On Fri, Jan 9, 2009 at 6:42 PM, John H Palmieri <jhpalmier...@gmail.com> wrote:
>
>
>
> On Jan 9, 3:40 pm, "William Stein" <wst...@gmail.com> wrote:
>> On Fri, Jan 9, 2009 at 3:38 PM, Mike Hansen <mhan...@gmail.com> wrote:
>>
>> > On Fri, Jan 9, 2009 at 3:36 PM, John H Palmieri <jhpalmier...@gmail.com>
>> > wrote:
>>
>> >> Is this a bug?
>>
>> >> sage: Set([])
>> >> {}
>> >> sage: Set(Set([]))
>> >> {}
>> >> sage: Set([]) == Set(Set([]))
>> >> True
>>
>> > This is because Set takes a list (iterable) for all the of the
>> > elements of the set. So, if you want to construct the set containing
>> > the empty set, you'd do the following:
>>
>> > sage: e = Set([])
>> > sage: ee = Set([e])
>> > sage: e
>> > {}
>> > sage: ee
>> > {{}}
>> > sage: e == ee
>> > False
>>
>> Yep. Think "coercion" -- Set(foo) makes foo into a set.
>> It doesn't make the set containing foo.
>
> Oh, I should have figured that out.
>
> Here's another question: what is the most efficient way of testing
> whether one Set is a subset of another? I can do
>
> S in list(T.subsets())
>
> -- and it's a bit frustrating that I can't do S in T.subsets() -- and
> I can also manipulate intersections, unions, differences, etc. I can
> also convert to python sets and use <=. Is there a preferred way?
>
There should be an is_subset method, but mysteriously there isn't. Implement
it and send a patch.
William
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