Python: See if one set contains another entirely?

Question

Is there a fast way to check if one set entirely contains another?

Something like:

>>>[1, 2, 3].containsAll([2, 1])
True

>>>[1, 2, 3].containsAll([3, 5, 9])
False

Answer 1

WRONG-ANSWER

This is an order comparison, not a set content comparison:

>>> [1,2,3] > [1,2]
True
>>> [1,2,3] > [3,5,9]
False
>>> 

As the above answer states, convert to a set first, then use set methods.

Answer 2

Those are lists, but if you really mean sets you can use the issubset method.

>>> s = set([1,2,3])
>>> t = set([1,2])
>>> t.issubset(s)
True
>>> s.issuperset(t)
True

For a list, you will not be able to do better than checking each element.

Answer 3

Try issubset:

>>> a=set([1,2,3])
>>> b=set([1,2])
>>> b.issubset(a)
True
>>> a.issubset(b)
False

Answer 4

For completeness: this is equivalent to issubset (although arguably a bit less explicit/readable):

>>> set([1,2,3]) >= set([2,1])
True
>>> set([1,2,3]) >= set([3,5,9])
False