如何按 Swift 中的数组元素分组

Question

假设我有这个代码：

class Stat {
   var statEvents : [StatEvents] = []
}

struct StatEvents {
   var name: String
   var date: String
   var hours: Int
}


var currentStat = Stat()

currentStat.statEvents = [
   StatEvents(name: "lunch", date: "01-01-2015", hours: 1),
   StatEvents(name: "dinner", date: "01-01-2015", hours: 1),
   StatEvents(name: "dinner", date: "01-01-2015", hours: 1),
   StatEvents(name: "lunch", date: "01-01-2015", hours: 1),
   StatEvents(name: "dinner", date: "01-01-2015", hours: 1)
]

var filteredArray1 : [StatEvents] = []
var filteredArray2 : [StatEvents] = []

为了让 2 个数组按 "same name" 分组，我可以多次手动调用下一个函数。

filteredArray1 = currentStat.statEvents.filter({[=12=].name == "dinner"})
filteredArray2 = currentStat.statEvents.filter({[=12=].name == "lunch"})

问题是我不知道变量值，在本例中为 "dinner" 和 "lunch"，所以我想按名称自动对这个 statEvents 数组进行分组，所以我得到名称不同的数组数。

我该怎么做？

Answer 1

Swift 4:

自 Swift 4 起，此功能已 added to the standard library。你可以像这样使用它：

Dictionary(grouping: statEvents, by: { [=10=].name })
[
  "dinner": [
    StatEvents(name: "dinner", date: "01-01-2015", hours: 1),
    StatEvents(name: "dinner", date: "01-01-2015", hours: 1),
    StatEvents(name: "dinner", date: "01-01-2015", hours: 1)
  ],
  "lunch": [
    StatEvents(name: "lunch", date: "01-01-2015", hours: 1),
    StatEvents(name: "lunch", date: "01-01-2015", hours: 1)
]

Swift 3:

public extension Sequence {
    func group<U: Hashable>(by key: (Iterator.Element) -> U) -> [U:[Iterator.Element]] {
        var categories: [U: [Iterator.Element]] = [:]
        for element in self {
            let key = key(element)
            if case nil = categories[key]?.append(element) {
                categories[key] = [element]
            }
        }
        return categories
    }
}

不幸的是，上面的 append 函数复制了底层数组，而不是就地改变它，这会更可取。 This causes a pretty big slowdown。您可以通过使用引用类型包装器来解决这个问题：

class Box<A> {
  var value: A
  init(_ val: A) {
    self.value = val
  }
}

public extension Sequence {
  func group<U: Hashable>(by key: (Iterator.Element) -> U) -> [U:[Iterator.Element]] {
    var categories: [U: Box<[Iterator.Element]>] = [:]
    for element in self {
      let key = key(element)
      if case nil = categories[key]?.value.append(element) {
        categories[key] = Box([element])
      }
    }
    var result: [U: [Iterator.Element]] = Dictionary(minimumCapacity: categories.count)
    for (key,val) in categories {
      result[key] = val.value
    }
    return result
  }
}

即使你遍历最终字典两次，这个版本在大多数情况下仍然比原来的更快。

Swift 2:

public extension SequenceType {

  /// Categorises elements of self into a dictionary, with the keys given by keyFunc

  func categorise<U : Hashable>(@noescape keyFunc: Generator.Element -> U) -> [U:[Generator.Element]] {
    var dict: [U:[Generator.Element]] = [:]
    for el in self {
      let key = keyFunc(el)
      if case nil = dict[key]?.append(el) { dict[key] = [el] }
    }
    return dict
  }
}

在你的例子中，你可以让 keyFunc 返回的 "keys" 是名字：

currentStat.statEvents.categorise { [=14=].name }
[  
  dinner: [
    StatEvents(name: "dinner", date: "01-01-2015", hours: 1),
    StatEvents(name: "dinner", date: "01-01-2015", hours: 1),
    StatEvents(name: "dinner", date: "01-01-2015", hours: 1)
  ], lunch: [
    StatEvents(name: "lunch", date: "01-01-2015", hours: 1),
    StatEvents(name: "lunch", date: "01-01-2015", hours: 1)
  ]
]

所以你会得到一个字典，其中每个键都是一个名称，每个值都是具有该名称的 StatEvents 的数组。

Swift 1

func categorise<S : SequenceType, U : Hashable>(seq: S, @noescape keyFunc: S.Generator.Element -> U) -> [U:[S.Generator.Element]] {
  var dict: [U:[S.Generator.Element]] = [:]
  for el in seq {
    let key = keyFunc(el)
    dict[key] = (dict[key] ?? []) + [el]
  }
  return dict
}

categorise(currentStat.statEvents) { [=15=].name }

给出输出：

extension StatEvents : Printable {
  var description: String {
    return "\(self.name): \(self.date)"
  }
}
print(categorise(currentStat.statEvents) { [=16=].name })
[
  dinner: [
    dinner: 01-01-2015,
    dinner: 01-01-2015,
    dinner: 01-01-2015
  ], lunch: [
    lunch: 01-01-2015,
    lunch: 01-01-2015
  ]
]

(swiftstub 是 here)

Answer 2

从中摘下一片叶子。将解决方案扩展到基于 class 名称 Demo & Source Code link.

的对象分组

根据 Class 姓名分组的代码段：

 func categorise<S : SequenceType>(seq: S) -> [String:[S.Generator.Element]] {
    var dict: [String:[S.Generator.Element]] = [:]
    for el in seq {
        //Assigning Class Name as Key
        let key = String(el).componentsSeparatedByString(".").last!
        //Generating a dictionary based on key-- Class Names
        dict[key] = (dict[key] ?? []) + [el]
    }
    return dict
}
//Grouping the Objects in Array using categorise
let categorised = categorise(currentStat)
print("Grouped Array :: \(categorised)")

//Key from the Array i.e, 0 here is Statt class type
let key_Statt:String = String(currentStat.objectAtIndex(0)).componentsSeparatedByString(".").last!
print("Search Key :: \(key_Statt)")

//Accessing Grouped Object using above class type key
let arr_Statt = categorised[key_Statt]
print("Array Retrieved:: ",arr_Statt)
print("Full Dump of Array::")
dump(arr_Statt)

Answer 3

对于Swift 3:

public extension Sequence {
    func categorise<U : Hashable>(_ key: (Iterator.Element) -> U) -> [U:[Iterator.Element]] {
        var dict: [U:[Iterator.Element]] = [:]
        for el in self {
            let key = key(el)
            if case nil = dict[key]?.append(el) { dict[key] = [el] }
        }
        return dict
    }
}

用法：

currentStat.statEvents.categorise { [=11=].name }
[  
  dinner: [
    StatEvents(name: "dinner", date: "01-01-2015", hours: 1),
    StatEvents(name: "dinner", date: "01-01-2015", hours: 1),
    StatEvents(name: "dinner", date: "01-01-2015", hours: 1)
  ], lunch: [
    StatEvents(name: "lunch", date: "01-01-2015", hours: 1),
    StatEvents(name: "lunch", date: "01-01-2015", hours: 1)
  ]
]

Answer 4

扩展已接受的答案以允许有序分组：

extension Sequence {
    func group<GroupingType: Hashable>(by key: (Iterator.Element) -> GroupingType) -> [[Iterator.Element]] {
        var groups: [GroupingType: [Iterator.Element]] = [:]
        var groupsOrder: [GroupingType] = []
        forEach { element in
            let key = key(element)
            if case nil = groups[key]?.append(element) {
                groups[key] = [element]
                groupsOrder.append(key)
            }
        }
        return groupsOrder.map { groups[[=10=]]! }
    }
}

然后它将对任何元组起作用：

let a = [(grouping: 10, content: "a"),
         (grouping: 20, content: "b"),
         (grouping: 10, content: "c")]
print(a.group { [=11=].grouping })

以及任何 struct 或 class:

struct GroupInt {
    var grouping: Int
    var content: String
}
let b = [GroupInt(grouping: 10, content: "a"),
         GroupInt(grouping: 20, content: "b"),
         GroupInt(grouping: 10, content: "c")]
print(b.group { [=12=].grouping })

Answer 5

对于 Swift 5，Dictionary 有一个名为 init(grouping:by:) 的初始化方法。 init(grouping:by:) 有以下声明：

init<S>(grouping values: S, by keyForValue: (S.Element) throws -> Key) rethrows where Value == [S.Element], S : Sequence

Creates a new dictionary where the keys are the groupings returned by the given closure and the values are arrays of the elements that returned each specific key.

以下 Playground 代码显示了如何使用 init(grouping:by:) 来解决您的问题：

struct StatEvents: CustomStringConvertible {
    
    let name: String
    let date: String
    let hours: Int
    
    var description: String {
        return "Event: \(name) - \(date) - \(hours)"
    }
    
}

let statEvents = [
    StatEvents(name: "lunch", date: "01-01-2015", hours: 1),
    StatEvents(name: "dinner", date: "01-01-2015", hours: 1),
    StatEvents(name: "lunch", date: "01-01-2015", hours: 1),
    StatEvents(name: "dinner", date: "01-01-2015", hours: 1)
]

let dictionary = Dictionary(grouping: statEvents, by: { (element: StatEvents) in
    return element.name
})
//let dictionary = Dictionary(grouping: statEvents) { [=11=].name } // also works  
//let dictionary = Dictionary(grouping: statEvents, by: \.name) // also works

print(dictionary)
/*
prints:
[
    "dinner": [Event: dinner - 01-01-2015 - 1, Event: dinner - 01-01-2015 - 1],
    "lunch": [Event: lunch - 01-01-2015 - 1, Event: lunch - 01-01-2015 - 1]
]
*/

Answer 6

嘿，如果你需要在分组元素而不是哈希字典时保持顺序，我使用了元组并在分组时保持列表的顺序。

extension Sequence
{
   func zmGroup<U : Hashable>(by: (Element) -> U) -> [(U,[Element])]
   {
       var groupCategorized: [(U,[Element])] = []
       for item in self {
           let groupKey = by(item)
           guard let index = groupCategorized.firstIndex(where: { [=10=].0 == groupKey }) else { groupCategorized.append((groupKey, [item])); continue }
           groupCategorized[index].1.append(item)
       }
       return groupCategorized
   }
}

Answer 7

这是我在使用 Swift 4 KeyPath's 作为组比较器时保持顺序的基于元组的方法：

extension Sequence{

    func group<T:Comparable>(by:KeyPath<Element,T>) -> [(key:T,values:[Element])]{

        return self.reduce([]){(accumulator, element) in

            var accumulator = accumulator
            var result :(key:T,values:[Element]) = accumulator.first(where:{ [=10=].key == element[keyPath:by]}) ?? (key: element[keyPath:by], values:[])
            result.values.append(element)
            if let index = accumulator.index(where: { [=10=].key == element[keyPath: by]}){
                accumulator.remove(at: index)
            }
            accumulator.append(result)

            return accumulator
        }
    }
}

使用示例：

struct Company{
    let name : String
    let type : String
}

struct Employee{
    let name : String
    let surname : String
    let company: Company
}

let employees : [Employee] = [...]
let companies : [Company] = [...]

employees.group(by: \Employee.company.type) // or
employees.group(by: \Employee.surname) // or
companies.group(by: \Company.type)

Answer 8

Swift 4: 你可以使用 init(grouping:by:) from apple developer site

示例：

let students = ["Kofi", "Abena", "Efua", "Kweku", "Akosua"]
let studentsByLetter = Dictionary(grouping: students, by: { [=10=].first! })
// ["E": ["Efua"], "K": ["Kofi", "Kweku"], "A": ["Abena", "Akosua"]]

所以在你的情况下

   let dictionary = Dictionary(grouping: currentStat.statEvents, by:  { [=11=].name! })

Answer 9

在 Swift 4 中，此扩展具有最佳性能并帮助链接您的运算符

extension Sequence {
    func group<U: Hashable>(by key: (Iterator.Element) -> U) -> [U:[Iterator.Element]] {
        return Dictionary.init(grouping: self, by: key)
    }
}

示例：

struct Asset {
    let coin: String
    let amount: Int
}

let assets = [
    Asset(coin: "BTC", amount: 12),
    Asset(coin: "ETH", amount: 15),
    Asset(coin: "BTC", amount: 30),
]
let grouped = assets.group(by: { [=11=].coin })

创建：

[
    "ETH": [
        Asset(coin: "ETH", amount: 15)
    ],
    "BTC": [
        Asset(coin: "BTC", amount: 12),
        Asset(coin: "BTC", amount: 30)
    ]
]

Answer 10

Swift 4

struct Foo {
  let fizz: String
  let buzz: Int
}

let foos: [Foo] = [Foo(fizz: "a", buzz: 1), 
                   Foo(fizz: "b", buzz: 2), 
                   Foo(fizz: "a", buzz: 3),
                  ]
// use foos.lazy.map instead of foos.map to avoid allocating an
// intermediate Array. We assume the Dictionary simply needs the
// mapped values and not an actual Array
let foosByFizz: [String: Foo] = 
    Dictionary(foos.lazy.map({ ([=10=].fizz, [=10=])}, 
               uniquingKeysWith: { (lhs: Foo, rhs: Foo) in
                   // Arbitrary business logic to pick a Foo from
                   // two that have duplicate fizz-es
                   return lhs.buzz > rhs.buzz ? lhs : rhs
               })
// We don't need a uniquing closure for buzz because we know our buzzes are unique
let foosByBuzz: [String: Foo] = 
    Dictionary(uniqueKeysWithValues: foos.lazy.map({ ([=10=].buzz, [=10=])})

Answer 11

Thr Dictionary(grouping: arr)就是这么简单！

 func groupArr(arr: [PendingCamera]) {

    let groupDic = Dictionary(grouping: arr) { (pendingCamera) -> DateComponents in
        print("group arr: \(String(describing: pendingCamera.date))")

        let date = Calendar.current.dateComponents([.day, .year, .month], from: (pendingCamera.date)!)

        return date
    }

    var cams = [[PendingCamera]]()

    groupDic.keys.forEach { (key) in
        print(key)
        let values = groupDic[key]
        print(values ?? "")

        cams.append(values ?? [])
    }
    print(" cams are \(cams)")

    self.groupdArr = cams
}

Answer 12

您也可以按 KeyPath 分组，像这样：

public extension Sequence {
    func group<Key>(by keyPath: KeyPath<Element, Key>) -> [Key: [Element]] where Key: Hashable {
        return Dictionary(grouping: self, by: {
            [=10=][keyPath: keyPath]
        })
    }
}

使用@duan 的加密示例：

struct Asset {
    let coin: String
    let amount: Int
}

let assets = [
    Asset(coin: "BTC", amount: 12),
    Asset(coin: "ETH", amount: 15),
    Asset(coin: "BTC", amount: 30),
]

然后用法如下：

let grouped = assets.group(by: \.coin)

产生相同的结果：

[
    "ETH": [
        Asset(coin: "ETH", amount: 15)
    ],
    "BTC": [
        Asset(coin: "BTC", amount: 12),
        Asset(coin: "BTC", amount: 30)
    ]
]

Answer 13

我的方式

extension Array {
    func group<T: Hashable>(by key: (_ element: Element) -> T) -> [[Element]] {
        var categories: [T: [Element]] = [:]
        for element in self {
            let key = key(element)
            categories[key, default: []].append(element)
        }
      return categories.values.map { [=10=] }
    }
}

Answer 14

你可以使用reduce

let result = currentStat.statEvents.reduce([String:[StatEvents]](), {
    var previous = [=10=]
    previous[.name] = (previous[.name] ?? []) + []
    return previous
})

let filteredArray1 = result["lunch"]
let filteredArray2 = result["dinner"]

或者您可以将其更改为

let result = currentStat.statEvents.reduce(["lunch":[StatEvents](), "dinner":[StatEvents]()], {
    var previous = [=11=]
    if let array = previous[.name] {
        previous[.name] = array + []
    }
    return previous
})

let filteredArray1 = result["lunch"]
let filteredArray2 = result["dinner"]

仅限于寻找午餐和晚餐

Answer 15

基于this，我这样做了：

func filterItems() -> Dictionary<Int, [YourDataType]> {
        return Dictionary(grouping: yourDataTypeVar, by: { (element: YourDataType) in
            return item.name
        })
    }

如何按 Swift 中的数组元素分组

How to group by the elements of an array in Swift

arrays

nsarray

ios

swift

Swift 4:

Swift 3:

Swift 2:

Swift 1