Sort Array of Objects With Some Conditions

Question

I have an array of Users objects:

Class Users {
    var firstName:String!
    var lastName:String!
    var status:Int!

    override init(fn: String, ln: String, s: Int) {
        self.firstName = fn
        self.lastName = ln
        self.status = s
    }
}

var users:[Users] = [Users]()
users.append(Users(fn:"a", ln:"b", s:1))
users.append(Users(fn:"c", ln:"d", s:3))
users.append(Users(fn:"e", ln:"f", s:2))
users.append(Users(fn:"g", ln:"h", s:1))
users.append(Users(fn:"i", ln:"j", s:1))
users.append(Users(fn:"k", ln:"l", s:2))

I know the method to sort this array on status like

users = users.sorted(by: {$0.status > $1.status})

But, how could I sort users array on status with 2 on top then 3 and in the last 1 i.e [2,2,3,1,1,1]

what is the purpose of sorting the array if you want a specific order ? — marshallino16
– marshallino16, Commented Nov 27, 2017 at 13:56
maybe you should define an enum that represents the status. Then you could assign an int value to each status type and order based on that value. — Oxthor
– Oxthor, Commented Nov 27, 2017 at 13:58

Hexfire · Accepted Answer · 2017-11-27 14:20:45Z

Consider refactoring:

Class to struct:

struct User {
    var firstName:String!
    var lastName:String!
    var status:Status
}

Status enum:

enum Status: Int {
    case online, busy, offline
}

Population:

var users: [User] = [
    User(firstName: "a", lastName: "b", status: .online),
    User(firstName: "c", lastName: "d", status: .offline),
    User(firstName: "e", lastName: "f", status: .busy),
    User(firstName: "g", lastName: "h", status: .online),
    User(firstName: "i", lastName: "j", status: .online),
    User(firstName: "k", lastName: "l", status: .busy)
]

Sort:

users = users.sorted(by:{$0.status.rawValue < $1.status.rawValue })

users.forEach { print($0) }

Output:

User(firstName: a, lastName: b, status: .online)
User(firstName: g, lastName: h, status: .online)
User(firstName: i, lastName: j, status: .online)
User(firstName: e, lastName: f, status: .busy)
User(firstName: k, lastName: l, status: .busy)
User(firstName: c, lastName: d, status: .offline)

Bilal · Accepted Answer · 2017-11-27 14:17:48Z

1

Try this. This will sort based on the order array. If any status is missing will be moved to bottom of the array.

let order = [2,3,1]
users = users.sorted(by: { order.index(of: $0.status) ?? Int.max < order.index(of: $1.status) ?? Int.max })
// 2,2,3,1,1,1

Lets say if 3 is missing in order array

let order = [2,1]
users = users.sorted(by: { order.index(of: $0.status) ?? Int.max < order.index(of: $1.status) ?? Int.max })
// 2,2,1,1,1,3

answered Nov 27, 2017 at 14:17

Bilal

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1 Comment

Noshaid Ali Over a year ago

Great! this is what I want. Thanks.

Josh Homann · Accepted Answer · 2017-11-27 14:16:22Z

1

Use a lookup for your sort order.

let sortPriority: [Int: Int] = [ 1: 3, 2: 1, 3: 2]

Also you can use the mutating sort to sort the array in place instead of sorted which returns a copy since it looks like you are discarding the original array:

users.sort(by: {sortPriority [$0.status, default: Int.max] < sortPriority [$1.status, default: Int.max]})
print(users.flatMap{$0.status})

output:

[2, 2, 3, 1, 1, 1]

answered Nov 27, 2017 at 14:16

Josh Homann

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1 Comment

Noshaid Ali Over a year ago

Thanks for your support, this solution is also fine I got it.

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