Compare and return a distinct swift array object based on a property

Question

I have an array of objects with some properties.

Ex: Restaurants {name, latitude, longitude, address}

I want to compare each object based on a particular property and return a list of distinct objects.

For example:

Array = [R1, R2, R3, R4, R5]

Let say, Restaurants R4 and R5 are closer to each other and they have the same latitude and longitude.

I need to compare and filter the array

($0.latitude == $1.latitude and $0.longitude == $1.longitude) rounded to 3 decimal points

so that the final result will have the following items [R1, R2, R3, R4]

Arthur · Accepted Answer · 2016-09-12 20:17:58Z

2

As @kandelvijaya said, if you want/have to compare elements pairwise I'm not sure you can do better than quadratic time (which is fine in many cases).

Here's a generic O(n) alternative. Two things:

Instead of comparing two elements directly you provide a key representing the property that you want to compare by.
This key needs to be hashable (wich many standard types are).

.

extension Sequence {
    typealias Element = Iterator.Element

    func uniqued<Key: Hashable>(by key: (Element) -> (Key)) -> [Element] {
        var seen = Set<Key>()
        return filter {
            seen.update(with: key($0)) == nil
        }
    }
}

Usage:

let numbers = [1, 1, 1, 2, 2, 3, 1, 2, 3]
let uniqueByValue = numbers.uniqued { $0 } 

let strings = ["a", "b", "aa", "bb", "ccc", "dddd"]
let uniqueByLength = strings.uniqued { $0.characters.count }

In your case this could be:

let restaurantes = [R1, R2, R3, R4, R5]
let uniques = restaurants.uniqued { 
    let roundedLat = ...
    let roundedLong = ...
    return "\(roundedLat), \(roundedLong)"
}

edited Sep 12, 2016 at 20:17

answered Sep 12, 2016 at 10:50

Arthur

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1 Comment

Nibin V Over a year ago

Thanks Arthur. Couldnt try out your solutions. Will check it later

kandelvijaya · Accepted Answer · 2016-09-12 18:20:12Z

1

class B {
    var a: String?
    var b: Int?
}

var arr:[B] = []

arr.filter{ index in
    var repeatedCount = 0
    arr.forEach{
        repeatedCount += $0.a == index.a ? 1 : 0
    }
    return repeatedCount == 1
}

The above solution will work but the complexity is o(arr.count^2) which is bad for large sets of items.

You can generalize the algorithm and make it efficient. Or make a extension function on Array distinct

In your case the code would be something like this:

arr.filter{ index in
    var repeatedCount = 0
    arr.forEach{
        if $0.latitude == index.latitude && $0.longitude == index.longitude {
            repeatedCount += 1
        }
    }
    return repeatedCount == 1
}

edited Sep 12, 2016 at 18:20

answered Sep 12, 2016 at 9:26

kandelvijaya

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1 Comment

Nibin V Over a year ago

Thanks @kandelvijaya. Minor update - instead of repeatedCount += 0 we need repeatedCount += 1 right?

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