Regex to replace substring (if not the start of a word)

Question

I'm handling tweets like @Alice @ home. I want to convert user mentions to normal words (e.g. @Alice => Alice) but keep the individual @ as a surrogate for at. So simply replacing all occurrences of @ won't work.

I found out about the concept of word boundaries, but I haven't been able to make them work for this. For one,

print re.sub(r'\b@\b', '', '@Alice @ home')

doesn't change anything, while

print re.sub(r'\bAlice\b', '', '@Alice @ home')

results in @ @ home. So I assume that the at-sign (@) does not count as part of a word.

In short, I'm basically looking for the pattern so that

print re.sub(pattern, '', '@Alice @ home')

outputs Alice @ home.

Thanks for any hints.

Avinash Raj · Accepted Answer · 2015-01-19 07:49:16Z

4

You may play with the boundaries \B and \b

>>> print re.sub(r'\B@\b', r'', '@Alice @ home')
Alice @ home
>>> print re.sub(r'\B@\b', r'', 'foo @Alice @ home')
foo Alice @ home

\B matches between two word characters or between two non-word characters. So \B@ in the above regex matches the @ before Alice and the another separate @ symbol. \b matches between a word character and a non-word character (vice versa). So the following \b makes the pattern to match only the first because the first @ itself followed by a word character ie, A.

edited Jan 19, 2015 at 7:49

answered Jan 19, 2015 at 7:43

Avinash Raj

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1 Comment

J0e3gan Over a year ago

Great answer. I upvoted it per the About-face in my answer. Thanks!

vks · Accepted Answer · 2015-01-19 09:14:05Z

3

(?:^|(?<=\s))@(?!\s)

Try this.This will replace @ only from the start of word.if @ is in the middle of word like as@sas it will save it.See demo.

https://regex101.com/r/tX2bH4/44

re.sub(r'(?:^|(?<=\s))@(?!\s)','',s)

answered Jan 19, 2015 at 9:14

vks

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Comments

J0e3gan · Accepted Answer · 2015-01-19 09:38:52Z

Initial Answer

Try the following regex:

@(?! )

Here are a couple examples of how it performs:

>>> print re.sub(r'@(?! )', '', '@Alice @ home')
Alice @ home
>>> print re.sub(r'@(?! )', '', 'Whatever @Alice @ home')
Whatever Alice @ home

You can also test it with a related regex fiddle.

Key points:

@ – the at-sign
(?! ) – a negative lookahead that matches anything but a space (i.e. not followed by a space)

Personally I find the zero-width word-boundary assertions (\b and \B) a bit distracting and prefer to use zero-width lookarounds for this sort of thing, but TMTOWTDI.

About-face

I thought about this more (as usual), and what I found is admittedly a compelling case for the zero-width word-boundary assertions' simplicity and start- and end-of-string matching.

Consider a fuller set of conceivable tweets:

@Alice @ home
Whatever @Alice @ home
What're you lookin' @
What're you lookin' @?

It turns out that to get these right requires a much more complicated negative lookahead, turning my initial regex into:

@(?![ \W]|$)

As before, here are examples of how it performs:

>>> print re.sub(r'@(?![ \W]|$)', '', '@Alice @ home')
Alice @ home
>>> print re.sub(r'@(?![ \W]|$)', '', 'Whatever @Alice @ home')
Whatever Alice @ home
>>> print re.sub(r'@(?![ \W]|$)', '', "What're you lookin' @")
What're you lookin' @
>>> print re.sub(r'@(?![ \W]|$)', '', "What're you lookin' @?")
What're you lookin' @?

And as before, you can also test it with a related regex fiddle.

But a word-boundary pattern like Avinash Raj employed gets this fuller set of conceivable tweets right...with much less fanfare:

>>> print re.sub(r'\B@\b', '', '@Alice @ home')
Alice @ home
>>> print re.sub(r'\B@\b', '', 'Whatever @Alice @ home')
Whatever Alice @ home
>>> print re.sub(r'\B@\b', '', "What're you lookin' @")
What're you lookin' @
>>> print re.sub(r'\B@\b', '', "What're you lookin' @?")
What're you lookin' @?

Test it out with another related regex fiddle if you like too.

Bottom line, this has been a cool learning experience for me to question what I tend to prefer using, and I hope you find it the same: onward on our word-boundary-assertion adventures! :)

Vasily Ryabov · Accepted Answer · 2015-01-19 07:52:09Z

1

The simplest way working for me:

>>> s = '@Alice @ home'
>>> re.sub('\s@\s', ' at ', s).replace('@', '')
'Alice at home'

answered Jan 19, 2015 at 7:52

Vasily Ryabov

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