Sorting Array in Swift3
Solution 1
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()
Solution 2
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})
Solution 3
You can still use shorthand for sorted
:
objNameSorted = objArray.sorted { $0 < $1 }
While less readable, it more closely mimics the sort
syntax.
Narusan
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Updated on June 09, 2020Comments
-
Narusan almost 4 years
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 orderBasically, there will be an array like so:
[Object1, Object2, Object3]
. Every Object has aname
andcount
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. -
Rob over 7 yearsNo, in Swift 3,
sorted
is the version that returns a new array.sort
is the version that sorts in place. -
Rob over 7 yearsThen you must not be sorting in place, but rather trying to return it as a new array, in which case you'd do
let sortedArray = array.sorted() { ... }
. -
Narusan over 7 yearsSorry that I can't do anything more than up vote your answer and comments, you deserve way more!
-
mfaani over 7 yearswhy did you write static func?
-
Rob over 7 yearsIn Swift 2, the equality and comparison operators were implemented as globals, but in Swift 3 they are defined as
static
methods of the type, which keeps your code a little better organized. See Improving operators in protocols. -
mfaani over 7 yearscorrect me if I'm wrong, you can still do it the Swift2 way in Swift 3, ie define it as a global func ? However it's not recommended?
-
Rob over 7 yearsIt's a poor design, IMHO, but, yes, you can do it the old way, too.
-
kakubei almost 7 yearsYou can still use shorthand for
sorted
: objNameSorted = objArray.sorted { $0 < $1 } While less readable, it more closely mimics thesort
syntax.