Steven D'Aprano, 08.04.2010 03:41:
On Wed, 07 Apr 2010 10:55:10 -0700, Raymond Hettinger wrote:
[Gustavo Nare]
In other words: The more different elements two collections have, the
faster it is to compare them as sets. And as a consequence, the more
equivalent elements two collections have, the faster it is to compare
them as lists.
Is this correct?
If two collections are equal, then comparing them as a set is always
slower than comparing them as a list. Both have to call __eq__ for
every element, but sets have to search for each element while lists can
just iterate over consecutive pointers.
If the two collections have unequal sizes, then both ways immediately
return unequal.
Perhaps I'm misinterpreting what you are saying, but I can't confirm that
behaviour, at least not for subclasses of list:
>>> class MyList(list):
... def __len__(self):
... return self.n
...
>>> L1 = MyList(range(10))
>>> L2 = MyList(range(10))
>>> L1.n = 9
>>> L2.n = 10
>>> L1 == L2
True
>>> len(L1) == len(L2)
False
This code incorrectly assumes that overriding __len__ has an impact on the
equality of two lists. If you want to influence the equality, you need to
override __eq__. If you don't, the original implementation is free to do
whatever it likes to determine if it is equal to another value or not. If
it uses __len__ for that or not is only an implementation detail that can't
be relied upon.
Stefan
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