Capturing emoticons using regular expression in python

Question

I would like to have a regex pattern to match smileys ":)" ,":(" .Also it should capture repeated smileys like ":) :)" , ":) :(" but filter out invalid syntax like ":( (" .

I have this with me, but it matches ":( ("

bool( re.match("(:\()",str) )

I maybe missing something obvious here, and I'd like some help for this seemingly simple task.

Then you should specify the entire string using `^` and `$`.. as it stands it successfully matches any similey *somewhere* in the string — zebediah49, Jan 28 '13 at 21:05

score 9 · Accepted Answer · answered Feb 01 '13 at 18:15

I think it finally "clicked" exactly what you're asking about here. Take a look at the below:

import re

smiley_pattern = '^(:\(|:\))+$' # matches only the smileys ":)" and ":("

def test_match(s):
    print 'Value: %s; Result: %s' % (
        s,
        'Matches!' if re.match(smiley_pattern, s) else 'Doesn\'t match.'
    )

should_match = [
    ':)',   # Single smile
    ':(',   # Single frown
    ':):)', # Two smiles
    ':(:(', # Two frowns
    ':):(', # Mix of a smile and a frown
]
should_not_match = [
    '',         # Empty string
    ':(foo',    # Extraneous characters appended
    'foo:(',    # Extraneous characters prepended
    ':( :(',    # Space between frowns
    ':( (',     # Extraneous characters and space appended
    ':(('       # Extraneous duplicate of final character appended
]

print('The following should all match:')
for x in should_match: test_match(x);

print('')   # Newline for output clarity

print('The following should all not match:')
for x in should_not_match: test_match(x);

The problem with your original code is that your regex is wrong: (:\(). Let's break it down.

The outside parentheses are a "grouping". They're what you'd reference if you were going to do a string replacement, and are used to apply regex operators on groups of characters at once. So, you're really saying:

( begin a group
- :\( ... do regex stuff ...
')' end the group

The : isn't a regex reserved character, so it's just a colon. The \ is, and it means "the following character is literal, not a regex operator". This is called an "escape sequence". Fully parsed into English, your regex says

( begin a group
- : a colon character
- \( a left parenthesis character
) end the group

The regex I used is slightly more complex, but not bad. Let's break it down: ^(:$|:$)+$.

^ and $ mean "the beginning of the line" and "the end of the line" respectively. Now we have ...

^ beginning of line
- (:$|:$)+ ... do regex stuff ...
$ end of line

... so it only matches things that comprise the entire line, not simply occur in the middle of the string.

We know that ( and ) denote a grouping. + means "one of more of these". Now we have:

^ beginning of line
( start a group
- :$|:$ ... do regex stuff ...
) end the group
+ match one or more of this
$ end of line

Finally, there's the | (pipe) operator. It means "or". So, applying what we know from above about escaping characters, we're ready to complete the translation:

^ beginning of line
( start a group
- : a colon character
- \( a left parenthesis character
| or
- : a colon character
- \) a right parenthesis character
) end the group
+ match one or more of this
$ end of line

I hope this helps. If not, let me know and I'll be happy to edit my answer with a reply.

Thank you. Even though I had managed to solve it on my own, your incredibly well explained answer helped to clear a lot of regex concepts for me. — coding_pleasures, Feb 03 '13 at 16:41
Good to hear you got it taken care of, and I'm glad my explanation helped :) — Lyndsy Simon, Feb 04 '13 at 18:10

woemler · Answer 2 · 2013-01-28T21:24:07.913

3

Maybe something like:

re.match('[:;][)(](?![)(])', str)

edited Jan 28 '13 at 21:24

answered Jan 28 '13 at 21:05

woemler

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urm..could you explain what exactly does it do? – coding_pleasures Jan 28 '13 at 21:11
This regex will match either a `;` or a `:`, followed by either a `)` or `(`, but only when it is then NOT followed by another `)` or `(`. This is probably not the perfect solution, but is at least another way to look at the problem. – woemler Jan 28 '13 at 21:15

root · Answer 3 · 2013-01-28T21:35:23.810

2

Try (?::|;|=)(?:-)?(?:\)|\(|D|P). Haven't tested it extensively, but does seem to match the right ones and not more...

In [15]: import re

In [16]: s = "Just: to :)) =) test :(:-(( ():: :):) :(:( :P ;)!"

In [17]: re.findall(r'(?::|;|=)(?:-)?(?:\)|\(|D|P)',s)
Out[17]: [':)', '=)', ':(', ':-(', ':)', ':)', ':(', ':(', ':P', ';)']

edited Jan 28 '13 at 21:35

answered Jan 28 '13 at 21:23

root

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thank you..whats the meaning of (? ..)? I read the documentation, but couldn't understand. I'd be happy if you could explain the pattern a lil bit.. – coding_pleasures Jan 28 '13 at 21:37

score 0 · Answer 4 · answered Jan 28 '13 at 22:01

0

I got the answer I was looking for from the comments and answers posted here.

re.match("^(:[)(])*$",str)

Thanks to all.

answered Jan 28 '13 at 22:01

coding_pleasures

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This regex will only work if `str` starts with `:` and only contains repetitions of `:)` and `:(` all the way to the end. Are you sure this is what you are looking for? – woemler Jan 28 '13 at 22:08
Yes, that's exactly what I was looking for. I'm sorry for not being able to post the question correctly enough. And thanks for answering. – coding_pleasures Jan 28 '13 at 22:20
Hmm... Are you *sure* you're sure? :) The following string would match this regex: `:):):):):(:(:):(:)`. That doesn't seem useful from this side of the monitor. – Lyndsy Simon Jan 28 '13 at 23:01
1

Also, an empty string matches. `*` Means "zero or more occurrences". `+` Means "one or more occurrences," and `?` means "Zero or one occurrences." – Lyndsy Simon Jan 28 '13 at 23:03
Yes, I agree it doesn't make sense generally. But I was using it to solve a programming contest problem. – coding_pleasures Feb 03 '13 at 16:55

Capturing emoticons using regular expression in python

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