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
我认为这是更好的理解逻辑的方法
var arrayOfInts = [2, 2, 4, 4]
var mainArray = [Int]()
for value in arrayOfInts {
if mainArray.contains(value) != true {
mainArray.append(value)
print("mainArray:\(mainArray)")
}}
其他回答
如果你把两个扩展都放在你的代码中,更快的Hashable版本将在可能的情况下使用,Equatable版本将用作备用版本。
public extension Sequence where Element: Hashable {
/// The elements of the sequence, with duplicates removed.
/// - Note: Has equivalent elements to `Set(self)`.
@available(
swift, deprecated: 5.4,
message: "Doesn't compile without the constant in Swift 5.3."
)
var firstUniqueElements: [Element] {
let getSelf: (Element) -> Element = \.self
return firstUniqueElements(getSelf)
}
}
public extension Sequence where Element: Equatable {
/// The elements of the sequence, with duplicates removed.
/// - Note: Has equivalent elements to `Set(self)`.
@available(
swift, deprecated: 5.4,
message: "Doesn't compile without the constant in Swift 5.3."
)
var firstUniqueElements: [Element] {
let getSelf: (Element) -> Element = \.self
return firstUniqueElements(getSelf)
}
}
public extension Sequence {
/// The elements of the sequences, with "duplicates" removed
/// based on a closure.
func firstUniqueElements<Hashable: Swift.Hashable>(
_ getHashable: (Element) -> Hashable
) -> [Element] {
var set: Set<Hashable> = []
return filter { set.insert(getHashable($0)).inserted }
}
/// The elements of the sequence, with "duplicates" removed,
/// based on a closure.
func firstUniqueElements<Equatable: Swift.Equatable>(
_ getEquatable: (Element) -> Equatable
) -> [Element] {
reduce(into: []) { uniqueElements, element in
if zip(
uniqueElements.lazy.map(getEquatable),
AnyIterator { [equatable = getEquatable(element)] in equatable }
).allSatisfy(!=) {
uniqueElements.append(element)
}
}
}
}
如果顺序不重要,那么你总是可以使用这个Set初始化式。
你可以很容易地转换为一个Set和一个数组:
let unique = Array(Set(originals))
这并不能保证保持数组的原始顺序。
为此,我做了一个尽可能简单的扩展。
extension Array where Element: Equatable {
func containsHowMany(_ elem: Element) -> Int {
return reduce(0) { $1 == elem ? $0 + 1 : $0 }
}
func duplicatesRemoved() -> Array {
return self.filter { self.containsHowMany($0) == 1 }
}
mutating func removeDuplicates() {
self = self.duplicatesRemoved(()
}
}
您可以使用duplicatesRemoved()来获取一个新数组,它的重复元素将被删除,或者使用removeduplicate()来改变自身。看到的:
let arr = [1, 1, 1, 2, 2, 3, 4, 5, 6, 6, 6, 6, 6, 7, 8]
let noDuplicates = arr.duplicatesRemoved()
print(arr) // [1, 1, 1, 2, 2, 3, 4, 5, 6, 6, 6, 6, 6, 7, 8]
print(noDuplicates) // [1, 2, 3, 4, 5, 6, 7, 8]
arr.removeDuplicates()
print(arr) // [1, 2, 3, 4, 5, 6, 7, 8]
Swift 3/ Swift 4/ Swift 5
只需要一行代码就可以省略数组副本而不影响顺序:
let filteredArr = Array(NSOrderedSet(array: yourArray))
你可以自己卷,比如这样:
func unique<S : Sequence, T : Hashable>(source: S) -> [T] where S.Iterator.Element == T {
var buffer = [T]()
var added = Set<T>()
for elem in source {
if !added.contains(elem) {
buffer.append(elem)
added.insert(elem)
}
}
return buffer
}
let vals = [1, 4, 2, 2, 6, 24, 15, 2, 60, 15, 6]
let uniqueVals = uniq(vals) // [1, 4, 2, 6, 24, 15, 60]
作为Array的扩展:
extension Array where Element: Hashable {
func uniqued() -> Array {
var buffer = Array()
var added = Set<Element>()
for elem in self {
if !added.contains(elem) {
buffer.append(elem)
added.insert(elem)
}
}
return buffer
}
}
或者更优雅一点(Swift 4/5):
extension Sequence where Element: Hashable {
func uniqued() -> [Element] {
var set = Set<Element>()
return filter { set.insert($0).inserted }
}
}
将被使用:
[1,2,4,2,1].uniqued() // => [1,2,4]