what i am getting

<verse osisID="lol">*broken character*</verse>a bunch of random text here.

Martijn Pieters · Accepted Answer · 2012-08-09 19:45:40Z

8

You should either escape the \ character or use a r'' raw string:

>>> re.sub('handle(.*?)', r'<verse osisID="lol">\1</verse>', 'handle a bunch of random text here.')
'<verse osisID="lol"></verse> a bunch of random text here.'

Without the r'' raw string literal, backslashes are interpreted as escape codes. You can double the backslash as well:

>>> '\1'
'\x01'
>>> '\\1'
'\\1'
>>> r'\1'
'\\1'
>>> print r'\1'
\1

Note that you replace just the word handle there, the .*? pattern matches 0 characters at minimum. Remove the question mark and it'll match your expected output:

>>> re.sub('handle(.*)', r'<verse osisID="lol">\1</verse>', 'handle a bunch of random text here.')
'<verse osisID="lol"> a bunch of random text here.</verse>'

answered Aug 9, 2012 at 19:45

Martijn Pieters

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3 Comments

user1442957 Over a year ago

you are a beautiful person :)

Andrew Cox Over a year ago

you might want to match the space after the handle but before the next word as well, as this will prevent the ...> a br... You could do this with handle *(.*) presuming you only have spaces (not other whitespace)

Martijn Pieters Over a year ago

@AndrewCox: I'd use \s* to match whitespace there instead; why limit only to spaces?

xiyurui · Accepted Answer · 2017-06-02 02:31:08Z

0

below code tested under python 3.6

import re 

test = 'a bunch of random text here.'
resp = re.sub(r'(.*)',r'<verse osisID="lol">\1</verse>',test)
print (resp)

<verse osisID="lol">a bunch of random text here.</verse>

answered Jun 2, 2017 at 2:31

xiyurui

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