Proposal: === and !=== operators
Ethan Furman
ethan at stoneleaf.us
Sat Jul 12 02:11:37 EDT 2014
On 07/11/2014 10:07 PM, Alan Bawden wrote:
> Steven D'Aprano <steve+comp.lang.python at 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~
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