CODEWITHSUNDEEP
pythonregexpython-3.4
Yotam
Yotam

Reputation: 10685

python3 regex returns empty tuple

I'm trying to apply regex to gather data from a file, and I only get empty results. For example, this little test (python3.4.3)

import re

a = 'abcde'
r = re.search('a',a)
print(r.groups())
exit()

Results with empty tuple (()). Clearly, I'm doing something wrong here, but what?

Comment: What I'm actually trying to do is to interpret expressions such as 0.7*sqrt(2), by finding the value inside the parenthesis.

Upvotes: 1

Views: 321

Answers (2)

Balthazar Rouberol
Balthazar Rouberol

Reputation: 7180

r.groups() returns an empty tuple, because your regular expression did not contain any group.

>>> import re
>>> a = 'abcde'
>>> re.search('a', a)
<_sre.SRE_Match object; span=(0, 1), match='a'>
>>> re.search('a', a).groups()
()
>>> re.search('(a)', a).groups()
('a',)

Have a look at the re module documentation:

(...) Matches whatever regular expression is inside the parentheses, and indicates the start and end of a group;

Edit: If you want to catch the bit between the brackets in the expression O.7*sqrt(2), you could use the following pattern:

>>> re.search('[\d\.]+\*sqrt\((\d)\)', '0.7*sqrt(2)').group(1)
'2'

Upvotes: 2

user6022341
user6022341

Reputation:

It happens because there are no groups in your regex. If you replace it with:

>>> r = re.search('(a)',a)

you'll get the groups:

>>> print(r.groups())
('a',)

Using group should work with the first option:

>>> print(re.search('a',a).group())
a

Upvotes: 4

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