Can't get an object's class name in Python

Question

I'm using isinstance to check argument types, but I can't find the class name of a regex pattern object:

>>> import re
>>> x = re.compile('test')
>>> x.__class__.__name__
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: __class__

...

>>> type(x)
<type '_sre.SRE_Pattern'>
>>> isinstance(x, _sre.SRE_Pattern)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name '_sre' is not defined
>>>
>>>
>>> isinstance(x, '_sre.SRE_Pattern')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: isinstance() arg 2 must be a class, type, or tuple of classes and types
>>>

Any ideas?

did you do import _sre? What surprises you about the NameError? — SilentGhost
– SilentGhost, Commented Jan 18, 2011 at 23:13
@SilentGhost: _sre is an internal module, and does not reveal SRE_Pattern to the outside. — poke
– poke, Commented Jan 18, 2011 at 23:25
@SilentGhost: It didn't -- and still doesn't -- make much sense that an object wouldn't know itself or that it'd be in an outside module. — zumjinger
– zumjinger, Commented Jan 18, 2011 at 23:46
@zum: x.__class__.__name__ works for me just fine in py3k. I'm not sure I'm following your object know itself charge. — SilentGhost
– SilentGhost, Commented Jan 19, 2011 at 0:18

mthurlin · Accepted Answer · 2011-01-18 23:16:26Z

4

You could do this:

import re

pattern_type = type(re.compile("foo"))

if isinstance(your_object, pattern_type):
   print "It's a pattern object!"

The idiomatic way would be to try to use it as a pattern object, and then handle the resulting exception if it is not.

answered Jan 18, 2011 at 23:16

mthurlin

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1 Comment

poke Over a year ago

+1, that should be the safest way to do it.. Everything else will be dangerous given that the actual regexp implementation is inside the C core..

eolo999 · Accepted Answer · 2011-01-18 23:27:24Z

0

In : x = re.compile('test')
In : isinstance(x, type(x))
Out: True

In [14]: type(type(x))
Out[14]: <type 'type'>

I think it relates to the type/object subtilities and to the re module implemetation. You can read a nice article here.

answered Jan 18, 2011 at 23:27

eolo999

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