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按两个值对数组排序Swift 4[重复]

swift arrays

RileyDev · 技术社区 · 6 年前

Contact 物体:

var contacts:[Contact] = [Contact]()

联系人类别:

Class Contact:NSOBject {
    var firstName:String!
    var lastName:String!
}

lastName 然后 firstName 以防有些联系人也这样 .

我可以按其中一个标准来排序,但不能同时按这两个标准来排序。

contacts.sortInPlace({$0.lastName < $1.lastName})

如何添加更多条件来排序此数组?

0 回复 | 直到 6 年前

Alexander 6 年前

想想“按多个标准排序”意味着什么。这意味着首先用一个标准比较两个对象。然后,如果这些条件相同,则领带将被下一个条件打断,以此类推,直到获得所需的顺序。

let sortedContacts = contacts.sort {
    if $0.lastName != $1.lastName { // first, compare by last names
        return $0.lastName < $1.lastName
    }
    /*  last names are the same, break ties by foo
    else if $0.foo != $1.foo {
        return $0.foo < $1.foo
    }
    ... repeat for all other fields in the sorting
    */
    else { // All other fields are tied, break ties by last name
        return $0.firstName < $1.firstName
    }
}

你现在看到的是 Sequence.sorted(by:) method

如果您的排序将在许多地方使用,最好使您的类型符合 Comparable protocol Sequence.sorted() method ,它将参考您对 Comparable.<(_:_:) operator 以确定元素的比较方式。这样,你可以把任何 Sequence 属于 Contact 无需复制排序代码。

Hamish 6 年前

按多个条件(即按一个比较排序,如果相等,则按另一个比较排序)执行排序的一种非常简单的方法是使用元组 ,作为 < 和 > 操作员对执行字典比较的重载。

/// Returns a Boolean value indicating whether the first tuple is ordered
/// before the second in a lexicographical ordering.
///
/// Given two tuples `(a1, a2, ..., aN)` and `(b1, b2, ..., bN)`, the first
/// tuple is before the second tuple if and only if
/// `a1 < b1` or (`a1 == b1` and
/// `(a2, ..., aN) < (b2, ..., bN)`).
public func < <A : Comparable, B : Comparable>(lhs: (A, B), rhs: (A, B)) -> Bool

例如:

struct Contact {
  var firstName: String
  var lastName: String
}

var contacts = [
  Contact(firstName: "Leonard", lastName: "Charleson"),
  Contact(firstName: "Michael", lastName: "Webb"),
  Contact(firstName: "Charles", lastName: "Alexson"),
  Contact(firstName: "Michael", lastName: "Elexson"),
  Contact(firstName: "Alex", lastName: "Elexson"),
]

contacts.sort {
  ($0.lastName, $0.firstName) <
    ($1.lastName, $1.firstName)
}

print(contacts)

// [
//   Contact(firstName: "Charles", lastName: "Alexson"),
//   Contact(firstName: "Leonard", lastName: "Charleson"),
//   Contact(firstName: "Alex", lastName: "Elexson"),
//   Contact(firstName: "Michael", lastName: "Elexson"),
//   Contact(firstName: "Michael", lastName: "Webb")
// ]

这将比较元素 lastName < 与他们比较。如果他们相等,然后它将移动到元组中的下一对元素,即比较 firstName 财产。

< 和 > 2到6个元素的元组的重载。

contacts.sort {
  ($1.lastName, $0.firstName) <
    ($0.lastName, $1.firstName)
}

// [
//   Contact(firstName: "Michael", lastName: "Webb")
//   Contact(firstName: "Alex", lastName: "Elexson"),
//   Contact(firstName: "Michael", lastName: "Elexson"),
//   Contact(firstName: "Leonard", lastName: "Charleson"),
//   Contact(firstName: "Charles", lastName: "Alexson"),
// ]

现在可以按 姓氏 名字 提升。

`sort(by:)` 接受多个谓词的重载

受到以下讨论的启发 Sorting Collections with map closures and SortDescriptors ,另一个选项是定义 排序(按:) sorted(by:) 它处理多个谓词,其中每个谓词依次被考虑以决定元素的顺序。

extension MutableCollection where Self : RandomAccessCollection {
  mutating func sort(
    by firstPredicate: (Element, Element) -> Bool,
    _ secondPredicate: (Element, Element) -> Bool,
    _ otherPredicates: ((Element, Element) -> Bool)...
  ) {
    sort(by:) { lhs, rhs in
      if firstPredicate(lhs, rhs) { return true }
      if firstPredicate(rhs, lhs) { return false }
      if secondPredicate(lhs, rhs) { return true }
      if secondPredicate(rhs, lhs) { return false }
      for predicate in otherPredicates {
        if predicate(lhs, rhs) { return true }
        if predicate(rhs, lhs) { return false }
      }
      return false
    }
  }
}

extension Sequence { mutating func sorted( by firstPredicate: (Element, Element) -> Bool, _ secondPredicate: (Element, Element) -> Bool, _ otherPredicates: ((Element, Element) -> Bool)... ) -> [Element] { return sorted(by:) { lhs, rhs in if firstPredicate(lhs, rhs) { return true } if firstPredicate(rhs, lhs) { return false } if secondPredicate(lhs, rhs) { return true } if secondPredicate(rhs, lhs) { return false } for predicate in otherPredicates { if predicate(lhs, rhs) { return true } if predicate(rhs, lhs) { return false } } return false } } }

^{(在

secondPredicate:

排序(按:)}

这样我们就可以说(使用 contacts 以前的数组):

contacts.sort(by: { $0.lastName > $1.lastName }, // first sort by lastName descending { $0.firstName < $1.firstName } // ... then firstName ascending // ... ) print(contacts) // [ // Contact(firstName: "Michael", lastName: "Webb") // Contact(firstName: "Alex", lastName: "Elexson"), // Contact(firstName: "Michael", lastName: "Elexson"), // Contact(firstName: "Leonard", lastName: "Charleson"), // Contact(firstName: "Charles", lastName: "Alexson"), // ] // or with sorted(by:)... let sortedContacts = contacts.sorted(by: { $0.lastName > $1.lastName }, // first sort by lastName descending { $0.firstName < $1.firstName } // ... then firstName ascending // ... )

符合 Comparable

如果你要定期做这些比较 @AMomchilov @appzYourLife 建议,你可以遵从 Contact 到 可比

extension Contact : Comparable { static func == (lhs: Contact, rhs: Contact) -> Bool { return (lhs.firstName, lhs.lastName) == (rhs.firstName, rhs.lastName) } static func < (lhs: Contact, rhs: Contact) -> Bool { return (lhs.lastName, lhs.firstName) < (rhs.lastName, rhs.firstName) } }

sort() 对于升序:

contacts.sort()

或 sort(by: >)

contacts.sort(by: >)

在嵌套类型中定义自定义排序顺序

如果要使用其他排序顺序,可以在嵌套类型中定义它们:

extension Contact { enum Comparison { static let firstLastAscending: (Contact, Contact) -> Bool = { return ($0.firstName, $0.lastName) < ($1.firstName, $1.lastName) } } }

contacts.sort(by: Contact.Comparison.firstLastAscending)

oyalhi 7 年前

下面显示了另一种使用两个标准进行排序的简单方法。

检查第一个字段,在本例中是 lastName 姓氏 姓氏 的值相等,然后按第二个字段排序,在本例中 firstName .

contacts.sort { $0.lastName == $1.lastName ? $0.firstName < $1.firstName : $0.lastName < $1.lastName  }

Jaime Allauca 8 年前

我在Swift 3中创建了一篇关于如何做到这一点的博客文章,并使代码简单易读。

你可以在这里找到它:

http://master-method.com/index.php/2016/11/23/sort-a-sequence-i-e-arrays-of-objects-by-multiple-properties-in-swift-3/

您还可以在这里找到带有以下代码的GitHub存储库:

https://github.com/jallauca/SortByMultipleFieldsSwift.playground

所有这些的要点,比如说,如果你有位置列表,你就可以做到:

struct Location {
    var city: String
    var county: String
    var state: String
}

var locations: [Location] {
    return [
        Location(city: "Dania Beach", county: "Broward", state: "Florida"),
        Location(city: "Fort Lauderdale", county: "Broward", state: "Florida"),
        Location(city: "Hallandale Beach", county: "Broward", state: "Florida"),
        Location(city: "Delray Beach", county: "Palm Beach", state: "Florida"),
        Location(city: "West Palm Beach", county: "Palm Beach", state: "Florida"),
        Location(city: "Savannah", county: "Chatham", state: "Georgia"),
        Location(city: "Richmond Hill", county: "Bryan", state: "Georgia"),
        Location(city: "St. Marys", county: "Camden", state: "Georgia"),
        Location(city: "Kingsland", county: "Camden", state: "Georgia"),
    ]
}

let sortedLocations =
    locations
        .sorted(by:
            ComparisonResult.flip <<< Location.stateCompare,
            Location.countyCompare,
            Location.cityCompare
        )

XueYu 7 年前

Sort Descriptors in Swift . 我们有几种方法来进行多标准排序。

class Person: NSObject {
    var first: String
    var last: String
    var yearOfBirth: Int
    init(first: String, last: String, yearOfBirth: Int) {
        self.first = first
        self.last = last
        self.yearOfBirth = yearOfBirth
    }

    override var description: String {
        get {
            return "\(self.last) \(self.first) (\(self.yearOfBirth))"
        }
    }
}

let people = [
    Person(first: "Jo", last: "Smith", yearOfBirth: 1970),
    Person(first: "Joe", last: "Smith", yearOfBirth: 1970),
    Person(first: "Joe", last: "Smyth", yearOfBirth: 1970),
    Person(first: "Joanne", last: "smith", yearOfBirth: 1985),
    Person(first: "Joanne", last: "smith", yearOfBirth: 1970),
    Person(first: "Robert", last: "Jones", yearOfBirth: 1970),
]

例如,在这里,我们要按姓氏排序,然后按姓氏排序,最后按出生年份排序。我们希望在不敏感的情况下使用用户区域设置。

let lastDescriptor = NSSortDescriptor(key: "last", ascending: true,
  selector: #selector(NSString.localizedCaseInsensitiveCompare(_:)))
let firstDescriptor = NSSortDescriptor(key: "first", ascending: true, 
  selector: #selector(NSString.localizedCaseInsensitiveCompare(_:)))
let yearDescriptor = NSSortDescriptor(key: "yearOfBirth", ascending: true)



(people as NSArray).sortedArray(using: [lastDescriptor, firstDescriptor, yearDescriptor]) 
// [Robert Jones (1970), Jo Smith (1970), Joanne smith (1970), Joanne smith (1985), Joe Smith (1970), Joe Smyth (1970)]

使用快速的姓/名排序方式。

let sortedPeople = people.sorted { p0, p1 in
    let left =  [p0.last, p0.first]
    let right = [p1.last, p1.first]

    return left.lexicographicallyPrecedes(right) {
        $0.localizedCaseInsensitiveCompare($1) == .orderedAscending
    }
}
sortedPeople // [Robert Jones (1970), Jo Smith (1970), Joanne smith (1985), Joanne smith (1970), Joe Smith (1970), Joe Smyth (1970)]

快速进入inmitate NSSortDescriptor。这使用了“函数是一类类型”的概念。SortDescriptor是函数类型,接受两个值,返回bool。假设sortByFirstName,我们取两个参数($0,$1)并比较它们的名字。combine函数使用一组sortdescriptor,比较所有这些函数并发出命令。

typealias SortDescriptor<Value> = (Value, Value) -> Bool

let sortByFirstName: SortDescriptor<Person> = {
    $0.first.localizedCaseInsensitiveCompare($1.first) == .orderedAscending
}
let sortByYear: SortDescriptor<Person> = { $0.yearOfBirth < $1.yearOfBirth }
let sortByLastName: SortDescriptor<Person> = {
    $0.last.localizedCaseInsensitiveCompare($1.last) == .orderedAscending
}

func combine<Value>
    (sortDescriptors: [SortDescriptor<Value>]) -> SortDescriptor<Value> {
    return { lhs, rhs in
        for isOrderedBefore in sortDescriptors {
            if isOrderedBefore(lhs,rhs) { return true }
            if isOrderedBefore(rhs,lhs) { return false }
        }
        return false
    }
}

let combined: SortDescriptor<Person> = combine(
    sortDescriptors: [sortByLastName,sortByFirstName,sortByYear]
)
people.sorted(by: combined)
// [Robert Jones (1970), Jo Smith (1970), Joanne smith (1970), Joanne smith (1985), Joe Smith (1970), Joe Smyth (1970)]

这很好,因为您可以将它与struct和class一起使用,甚至可以扩展它以与nils进行比较。

original article 强烈建议。它有更多的细节和很好的解释。

Community ahmed 7 年前

我建议你用 Hamish's tuple solution

如果你想要像这样的东西 if statements 但是简化了分支逻辑,您可以使用此解决方案,它允许您执行以下操作:

animals.sort {
  return comparisons(
    compare($0.family, $1.family, ascending: false),
    compare($0.name, $1.name))
}

func compare<C: Comparable>(_ value1Closure: @autoclosure @escaping () -> C, _ value2Closure: @autoclosure @escaping () -> C, ascending: Bool = true) -> () -> ComparisonResult {
  return {
    let value1 = value1Closure()
    let value2 = value2Closure()
    if value1 == value2 {
      return .orderedSame
    } else if ascending {
      return value1 < value2 ? .orderedAscending : .orderedDescending
    } else {
      return value1 > value2 ? .orderedAscending : .orderedDescending
    }
  }
}

func comparisons(_ comparisons: (() -> ComparisonResult)...) -> Bool {
  for comparison in comparisons {
    switch comparison() {
    case .orderedSame:
      continue // go on to the next property
    case .orderedAscending:
      return true
    case .orderedDescending:
      return false
    }
  }
  return false // all of them were equal
}

如果要测试它,可以使用以下额外代码:

enum Family: Int, Comparable {
  case bird
  case cat
  case dog

  var short: String {
    switch self {
    case .bird: return "B"
    case .cat: return "C"
    case .dog: return "D"
    }
  }

  public static func <(lhs: Family, rhs: Family) -> Bool {
    return lhs.rawValue < rhs.rawValue
  }
}

struct Animal: CustomDebugStringConvertible {
  let name: String
  let family: Family

  public var debugDescription: String {
    return "\(name) (\(family.short))"
  }
}

let animals = [
  Animal(name: "Leopard", family: .cat),
  Animal(name: "Wolf", family: .dog),
  Animal(name: "Tiger", family: .cat),
  Animal(name: "Eagle", family: .bird),
  Animal(name: "Cheetah", family: .cat),
  Animal(name: "Hawk", family: .bird),
  Animal(name: "Puma", family: .cat),
  Animal(name: "Dalmatian", family: .dog),
  Animal(name: "Lion", family: .cat),
]

主要区别在于 Jamie's solution $0.family 而不是 Animal.familyCompare . 升序/降序由参数而不是重载运算符控制。Jamie的解决方案在数组上添加了一个扩展,而我的解决方案使用内置的 sort sorted 方法,但需要定义另外两个方法: compare comparisons .

为了完整起见,下面是我的解决方案与 . 为了证明这一点,我将使用一个疯狂的例子 (name, address, profileViews) 在比较开始之前,Hamish的解将精确地计算6个属性值中的每一个。这可能不需要,也可能不需要。例如,假设 profileViews 除非绝对必要。我的解决方案将避免评估 纵断面图 $0.name == $1.name 和 $0.address == $1.address 纵断面图 它可能会被多次评估。

Lou Zell 8 年前

contacts.sort() { [$0.last, $0.first].lexicographicalCompare([$1.last, $1.first]) }

-1

pkamb santoshIOS 6 年前

在Swift 3中对我的数组[字符串]起作用,在Swift 4中似乎没问题

array = array.sorted{$0.compare($1, options: .numeric) == .orderedAscending}

按两个值对数组排序Swift 4[重复]

sort(by:) 接受多个谓词的重载

符合 Comparable

在嵌套类型中定义自定义排序顺序

`sort(by:)` 接受多个谓词的重载

符合 `Comparable`