In this tutorial, we’ll be discussing the use cases of Swift Set. We’ll look at the basic syntax and some important operations performed using Swift sets.
1. Swift Set
- Swift Sets are a data structure type that holds unique values of the same type in an unordered fashion.
- Set can only take types that conform to the
Hashableprotocol. - Sets store the values based on the
hashValue()thanks to the hashable protocol. Hence the access time is constant in sets. - A hashValue is an Integer such that it is the same for all objects that compare equally. If
a == bis true thena.hashValue == b.hashValuemust be true.
2. Swift Set vs Array
- Sets store unique values. Arrays can store repeated.
- Sets are unordered. Printing a set won’t give the same order as it was defined. Arrays are ordered.
- Sets store values based on hash values. Arrays don’t.
- Sets have a constant lookup time thanks to storing by hash values. Arrays don’t. Hence Sets are faster than Array.
3. Swift Set vs Dictionary
Both are unordered, but dictionaries can have same values in two different keys. Sets have unique values always.
Let’s launch our Swift Playground in XCode and get started.
4. Declaring a set
We can declare a set in the following manner(s):
var emptySet = Set<String>()
emptySet = []
var filledSet : Set<Int> = [3,1,2]
var typeInferredSet : Set = ["Red","Yellow","Green"]
print(filledSet)
print(emptySet)
print(typeInferredSet)
To declare a set we must use the keyword Set followed by the type inside .
Swift can infer the type as well from the values set on the right-hand side.
The above code prints the following in the console.
5. Inserting and Removing Elements from a Set
We can insert and remove elements in a set in the following manner:
let typeInferredSet : Set = ["Red","Yellow","Green"]
//typeInferredSet.insert("Wednesday") //compilation error. Cannot insert in immutable Sets
var emptySet = Set<String>()
emptySet.insert("One")
emptySet.insert("Two")
emptySet.insert("One")
//emptySet.insert(3) //compilation error
var filledSet : Set<Int> = [3,1,2]
var x = filledSet.remove(2)
var y = filledSet.remove(0)
let storeIndex = filledSet.index(of: 1)
if let unwrapped = storeIndex {
filledSet.remove(at: unwrapped)
}
filledSet.removeAll() //removes all the elements
typeInferredSet.contains("Red") //returns true
print("empty set: \(emptySet) x is: \(x ?? -1) y is: \(y ?? -1)")
//Prints
//empty set: ["One", "Two"] x is: 2 y is: -1
- We cannot insert elements in Sets with
letconstant. - The removed elements can be retrieved in a var/let variable.
- The removed elements are returned as optionals.
- If the removed element doesn’t exist, it returns a nil which is wrapped in the optional.
- Duplicate elements that are inserted get ignored.
- Inserting an element with a different type would give a compilation error.
- To remove an element using it’s index we need to pass the
Set<Type>.Indexof the element intoremove(at:). For this we need to retrieve the index usingindex(of:)and pass the value by optional unwrapping usingif letstatement x ??checks whether the Optional is is null or not, if it is, it prints the value on the right of ?? else it prints the value from the optional after unwrapping.
6. Find an element in a set
contains() method checks whether the given element is present in the set and returns a boolean value.
let typeInferredSet : Set = ["Red","Yellow","Green"]
typeInferredSet.contains("Red") //returns true
7. Capacity of the set
The count property returns the number of elements present in the set.
isEmpty checks whether the set is empty or not and returns a boolean.
var filledSet : Set<Int> = [3,1,2]
filledSet.removeAll()
filledSet.count //returns 0
fillSet.isEmpty //returns true
typeInferredSet.count returns 3
8. Creating a Set from an Array
We can create a set from an array using the following syntax:
let intSet = Set([5,1,4,6])
let myArray = ["Monday","Tuesday"]
let anotherSet = Set(myArray)
let sequenceSet = Set<Int>(1...5)
print(sequenceSet) // [5, 2, 3, 1, 4]
We can create a set from an array by passing the array inside the Set(). We can also create a set using Ranges in Swift.
9. Iterating over a set
We can iterate over a set in the following manner using a for in loop.
var newSet : Set<String> = ["Yay","Nay","Meh"]
for words in newSet{
print(words)
}
//prints
//Nay
//Meh
//Yay
10. Sorting the set
Use the sorted() method to keep the set in an ordered way.
var newSet : Set<Int> = [5,4,1,2,3]
print(newSet)
for words in newSet.sorted(){
print(words)
}
The above code prints 1 to 5. It sorts the set in ascending order by default.
11. Swift Set Operations
Swift Set operations are useful for comparing two data sets. Following venn diagrams illustrate the commonly used operations.
- subset and superset
var setA : Set<String> = ["A","B","C","D","E"] var setB : Set<String> = ["C","D","E"] var setC : Set<String> = ["A","B","C","D","E"] setB.isSubset(of: setA) //true setB.isStrictSubset(of: setA) //true setC.isSubset(of: setA) //true setC.isStrictSubset(of: setA) //false setA.isSuperset(of: setB) //true setA.isStrictSuperset(of: setB) //trueisStrictSubset()returns true if the set is a subset of the passed in superset but NOT an exact copy.
isStrictSuperset()returns true if the set is a superset of the passed in subset but NOT an exact copy. - disjoint and union
Two sets are disjointed if no elements are common. Union of two sets joins both of them (ignoring duplicates).
var setA : Set<String> = ["A","B","C","D","E"] var setB : Set<String> = ["C","D","E"] var setC : Set<String> = ["A","B","C","D","E"] var setD : Set<String> = ["A","F"] setD.isDisjoint(with: setB)//true setD.isDisjoint(with: setA)//false var unionSet = setD.union(setA) // {"B", "A", "F", "C", "D} var inter = setD.intersection(setA) // {"A"}unionneeds to store the result in a new variable.intersectionmethod would get the common elements from both the sets.setA.substract(setB)would remove the elements from A that are their in B too and return a new set. setA won’t be changed unless we reassign the new set as setA only.
That’s all for a quick roundup on Swift Set.
Reference: Apple Docs
