Sorting Array in Swift3

Question

In my code, I have a struct like the following:

struct Object {
    var name: String
    var count: Int

I am now creating an array of 10 Objects with random names and random counts.

Is there an easy way to
a) sort them alphabetically
b) sort them numerically in ascending order

Basically, there will be an array like so: [Object1, Object2, Object3]. Every Object has a name and count attribute, and I want the objects in that list be sorted via these two attributes.

Solution in Swift2 (using this solution: StackOverflow):

Object.sort{
        if $0.name != $1.name {
            return $0.name < $1.name
        }
        else {
            //suits are the same
            return $0.count < $1.count
        }
    }

However, this has been renamed to sorted(by: ) in Swift3, and I don't quit get how to do that.

Answer 1

Narusan, maybe this will help you. Let's say you have an array with your struct objects called objArray, then you can order it by the code bellow:

var objArray = [Object]()
objArray.append(Object(name:"Steve", count:0))
objArray.append(Object(name:"Alex", count:1))

objNameSorted = objArray.sorted (by: {$0.name < $1.name})
objNCountSorted = objArray.sorted (by: {$0.count < $1.count})

Answer 2

If you want to sort alphabetically and then numerically, you can:

var array = ["A2", "B7", "A4", "C3", "A1", "A10"]
array.sort { $0.compare($1, options: .numeric) == .orderedAscending }

That produces:

["A1", "A2", "A4", "A10", "B7", "C3"]

I added A10 to your array, because without it, a simple alphabetic sort would have been sufficient. But I'm assuming you wanted A10 after A4, in which case the numeric comparison will do the job for you.

---

You changed the example to be a struct with two properties. In that case, you can do something like:

struct Foo {
    var name: String
    var count: Int
}

var array = [
    Foo(name:"A", count: 2),
    Foo(name:"B", count: 7),
    Foo(name:"A", count: 7),
    Foo(name:"C", count: 3),
    Foo(name:"A", count: 1),
    Foo(name:"A", count: 10)
]

array.sort { (object1, object2) -> Bool in
    if object1.name == object2.name {
        return object1.count < object2.count
    } else {
        return object1.name < object2.name
    }
}

Or, more concisely:

array.sort { $0.name == $1.name ? $0.count < $1.count : $0.name < $1.name }

Or

array.sort { ($0.name, $0.count) < ($1.name, $1.count) }

Note, rather than putting this logic in the closure, I'd actually make Foo conform to Comparable:

struct Foo {
    var name: String
    var count: Int
}

extension Foo: Equatable {
    static func ==(lhs: Foo, rhs: Foo) -> Bool {
        return (lhs.name, lhs.count) == (rhs.name, rhs.count)
    }
}

extension Foo: Comparable {
    static func <(lhs: Foo, rhs: Foo) -> Bool {
        return (lhs.name, lhs.count) < (rhs.name, rhs.count)
    }
}

This keeps the comparison logic nicely encapsulated within the Foo type, where it belongs.

Then you can just do the following to sort in place:

var array = ...
array.sort()

Or, alternatively, you can return a new array if you don't want to sort the original one in place:

let array = ...
let sortedArray = array.sorted()

Answer 3

You can still use shorthand for sorted:

objNameSorted = objArray.sorted { $0 < $1 }

While less readable, it more closely mimics the sort syntax.