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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)})