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Cómo agrupar por los elementos de una matriz en Swift

Question

Digamos que tengo este código:

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] = []

Podría llamar tantas veces manualmente la siguiente función para tener 2 matrices agrupadas por "mismo nombre".

filteredArray1 = currentStat.statEvents.filter({$0.name == "dinner"})
filteredArray2 = currentStat.statEvents.filter({$0.name == "lunch"})

El problema es que no sabré el valor de la variable, en este caso "cena" y "almuerzo", por lo que me gustaría agrupar esta matriz de statEvents automáticamente por nombre, por lo que obtengo tantas matrices como el nombre es diferente.

¿Cómo podría hacer eso?

Answer 1

Swift 4:

Desde Swift 4, esta funcionalidad se ha agregado a la biblioteca estándar . Puedes usarlo así:

Dictionary(grouping: statEvents, by: { $0.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
    }
}

Desafortunadamente, la función append anterior copia la matriz subyacente, en lugar de mutarla en su lugar, lo que sería preferible. Esto causa una desaceleración bastante grande . Puede solucionar el problema utilizando un contenedor de tipo de referencia:

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
  }
}

Aunque atraviese el diccionario final dos veces, esta versión sigue siendo más rápida que la original en la mayoría de los casos.

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
  }
}

En su caso, podría hacer que las "claves" devueltas por keyFunc sean los nombres:

currentStat.statEvents.categorise { $0.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)
  ]
]

Entonces obtendrá un diccionario, donde cada clave es un nombre, y cada valor es una matriz de los eventos de estadísticas con ese nombre.

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) { $0.name }

Lo que da la salida:

extension StatEvents : Printable {
  var description: String {
    return "/(self.name): /(self.date)"
  }
}
print(categorise(currentStat.statEvents) { $0.name })
[
  dinner: [
    dinner: 01-01-2015,
    dinner: 01-01-2015,
    dinner: 01-01-2015
  ], lunch: [
    lunch: 01-01-2015,
    lunch: 01-01-2015
  ]
]

(El swiftstub está here )

Answer 2

¡Thr Dictionary (agrupación: arr) es tan fácil!

 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 3

Aquí está mi enfoque basado en tuplas para mantener el orden mientras uso Swift 4 KeyPath''s como comparador de grupo:

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:{ $0.key == element[keyPath:by]}) ?? (key: element[keyPath:by], values:[])
            result.values.append(element)
            if let index = accumulator.index(where: { $0.key == element[keyPath: by]}){
                accumulator.remove(at: index)
            }
            accumulator.append(result)

            return accumulator
        }
    }
}

Ejemplo de cómo usarlo:

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 4

Con Swift 5, Dictionary tiene un método inicializador llamado init(grouping:by:) . init(grouping:by:) tiene la siguiente declaración:

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

Crea un nuevo diccionario donde las claves son las agrupaciones devueltas por el cierre dado y los valores son matrices de los elementos que devolvieron cada clave específica.

El siguiente código de Playground muestra cómo usar init(grouping:by:) para resolver su problema:

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) { $0.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 5

En Swift 4, esta extensión tiene el mejor rendimiento y ayuda a encadenar a sus operadores

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

Ejemplo:

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: { $0.coin })

crea:

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

Answer 6

Extender sobre la respuesta aceptada para permitir la agrupación ordenada :

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[$0]! }
    }
}

Entonces funcionará en cualquier tupla :

let a = [(grouping: 10, content: "a"),
         (grouping: 20, content: "b"),
         (grouping: 10, content: "c")]
print(a.group { $0.grouping })

Además de cualquier estructura o clase :

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 { $0.grouping })

Answer 7

Hola, si necesita mantener el orden al agrupar elementos en lugar del diccionario hash, he usado tuplas y he mantenido el orden de la lista durante la agrupación.

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.index(where: { $0.0 == groupKey }) else { groupCategorized.append((groupKey, [item])); continue }
           groupCategorized[index].1.append(item)
       }
       return groupCategorized
   }
}

Answer 8

Para 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
    }
}

Uso:

currentStat.statEvents.categorise { $0.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 9

Sacando una hoja del ejemplo "oisdk" . Extender la solución para agrupar objetos en función del nombre de la clase Enlace de demostración y código fuente .

Fragmento de código para agrupar basado en el Nombre de clase:

 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 10

Swift 4: puede usar init (agrupación: por :) del sitio de desarrolladores de Apple

Ejemplo :

let students = ["Kofi", "Abena", "Efua", "Kweku", "Akosua"]
let studentsByLetter = Dictionary(grouping: students, by: { $0.first! })
// ["E": ["Efua"], "K": ["Kofi", "Kweku"], "A": ["Abena", "Akosua"]]

Entonces en tu caso

   let dictionary = Dictionary(grouping: currentStat.statEvents, by:  { $0.name! })

Answer 11

También puede agrupar por KeyPath , así:

public extension Sequence {
    func group<Key>(by keyPath: KeyPath<Element, Key>) -> [Key: [Element]] where Key: Hashable {
        return Dictionary(grouping: self, by: {
            $0[keyPath: keyPath]
        })
    }
}

Usando el ejemplo de cifrado de @ duan:

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),
]

Entonces el uso se ve así:

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

Produciendo el mismo resultado:

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

Answer 12

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({ ($0.fizz, $0)}, 
               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({ ($0.buzz, $0)})