How can I obtain pattern string from compiled regexp pattern in python?

Question:

I have some code like this one:

>>> import re
>>> p = re.compile('my pattern')
>>> print p
_sre.SRE_Pattern object at 0x02274380

Is it possible to get string "my pattern" from p variable?

Asked By: Mykola Kharechko

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Answers:

p.pattern

Read more about re module here:
http://docs.python.org/library/re.html

Answered By: Mikhail Churbanov

Yes:

print p.pattern

hint, use the dir function in python to obtain a list of members:

dir(p)

this lists:

['__class__', '__copy__', '__deepcopy__', '__delattr__', '__doc__', '__eq__',
'__format__', '__ge__', '__getattribute__', '__gt__', '__hash__', '__init__',
'__le__', '__lt__', '__ne__', '__new__', '__reduce__', '__reduce_ex__',
'__repr__', '__setattr__', '__sizeof__', '__str__', '__subclasshook__',
'findall', 'finditer', 'flags', 'groupindex', 'groups', 'match', 'pattern',
'scanner', 'search', 'split', 'sub', 'subn']
Answered By: Lasse V. Karlsen

From the “Regular Expression Objects” section of the re module documentation:

RegexObject.pattern

The pattern string from which the RE object was compiled.

For example:

>>> import re
>>> p = re.compile('my pattern')
>>> p
<_sre.SRE_Pattern object at 0x1001ba818>
>>> p.pattern
'my pattern'

With the re module in Python 3.0 and above, you can find this by doing a simple dir(p):

>>> print(dir(p))
['__class__', '__copy__', '__deepcopy__', '__delattr__', '__doc__',
'__eq__', '__format__', '__ge__', '__getattribute__', '__gt__',
'__hash__', '__init__', '__le__', '__lt__', '__ne__', '__new__',
'__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__',
'__str__', '__subclasshook__', 'findall', 'finditer', 'flags',
'groupindex', 'groups', 'match', 'pattern', 'scanner', 'search',
'split', 'sub', 'subn']

This however does not work on Python 2.6 (or 2.5) – the dir command isn’t perfect, so it’s always worth checking the docs!

>>> print dir(p)
['__copy__', '__deepcopy__', 'findall', 'finditer', 'match', 'scanner',
'search', 'split', 'sub', 'subn']
Answered By: dbr
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