Comparison with unicode string in if-condition (python) [duplicate]

Question

Possible Duplicate:
Python '==' vs 'is' comparing strings, 'is' fails sometimes, why?

I am going to skip the part where i tell you how i tested my code and jump straight to the problem.

Python seems to be having some problem matching split of a unicode string to another inline unicode string in an if statement.

>>>zone = u'domain.com.'
>>>zone[-1:]
u'.'

>>>u'.' is u'.' #works fine
True
>>> z[-1:] == u'.' #works fine
True
>>> zone[-1:] is u'.' # FAILS !
False

here is my actual code snippet

>>>if zone[-1:] is not u'.':
>>>    #this line will always run !

if i change 'is not' to != the code works fine !

Does anyone know why "is" caused the comparison to fail ?

See stackoverflow.com/questions/1150765/…, stackoverflow.com/questions/4165688/…, and stackoverflow.com/questions/2988017/… to name but three. — Martijn Pieters
– Martijn Pieters, Commented Jun 6, 2011 at 20:33
@MartijnPieters stackoverflow.com/questions/4165688/… this almost nails it i guess — guron
– guron, Commented Jun 6, 2011 at 21:09

Jasmijn · Accepted Answer · 2011-06-06 20:28:53Z

2

It's because strings are objects in Python --- when you slice a string, you create a new one.

It's slightly more complicated than that, but that's the gist of it.

Solution: use == and != instead of is and is not.

answered Jun 6, 2011 at 20:28

Jasmijn

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