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Date: Fri, 11 Jul 2014 23:11:37 -0700
From: Ethan Furman <ethan@stoneleaf.us>
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To: python-list@python.org
Subject: Re: Proposal: === and !=== operators
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Xref: csiph.com comp.lang.python:74373

On 07/11/2014 10:07 PM, Alan Bawden wrote:
> Steven D'Aprano <steve+comp.lang.python@pearwood.info> writes:
>
>> But perhaps we only care about changes in value, not type. NAN or no NAN,
>> list equality works fine:
>>
>> py> data = [1.0, 2.0, float('nan'), 4.0]
>> py> old = data[:]
>> py> old == data  # No changes made yet, should return True
>> True
>
> You lost me right here.  If list equality is determined by comparing
> lists element-by-element, and the second element of old is _not_ equal
> to the second element of data, then why should old and data be equal?
>
> In fact, I find myself puzzled about exactly how list equality is
> actually defined.  Consider:
>
>    >>> a = float('nan')
>    >>> x = [1, a, 9]
>    >>> y = [1, a, 9.0]
>    >>> x == y
>    True
>
> So is there some equality predicate where corresponding elements of x
> and y are equal?
>
>    >>> map(operator.eq, x, y)
>    [True, False, True]
>
> It's not "==".
>
>    >>> map(operator.is_, x, y)
>    [True, True, False]
>
> And it's not "is".

class list:
     def __eq__(self, other):
         if len(self) != len(other):
             return False
         for this, that in zip(self, other):
              if this is that or this == that:
                  continue
              break
         else:
             return True
         return False

--
~Ethan~