Remove duplicate values in a tuple array

Question

I'm working on a project in Swift 3 where I have a tuple array with duplicate values is there a way to save it to a NSSet or to avoid replicating the same value. The structure of my tuple array as follow.

var selectedSongList : [(title: String, healerName: String, trackUrl: String, trackID: String, imageUrl: String)] = []

Thus later I'm using this to poplate my UITableView

• Hieu Dinh: Take a look at this: stackoverflow.com/questions/25738817/…
• danu: a code snippet would much appreciate
• Leo Dabus: make a struct that conforms to equatable
• danu: how do you do it with the code. Can you share a code snippet

Answer 1

There are two ways to do it.

Solution 1

You can create a structure and it should confirm to Hashable and equatable some thing like this

struct Post: Hashable, Equatable {
let id: String
var hashValue: Int { get { return id.hashValue } }
}
func ==(left:Post, right:Post) -> Bool {
return left.id == right.id
}

and to remove your object you can do like this

let uniquePosts = Array(Set(posts))

Solution 2

Create a set out of your array and then make it back to array.

Answer 2

Edit func static func == in Song struct if you have different definition about the same song

struct Song: Equatable {
 var title: String?
 var healerName: String?
 var trackUrl:String?
 var trackID:String?
 var imageUrl: String?
 init(_ title: String, _ healerName:String, _ trackUrl:String, _ trackID:String,_ imageUrl: String) {
 self.title = title
 self.healerName = healerName
 self.trackUrl = trackUrl
 self.trackID = trackID
 self.imageUrl = imageUrl
 }
 static func == (lhs: Song, rhs: Song) -> Bool {
 return lhs.title == rhs.title &&
 lhs.healerName == rhs.healerName && 
 lhs.trackUrl == rhs.trackUrl &&
 lhs.trackID == rhs.trackID &&
 lhs.imageUrl == rhs.imageUrl
 }
}
extension Array where Element:Equatable {
 func removeDuplicates() -> [Element] {
 var result = [Element]()
 for value in self {
 if result.contains(value) == false {
 result.append(value)
 }
 }
 return result
 }
}