Swift 1.2: Let's Talk About Sets, Baby!
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Join For Freelet odd = Set([1, 3, 3, 3, 5, 7, 9]) // {5, 7, 3, 1, 9} let even = Set([2, 2, 2, 2, 2, 2, 4, 6, 8, 8, 10]) // {6, 10, 2, 4, 8}
let oddDisjointWithEven = odd.isDisjointWith(even) // true
To combine these sets together, we’ll use the union() method to give us a new Set instance named allNumbers:
let allNumbers = odd.union(even) // {10, 2, 9, 4, 5, 7, 6, 3, 1, 8}
To do some more tinkering, I’ll create my own set named xyzzyNumbers:
let xyzzyNumbers = Set([1, 2, 3, 4]) // {2, 3, 1, 4}
…this new set does contain some odd numbers, so I’d expect isDisjoint() again my odd numbers to be false:
let xyzzyDisjointWithOdd = xyzzyNumbers.isDisjointWith(odd) // false
To get the odd numbers in xyzzy that are below ten, I can use the intersect() method:
let xyzzyOddNumbers = xyzzyNumbers.intersect(odd) // {3, 1}
…and likewise with even:
let xyzzyEvenNumbers = xyzzyNumbers.intersect(even) // {2, 4}
The isSubset() method returns true if one set is a subset of the other. xyzzy isn’t a subset of odd, because it contains even numbers, however it is a subset of allNumbers:
let xyzzyIsSubsetOfOdd = xyzzyNumbers.isSubsetOf(odd) // false let xyzzyIsSubSetOfAll = xyzzyNumbers.isSubsetOf(allNumbers) // true
…and this implies that allNumbers is a superset of xyzzy:
let allIsSupersetOfXyzzy = allNumbers.isSupersetOf(xyzzyNumbers) // true
I can also subtract one set from another. Here, I remove all of the numbers from xyzzy from my odd numbers:
let oddNumbersSubtractXyzzy = odd.subtract(xyzzyNumbers) // {5, 7, 9}
Sets aren’t limited to integers, they can be created from any hashable types, such as strings:
let colors = Set(["red", "orange", "purple"]) // {"orange", "purple", "red"} let fruit = Set(["apple", "banana", "orange"]) // {"banana", "apple", "orange"} let fruityColors = colors.intersect(fruit) // {"orange"} let nonFruityColors = colors.exclusiveOr(fruit) // {"banana", "purple", "red", "apple"}
But types can’t be mixed, so this will not compile:
let evenColors = colors.intersect(even)
Addendum - Creating Sets From Your Own Data Types
If you want to create sets of your own data type, you need to ensure your struct or class conforms to the Hashable protocol which extends the Equatable protocol. This means you’ll need your data data to implement an equality function (‘==‘) and be able to return a hashed value of itself.
I’ve created a simple Creature struct with a handful of properties that describe a creature:
let name: String let swims: Bool let flies: Bool let walks: Bool let legCount: Int
To conform to Hashable, Creature’s hashValue returns the integer hash value of its name property (String is already Hashable, so that’s built in) and adds increasing powers of ten for the three Boolean properties and finally adds the number of legs:
var hashValue: Int { get { return name.hashValue + (swims ? 10 : 0) + (flies ? 100 : 0) + (walks ? 1000 : 0) + legCount } }
To conform to Equatable, we implement an equality function (thanks to Davide De Franceschi for pointing out hash value is not a good way to equate two objects!):
func == (lhs: Creature, rhs: Creature) -> Bool { return lhs.name == rhs.name && lhs.swims == rhs.swims && lhs.flies == rhs.flies && lhs.walks == rhs.walks && lhs.legCount == rhs.legCount }
The final Creature definition looks like:
struct Creature: Hashable { init(name: String, swims: Bool, flies: Bool, walks: Bool, legCount: Int) { self.name = name self.swims = swims self.flies = flies self.walks = walks self.legCount = legCount } let name: String let swims: Bool let flies: Bool let walks: Bool let legCount: Int var hashValue: Int { get { return name.hashValue + (swims ? 10 : 0) + (flies ? 100 : 0) + (walks ? 1000 : 0) + legCount } } }
We can now create an array of a handful of different creatures:
…and then create some sets of creatures with different special skills:
let ant = Creature(name: "Ant", swims: false, flies: false, walks: true, legCount: 6) let bumbleBee = Creature(name: "Bumble Bee", swims: false, flies: true, walks: false, legCount: 6) let cat = Creature(name: "Cat", swims: false, flies: false, walks: true, legCount: 4) let flyingFish = Creature(name: "Flying Fish", swims: true, flies: true, walks: false, legCount: 0) let human = Creature(name: "Human", swims: true, flies: false, walks: true, legCount: 2) let penguin = Creature(name: "Penguin", swims: true, flies: false, walks: true, legCount: 2) let swift = Creature(name: "Swift", swims: false, flies: true, walks: true, legCount: 2) let creaturesArray = [ant, ant, ant, bumbleBee, cat, flyingFish, human, penguin, swift] let swimmers = Set(creaturesArray.filter({ $0.swims })) let walkers = Set(creaturesArray.filter({ $0.walks })) let fliers = Set(creaturesArray.filter({ $0.flies }))
Using the Set syntax, we can find the creatures that can fly and walk:
let flyingWalkers = fliers.intersect(walkers)
We can find creatures that can walk but can’t fly or swim by chaining two set methods together:
let walkingNonSwimmersNonFliers = walkers.subtract(swimmers).subtract(fliers)
…or we can find the creatures that can either fly or swim:
let fliersAndSwimmers = fliers.union(swimmers)
Again, although the flying fish can do both, because we’re dealing with sets, it only appears once.
Published at DZone with permission of Simon Gladman, DZone MVB. See the original article here.
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