John Nagle <na...@animats.com> writes:
> d1 = set('monday','tuesday')
> days_off = set('saturday','sunday')
> if not d1.isdisjoint(days_off) :...
> This is cheaper than intersection, since it doesn't have to
> allocate and construct a set. It just tests whether any element in the
> smaller of the two sets is in the larger one.
I wonder what the complexity is, since the simplest implementation using
the dict-like features of sets is quadratic. There is of course an
obvious n log n implementation involving sorting.
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