如何在Swift中按数组元素分组

Let's say that I have this code:

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] = []

I could call as many times manually the next function in order to have 2 arrays grouped by "same name".

我可以手动调用下一个函数，以使两个数组按“相同名称”分组。

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

The problem is that I won't know the variable value, in this case "dinner" and "lunch", so I would like to group this array of statEvents automatically by name, so I get as many arrays as the name gets different.

问题是我不知道变量值，在这个例子中是“dinner”和“lunch”，所以我想用name来自动分组这一系列的statEvents，所以我得到的数组和名字的不同。

How could I do that?

我怎么做呢?

10 个解决方案

#1

Swift 3:

斯威夫特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
    }
}

Unfortunately, the append function above copies the underlying array, instead of mutating it in place, which would be preferable. This causes a pretty big slowdown. You can get around the problem by using a reference type wrapper:

不幸的是，上面的append函数复制了底层数组，而不是在适当的地方对其进行突变，这是更好的选择。这导致了相当大的放缓。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
  }
}

Even though you traverse the final dictionary twice, this version is still faster than the original in most cases.

即使您两次遍历最终的字典，这个版本在大多数情况下仍然比原始版本快。

Swift 2:

斯威夫特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
  }
}

In your case, you could have the "keys" returned by keyFunc be the names:

在您的情况下，您可以将keyFunc返回的“key”作为名称:

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)
  ]
]

So you'll get a dictionary, where every key is a name, and every value is an array of the StatEvents with that name.

你会得到一个字典，每个键都是一个名字，每个值都是一个有这个名字的StatEvents数组。

The Swift 1 version would be:

Swift 1版本将是:

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 }

Which gives the output:

使输出:

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
  ]
]

(The swiftstub is here)

(swiftstub在这里)

#2

With Swift 4, Dictionary has an initializer method called init(grouping:by:). init(grouping:by:) has the following declaration:

对于Swift 4，字典有一个初始化方法称为init(分组:by:)。init(分组: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.

创建一个新字典，其中键是给定闭包返回的分组，值是返回每个特定键的元素的数组。

---

The following Playground code shows how to use init(grouping:by:) in order to solve your problem:

下面的操场代码展示了如何使用init(分组: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 predicate = { (element: StatEvents) in
    return element.name
}

let dictionary = Dictionary(grouping: statEvents, by: predicate)
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]
]
*/

#3

For Swift 3:

为迅速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
    }
}

Usage:

用法:

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)
  ]
]

#4

Swift 4: you can use init(grouping:by:) from apple developer site

Swift 4:您可以使用来自apple developer站点的init(分组:by:)

Example:

例子:

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

So in your case

所以在你的情况下

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

#5

In Swift4, this extension has the best performance and help chain your operations

在Swift4中，这个扩展具有最好的性能，并有助于将操作链连接起来

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

Example:

例子:

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 })

creates:

创建:

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

#6

Swift 4

斯威夫特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)})

#7

Extending on accepted answer to allow ordered grouping:

扩展已接受的答案，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[$0]! }
    }
}

Then it will work on any tuple:

然后它将在任何元组上工作:

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

As well as any struct or class:

以及任何结构体或类:

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 })

#8

Here is my tuple based approach for keeping order while using Swift 4 KeyPath's as group comparator:

下面是我使用Swift 4 KeyPath作为组比较器的基于tuple的排序方法:

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:{ $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
        }
    }
}

Example of how to use it:

如何使用它的例子:

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)

#9

Hey if you need to keep order while grouping elements instead of hash dictionary i have used tuples and kept the order of the list while grouping.

如果您需要在分组元素而不是散列字典时保持顺序，我使用了元组并在分组时保留列表的顺序。

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

#10

Taking a leaf out of "oisdk" example. Extending the solution to group objects based on class name Demo & Source Code link.

借鉴oisdk的例子。基于类名Demo和源代码链接将解决方案扩展到组对象。

Code snippet for grouping based on Class Name:

基于类名分组的代码片段:

 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)