Erlang Thursday – ordsets:subtract/2

Today’s Erlang Thursday is on ordsets:subtract/2.

ordsets:subtract/2 takes two ordered sets as its arguments, and returns a ordered set containing the items of the first ordered set that are not in the second ordered set.

OrderedSetA = ordsets:from_list([5, 4, 3, 2, 1]).
# [1,2,3,4,5]
OrderedSetB = ordsets:from_list([1, 1, 2, 3, 5, 8, 13]).
# [1,2,3,5,8,13]
OrderedSetC = ordsets:from_list([2, -2, 4, -4, 16, -16]).
# [-16,-4,-2,2,4,16]
EmptySet = ordsets:new().
# []
ordsets:subtract(OrderedSetA, OrderedSetB).
# [4]
ordsets:subtract(OrderedSetA, EmptySet).
# [1,2,3,4,5]
ordsets:subtract(OrderedSetB, EmptySet).
# [1,2,3,5,8,13]
ordsets:subtract(EmptySet, OrderedSetA).
# []
ordsets:subtract(OrderedSetB, OrderedSetC).
# [1,3,5,8,13]

And note that ordsets:subtract/2 is not commutative, unlike ordsets:union/2 or ordsets:intersection/2.

ordsets:subtract(OrderedSetA, OrderedSetC).
# [1,3,5]
ordsets:subtract(OrderedSetC, OrderedSetA).
# [-16,-4,-2,16]

And again, your friendly reminder if you haven’t been following along, just because Ordered Sets in Erlang are represented as a List, doesn’t mean that Lists are Ordered Sets.

–Proctor

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