我可能有一个像下面这样的数组:
[1, 4, 2, 2, 6, 24, 15, 2, 60, 15, 6]
或者,实际上,任何类似类型的数据部分的序列。我要做的是确保每个相同的元素只有一个。例如,上面的数组将变成:
[1, 4, 2, 6, 24, 15, 60]
请注意,删除了2、6和15的重复项,以确保每个相同的元素中只有一个。Swift是否提供了一种容易做到这一点的方法,还是我必须自己做?
我可能有一个像下面这样的数组:
[1, 4, 2, 2, 6, 24, 15, 2, 60, 15, 6]
或者,实际上,任何类似类型的数据部分的序列。我要做的是确保每个相同的元素只有一个。例如,上面的数组将变成:
[1, 4, 2, 6, 24, 15, 60]
请注意,删除了2、6和15的重复项,以确保每个相同的元素中只有一个。Swift是否提供了一种容易做到这一点的方法,还是我必须自己做?
当前回答
在Swift 5中
var array: [String] = ["Aman", "Sumit", "Aman", "Sumit", "Mohan", "Mohan", "Amit"]
let uniq = Array(Set(array))
print(uniq)
输出将是
["Sumit", "Mohan", "Amit", "Aman"]
其他回答
最简单的方法是使用NSOrderedSet,它存储唯一的元素并保持元素的顺序。如:
func removeDuplicates(from items: [Int]) -> [Int] {
let uniqueItems = NSOrderedSet(array: items)
return (uniqueItems.array as? [Int]) ?? []
}
let arr = [1, 4, 2, 2, 6, 24, 15, 2, 60, 15, 6]
removeDuplicates(from: arr)
这里有一个解决方案
不使用传统NS类型 对于O(n)来说是相当快的 简洁 保持元素顺序
extension Array where Element: Hashable {
var uniqueValues: [Element] {
var allowed = Set(self)
return compactMap { allowed.remove($0) }
}
}
这里我对对象做了O(n)解。不是少行解决方案,而是……
struct DistinctWrapper <T>: Hashable {
var underlyingObject: T
var distinctAttribute: String
var hashValue: Int {
return distinctAttribute.hashValue
}
}
func distinct<S : SequenceType, T where S.Generator.Element == T>(source: S,
distinctAttribute: (T) -> String,
resolution: (T, T) -> T) -> [T] {
let wrappers: [DistinctWrapper<T>] = source.map({
return DistinctWrapper(underlyingObject: $0, distinctAttribute: distinctAttribute($0))
})
var added = Set<DistinctWrapper<T>>()
for wrapper in wrappers {
if let indexOfExisting = added.indexOf(wrapper) {
let old = added[indexOfExisting]
let winner = resolution(old.underlyingObject, wrapper.underlyingObject)
added.insert(DistinctWrapper(underlyingObject: winner, distinctAttribute: distinctAttribute(winner)))
} else {
added.insert(wrapper)
}
}
return Array(added).map( { return $0.underlyingObject } )
}
func == <T>(lhs: DistinctWrapper<T>, rhs: DistinctWrapper<T>) -> Bool {
return lhs.hashValue == rhs.hashValue
}
// tests
// case : perhaps we want to get distinct addressbook list which may contain duplicated contacts like Irma and Irma Burgess with same phone numbers
// solution : definitely we want to exclude Irma and keep Irma Burgess
class Person {
var name: String
var phoneNumber: String
init(_ name: String, _ phoneNumber: String) {
self.name = name
self.phoneNumber = phoneNumber
}
}
let persons: [Person] = [Person("Irma Burgess", "11-22-33"), Person("Lester Davidson", "44-66-22"), Person("Irma", "11-22-33")]
let distinctPersons = distinct(persons,
distinctAttribute: { (person: Person) -> String in
return person.phoneNumber
},
resolution:
{ (p1, p2) -> Person in
return p1.name.characters.count > p2.name.characters.count ? p1 : p2
}
)
// distinctPersons contains ("Irma Burgess", "11-22-33") and ("Lester Davidson", "44-66-22")
这只是一个非常简单和方便的实现。具有相等元素的数组扩展中的计算属性。
extension Array where Element: Equatable {
/// Array containing only _unique_ elements.
var unique: [Element] {
var result: [Element] = []
for element in self {
if !result.contains(element) {
result.append(element)
}
}
return result
}
}
使用Set或NSOrderedSet删除重复项,然后转换回数组:
let uniqueUnordered = Array(Set(array))
let uniqueOrdered = Array(NSOrderedSet(array: array))