On Apr 7, 8:41 pm, Steven D'Aprano <ste...@remove.this.cybersource.com.au> wrote: > 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 > > -- > Steven
I think what he is saying is that the list __eq__ method will look at the list lengths first. This may or may not be considered a subtle bug for the edge case you are showing. If I do the following: >>> L1 = range(10000000) >>> L2 = range(10000000) >>> L3 = range(10000001) >>> L1 == L2 True >>> L1 == L3 False >>> I don't even need to run timeit -- the "True" takes awhile to print out, while the "False" prints out immediately. Regards, Pat -- http://mail.python.org/mailman/listinfo/python-list
