将Unicode字符串转换为ascii字符串，然后将其与ascii字符串进行比较始终返回false。 （Python 2.7.6）

Question

This seems like a simple question, but I can't find an answer anywhere.

I have a function that gets a Unicode string as an argument, and looks like this:

def foo(arg):
    if str(arg) is 'wxyz':
        print 'it is equal'

Given the input u'wxyz', the function doesn't print anything. I did some more testing and I have come up with a question.

Why does this not work:

>>> u = unicode('wxyz')
>>> str(u) is 'wxyz'
False

But this does work:

>>> str(u) == 'wxyz'
True

Here's what I have tried already:

>>> u = unicode('wxyz')
>>> s = str(u)
>>> a = u.encode('ascii')
>>> type(u)
<type 'unicode'>
>>> type(s)
<type 'str'>
>>> type(a)
<type 'str'>
>>> type('wxyz')
<type 'str'>
>>> u is 'wxyz'
False             # Should be False
>>> u == 'wxyz'
False             # Should be False
>>> s is 'wxyz'
False             # Should be True
>>> s == 'wxyz'
True              # Should be True
>>> a is 'wxyz'
False             # Should be True
>>> a == 'wxyz'
True              # Should be True
>>> u is u'wxyz'
False             # Should be True
>>> u == u'wxyz'
True              # Should be True

I guess that I could change the 'is' to a '==', but I've been using 'is' everywhere else in the code, and it doesn't seem very Python-esque to switch to using '=='. If someone could help me understand this, I would be very appreciative. Also, if you need me to be more specific, please ask.

I seriously apologize if this has been asked anywhere else. I read the Python documentation on Unicode and looked for similar questions here, but I couldn't find anything that answered my question.

Answer 1

The operator a is b returns True if a and b are bound to the same object . So is is the wrong operator to be using here. That probably means you need to fix most places you've used is in your code.

a = []
b = a
a is b # true
a == b # true

a = []
b = []
a is b # false
a == b # true