Regex expression after quotation on python

Question

I am trying to develop a program on Python that would get the name of the artists from a twit from Pandora. Like for example if I have this twitter:

I'm listening to "I Can Make It Better" by Luther Vandross on Pandora #pandora http://t.co/ieDbLC393F.

I would like to get only the name Luther Vandross back. I do not know much about regex, so I tried to do the following code:

print  re.findall('".+?" by [\w+]+',  text)

But the result was "I can Make it Better" by Luther

Do you have any idea on how I would be able to develop a regular expression on python to get it?

Federico Piazza · Accepted Answer · 2015-06-21 16:46:01Z

3

Your regex is near, but you can change the delimiters to use " by and on. However, you need to use capturing groups by using parentheses.

You can use a regex like this:

" by (.+?) on

Working demo

Regular expression visualization

The idea behind this regex is to capture the content between the " by and on, using a simple nongreedy regex.

Match information

MATCH 1
1.  [43-58] `Luther Vandross`

Code

import re
p = re.compile(ur'" by (.+?) on')
test_str = u"I'm listening to \"I Can Make It Better\" by Luther Vandross on Pandora #pandora http://t.co/ieDbLC393F.\n"

re.search(p, test_str)

edited Jun 21, 2015 at 16:46

answered Jun 21, 2015 at 16:39

Federico Piazza

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1 Comment

Filipe Over a year ago

Thank you for your help =), I was having some difficulties understanding on how regex works, but this answer made it way more clear.

poke · Accepted Answer · 2015-06-21 16:52:38Z

2

>>> s = '''I'm listening to "I Can Make It Better" by Luther Vandross on Pandora #pandora http://t.co/ieDbLC393F.'''

>>> import re
>>> m = re.search('to "?(.*?)"? by (.*?) on #?Pandora', s)
>>> m
<_sre.SRE_Match object; span=(14, 69), match='to "I Can Make It Better" by Luther Vandross on P>
>>> m.groups()
('I Can Make It Better', 'Luther Vandross')

More test cases:

>>> tests = [
    '''I'm listening to "Don't Turn Out The Lights (D.T.O.T.L.)" by NKOTBSB on #Pandora''',
    '''I'm listening to G.O.D. Remix by Canton Jones on #Pandora''',
    '''I'm listening to "It's Been Awhile" by @staindmusic on Pandora #pandora http://pdora.co/R1OdxE''',
    '''I'm listening to "Everlong" by @foofighters on #Pandora http://pdora.co/1eANfI0''',
    '''I'm listening to "El Preso (2000)" by Fruko Y Sus Tesos on #Pandora http://pdora.co/1GtOHC1'''
    '''I'm listening to "Cat Daddy" by Rej3ctz on #Pandora http://pdora.co/1eALNpc''',
    '''I'm listening to "Space Age Pimpin'" by 8 Ball & MJG on Pandora #pandora http://pdora.co/1h8swun'''
]
>>> expr = re.compile('to "?(.*?)"? by (.*?) on #?Pandora')
>>> for s in tests:
        print(expr.search(s).groups())

("Don't Turn Out The Lights (D.T.O.T.L.)", 'NKOTBSB')
('G.O.D. Remix', 'Canton Jones')
("It's Been Awhile", '@staindmusic')
('Everlong', '@foofighters')
('El Preso (2000)', 'Fruko Y Sus Tesos')
("Space Age Pimpin'", '8 Ball & MJG')

edited Jun 21, 2015 at 16:52

answered Jun 21, 2015 at 16:35

poke

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1 Comment

poke Over a year ago

I scanned the #Pandora hashtag on Twitter for a few more examples and adjusted the expression to make it work with all those patters.

Avinash Raj · Accepted Answer · 2015-06-21 16:37:22Z

2

You need to use capturing group.

print re.findall(r'"[^"]*" by ([A-Z][a-z]+(?: [A-Z][a-z]+){0,2})',  text)

I used the repeatation quantifier, since the name may contain only first name or first, lastname or first,middle,last name.

answered Jun 21, 2015 at 16:37

Avinash Raj

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Comments

vks · Accepted Answer · 2015-06-21 16:37:58Z

1

print  re.findall('".+?" by ((?:[A-Z][a-z]+ )+)',  text)

You can try this.See demo.

https://regex101.com/r/vH0iN5/5

answered Jun 21, 2015 at 16:37

vks

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Comments

anubhava · Accepted Answer · 2015-06-21 16:38:48Z

1

You can use this lookaround based regex:

str = 'I\'m listening to "I Can Make It Better" by Luther Vandross on Pandora #pandora http://t.co/ieDbLC393F.';
print re.search(r'(?<=by ).+?(?= on)', str).group()
Luther Vandross

answered Jun 21, 2015 at 16:38

anubhava

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