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February 28, 2018 22:13
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Removing duplicates in swift, keeping order
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| var arr = [1, 1, 1, 2, 2, 3, 4, 5, 6, 6, 6, 6, 6, 7, 8] | |
| var list = ["A","A","A","A","A","A","B","C","D","E","E","E","E"] | |
| let set = Set(arr) // no duplicates but no order either | |
| let orderedSet = NSOrderedSet(array: arr) // no duplicates and ordered, but not an array | |
| let arrayFromOrderedSet = Array(NSOrderedSet(array: arr)) // is array, ordered, no duplicates, but loses type info | |
| let sorted = Array(Set(arr)).sorted() // is array, ordered, no duplicates, keeps type info, but limited to ascending order | |
| extension Array where Element: Equatable { | |
| func duplicatesRemoved() -> [Element] { | |
| return reduce([]) { $0.contains($1) ? $0 : $0 + [$1] } | |
| } | |
| mutating func removeDuplicates() { | |
| self = duplicatesRemoved() | |
| } | |
| } | |
| list.duplicatesRemoved() // new list -> ["A", "B", "C", "D", "E"] | |
| list.removeDuplicates() // same list -> ["A", "B", "C", "D", "E"] | |
| arr.duplicatesRemoved() // new list -> [1, 2, 3, 4, 5, 6, 7, 8] | |
| arr.removeDuplicates() // same list -> [1, 2, 3, 4, 5, 6, 7, 8] |
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