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...).
Okay. My point was that the subset routine in python is an order of
magnitude faster than the above attempts at a Sage subset routine, so
hopefully we could do better than the above attempts at a subset routine
for Sage subsets. Sorry, I should have made that clearer.
Thanks,
Jason
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