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
下面是SequenceType上的一个类别,它保留了数组的原始顺序,但使用Set来进行contains查找,以避免数组的contains(_:)方法上的O(n)代价。
public extension Sequence where Element: Hashable {
/// Return the sequence with all duplicates removed.
///
/// i.e. `[ 1, 2, 3, 1, 2 ].uniqued() == [ 1, 2, 3 ]`
///
/// - note: Taken from stackoverflow.com/a/46354989/3141234, as
/// per @Alexander's comment.
func uniqued() -> [Element] {
var seen = Set<Element>()
return self.filter { seen.insert($0).inserted }
}
}
如果你不是Hashable或Equatable,你可以传入一个谓词来进行相等性检查:
extension Sequence {
/// Return the sequence with all duplicates removed.
///
/// Duplicate, in this case, is defined as returning `true` from `comparator`.
///
/// - note: Taken from stackoverflow.com/a/46354989/3141234
func uniqued(comparator: @escaping (Element, Element) throws -> Bool) rethrows -> [Element] {
var buffer: [Element] = []
for element in self {
// If element is already in buffer, skip to the next element
if try buffer.contains(where: { try comparator(element, $0) }) {
continue
}
buffer.append(element)
}
return buffer
}
}
现在,如果你没有Hashable,但是是Equatable,你可以使用这个方法:
extension Sequence where Element: Equatable {
/// Return the sequence with all duplicates removed.
///
/// i.e. `[ 1, 2, 3, 1, 2 ].uniqued() == [ 1, 2, 3 ]`
///
/// - note: Taken from stackoverflow.com/a/46354989/3141234
func uniqued() -> [Element] {
return self.uniqued(comparator: ==)
}
}
最后,你可以添加一个unique的关键路径版本,如下所示:
extension Sequence {
/// Returns the sequence with duplicate elements removed, performing the comparison using the property at
/// the supplied keypath.
///
/// i.e.
///
/// ```
/// [
/// MyStruct(value: "Hello"),
/// MyStruct(value: "Hello"),
/// MyStruct(value: "World")
/// ].uniqued(\.value)
/// ```
/// would result in
///
/// ```
/// [
/// MyStruct(value: "Hello"),
/// MyStruct(value: "World")
/// ]
/// ```
///
/// - note: Taken from stackoverflow.com/a/46354989/3141234
///
func uniqued<T: Equatable>(_ keyPath: KeyPath<Element, T>) -> [Element] {
self.uniqued { $0[keyPath: keyPath] == $1[keyPath: keyPath] }
}
}
你可以把这两个都放在你的应用程序中,Swift会根据你的序列的迭代器选择正确的一个。元素类型。
对于El Capitan,您可以扩展此方法以包括多个键盘,如下所示:
/// Returns the sequence with duplicate elements removed, performing the comparison using the property at
/// the supplied keypaths.
///
/// i.e.
///
/// ```
/// [
/// MyStruct(value1: "Hello", value2: "Paula"),
/// MyStruct(value1: "Hello", value2: "Paula"),
/// MyStruct(value1: "Hello", value2: "Bean"),
/// MyStruct(value1: "World", value2: "Sigh")
/// ].uniqued(\.value1, \.value2)
/// ```
/// would result in
///
/// ```
/// [
/// MyStruct(value1: "Hello", value2: "Paula"),
/// MyStruct(value1: "Hello", value2: "Bean"),
/// MyStruct(value1: "World", value2: "Sigh")
/// ]
/// ```
///
/// - note: Taken from stackoverflow.com/a/46354989/3141234
///
func uniqued<T: Equatable, U: Equatable>(_ keyPath1: KeyPath<Element, T>, _ keyPath2: KeyPath<Element, U>) -> [Element] {
self.uniqued {
$0[keyPath: keyPath1] == $1[keyPath: keyPath1] && $0[keyPath: keyPath2] == $1[keyPath: keyPath2]
}
}
但是(恕我直言)你最好把你自己的block传递给self.unique。
其他回答
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]
如果你想保持秩序,那么就用这个
let fruits = ["apple", "pear", "pear", "banana", "apple"]
let orderedNoDuplicates = Array(NSOrderedSet(array: fruits).map({ $0 as! String }))
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]
var numbers = [1,2,3,4,5,10,10, 12, 12, 6,6,6,7,8,8, 8, 8, 8 , 7 , 1 , 1, 2 , 9]
var newArr : [Int] = []
for n in numbers {
if !newArr.contains(n) {
newArr.append(n)
}
}
输出- [1,2,3,4,5,10,12,6,7,8,9]
上面的解决方案保持了顺序,但是很慢,因为.contains一次又一次地迭代。 因此使用有序集。
这将打印有序数组。
Array(NSOrderedSet.init(array: numbers))
输出- [1,2,3,4,5,10,12,6,7,8,9]
这将打印一个无序数组。
let uniqueUnordered = Array(Set(numbers))
输出- [4,2,1,9,10,3,5,6,8,12,7]
正如WWDC 2021所指出的那样,Swift有社区开发的算法、集合和数字包。算法包有一个唯一的()算法。
这些还不是Swift标准库的一部分。你目前可以从苹果的Github页面下载它们和/或通过Swift软件包管理器安装它们。
WWDC视频:
https://developer.apple.com/videos/play/wwdc2021/10256/
Github页面:
https://github.com/apple/swift-algorithms
unique()和unique (on:)文档:
https://github.com/apple/swift-algorithms/blob/main/Guides/Unique.md