On Sat, Jan 10, 2009 at 9:22 AM, Jason Grout
<[email protected]> wrote:
>
> John H Palmieri wrote:
>> sage: timeit('set(S).issubset(set(T))')
>>
>> gives me very similar times to the first option (all(s in T for s in
>> S)). So if I start with Sage sets, I don't seem to gain much by
>> converting back to python sets for this (not to mention that if S = Set
>> (ZZ), then set(S) runs into problems...).
The Sage enumerated Set type is just a light wrapper around Python's
"frozenset" type with more sage/mathematical semantics. Just delegate
to that for implementing is_subset:
sage: X = Set([1..10])
sage: Y = Set([1..20])
sage: X._Set_object__object.issubset(Y._Set_object__object)
True
sage: Y._Set_object__object.issubset(X._Set_object__object)
False
sage: timeit('X._Set_object__object.issubset(Y._Set_object__object)')
625 loops, best of 3: 2.07 µs per loop
sage: X._Set_object__object
frozenset([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
sage: type(X)
<class 'sage.sets.set.Set_object_enumerated'>
In the actual code for is_subset, you'll have
self.__object
instead of self._Set_object__object, since of course the above example
uses the command line so __ methods are mangled.
The one interesting wrinkle will be doing, e.g.,
sage: X.is_subset(Set(ZZ))
where the Y = Set(ZZ) isn't an enumerated set. There you have to
check each X for membership in Y.
-- William
William
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