我有一个对象数组,每个对象都有一个date
属性。
我想创建一个新的数组,该数组包含多个子数组,每个子数组都包含具有相同日期的对象。
我知道需要使用.filter
函数过滤对象,然后使用.map
函数将对象添加到数组中。
但是如何告诉.map
为每个分组创建一个单独的数组,并将此数组添加到"全局"数组中?同时如何告诉.filter
只选择具有相同日期的对象?
我有一个对象数组,每个对象都有一个date
属性。
我想创建一个新的数组,该数组包含多个子数组,每个子数组都包含具有相同日期的对象。
我知道需要使用.filter
函数过滤对象,然后使用.map
函数将对象添加到数组中。
但是如何告诉.map
为每个分组创建一个单独的数组,并将此数组添加到"全局"数组中?同时如何告诉.filter
只选择具有相同日期的对象?
可能有点晚了,但新的Xcode 9 SDK字典有新的初始化方法。
init<S>(grouping values: S, by keyForValue: (S.Element) throws -> Key) rethrows where Value == [S.Element], S : Sequence
文档中提供了这个方法的简单示例。下面是我贴出的示例:
let students = ["Kofi", "Abena", "Efua", "Kweku", "Akosua"]
let studentsByLetter = Dictionary(grouping: students, by: { $0.first! })
结果将是:
["E": ["Efua"], "K": ["Kofi", "Kweku"], "A": ["Abena", "Akosua"]]
改进东方解决方案,以允许对任何内容进行有序分组:
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]! }
}
}
然后它将适用于任何元组、结构体或类,并适用于任何属性:
let a = [(grouping: 10, content: "a"),
(grouping: 20, content: "b"),
(grouping: 10, content: "c")]
print(a.group { $0.grouping })
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 })
Dictionary
的init(grouping:by:)
初始化器将一个数组的元素按照其属性之一分组成字典。完成后,你可以使用Dictionary
的values
属性和Array
的init(_:)
初始化器,创建一个由数组组成的新数组。
import Foundation
struct Purchase: CustomStringConvertible {
let id: Int
let date: Date
var description: String {
return "Purchase #\(id) (\(date))"
}
}
let date1 = Calendar.current.date(from: DateComponents(year: 2010, month: 11, day: 22))!
let date2 = Calendar.current.date(from: DateComponents(year: 2015, month: 5, day: 1))!
let date3 = Calendar.current.date(from: DateComponents(year: 2012, month: 8, day: 15))!
let purchases = [
Purchase(id: 1, date: date1),
Purchase(id: 2, date: date1),
Purchase(id: 3, date: date2),
Purchase(id: 4, date: date3),
Purchase(id: 5, date: date3)
]
let groupingDictionary = Dictionary(grouping: purchases, by: { $0.date })
print(groupingDictionary)
/*
[
2012-08-14 22:00:00 +0000: [Purchase #4 (2012-08-14 22:00:00 +0000), Purchase #5 (2012-08-14 22:00:00 +0000)],
2010-11-21 23:00:00 +0000: [Purchase #1 (2010-11-21 23:00:00 +0000), Purchase #2 (2010-11-21 23:00:00 +0000)],
2015-04-30 22:00:00 +0000: [Purchase #3 (2015-04-30 22:00:00 +0000)]
]
*/
let groupingArray = Array(groupingDictionary.values)
print(groupingArray)
/*
[
[Purchase #3 (2015-04-30 22:00:00 +0000)],
[Purchase #4 (2012-08-14 22:00:00 +0000), Purchase #5 (2012-08-14 22:00:00 +0000)],
[Purchase #1 (2010-11-21 23:00:00 +0000), Purchase #2 (2010-11-21 23:00:00 +0000)]
]
*/
抽象化一步,您想要按某个属性对数组元素进行分组。您可以让一个映射来为您进行分组,如下所示:
protocol Groupable {
associatedtype GroupingType: Hashable
var grouping: GroupingType { get set }
}
extension Array where Element: Groupable {
typealias GroupingType = Element.GroupingType
func grouped() -> [[Element]] {
var groups = [GroupingType: [Element]]()
for element in self {
if let _ = groups[element.grouping] {
groups[element.grouping]!.append(element)
} else {
groups[element.grouping] = [element]
}
}
return Array<[Element]>(groups.values)
}
}
我将使用整数来举例说明;对于任何(可哈希)类型的T
,包括Date
,都应该很清楚如何使用。
struct GroupInt: Groupable {
typealias GroupingType = Int
var grouping: Int
var content: String
}
var a = [GroupInt(grouping: 1, content: "a"),
GroupInt(grouping: 2, content: "b") ,
GroupInt(grouping: 1, content: "c")]
print(a.grouped())
// > [[GroupInt(grouping: 2, content: "b")], [GroupInt(grouping: 1, content: "a"), GroupInt(grouping: 1, content: "c")]]
Rapheal的解决方案确实有效。然而,我建议修改解决方案以支持分组实际上是稳定的这一声明。
目前情况下,调用grouped()
将返回一个已分组的数组,但是后续的调用可能会返回一个顺序不同的组数组,尽管每个组的元素都按预期顺序排列。
internal protocol Groupable {
associatedtype GroupingType : Hashable
var groupingKey : GroupingType? { get }
}
extension Array where Element : Groupable {
typealias GroupingType = Element.GroupingType
func grouped(nilsAsSingleGroup: Bool = false) -> [[Element]] {
var groups = [Int : [Element]]()
var groupsOrder = [Int]()
let nilGroupingKey = UUID().uuidString.hashValue
var nilGroup = [Element]()
for element in self {
// If it has a grouping key then use it. Otherwise, conditionally make one based on if nils get put in the same bucket or not
var groupingKey = element.groupingKey?.hashValue ?? UUID().uuidString.hashValue
if nilsAsSingleGroup, element.groupingKey == nil { groupingKey = nilGroupingKey }
// Group nils together
if nilsAsSingleGroup, element.groupingKey == nil {
nilGroup.append(element)
continue
}
// Place the element in the right bucket
if let _ = groups[groupingKey] {
groups[groupingKey]!.append(element)
} else {
// New key, track it
groups[groupingKey] = [element]
groupsOrder.append(groupingKey)
}
}
// Build our array of arrays from the dictionary of buckets
var grouped = groupsOrder.flatMap{ groups[$0] }
if nilsAsSingleGroup, !nilGroup.isEmpty { grouped.append(nilGroup) }
return grouped
}
}
现在我们追踪发现新分组的顺序,因此可以更加一致地返回一个分组数组,而不仅仅依靠字典的无序 values
属性。
struct GroupableInt: Groupable {
typealias GroupingType = Int
var grouping: Int?
var content: String
}
var a = [GroupableInt(groupingKey: 1, value: "test1"),
GroupableInt(groupingKey: 2, value: "test2"),
GroupableInt(groupingKey: 2, value: "test3"),
GroupableInt(groupingKey: nil, value: "test4"),
GroupableInt(groupingKey: 3, value: "test5"),
GroupableInt(groupingKey: 3, value: "test6"),
GroupableInt(groupingKey: nil, value: "test7")]
print(a.grouped())
// > [[GroupableInt(groupingKey: 1, value: "test1")], [GroupableInt(groupingKey: 2, value: "test2"),GroupableInt(groupingKey: 2, value: "test3")], [GroupableInt(groupingKey: nil, value: "test4")],[GroupableInt(groupingKey: 3, value: "test5"),GroupableInt(groupingKey: 3, value: "test6")],[GroupableInt(groupingKey: nil, value: "test7")]]
print(a.grouped(nilsAsSingleGroup: true))
// > [[GroupableInt(groupingKey: 1, value: "test1")], [GroupableInt(groupingKey: 2, value: "test2"),GroupableInt(groupingKey: 2, value: "test3")], [GroupableInt(groupingKey: nil, value: "test4"),GroupableInt(groupingKey: nil, value: "test7")],[GroupableInt(groupingKey: 3, value: "test5"),GroupableInt(groupingKey: 3, value: "test6")]]
init<S>(grouping values: S, by keyForValue: (S.Element) throws -> Key) rethrows where Value == [S.Element], S : Sequence
enum Parity {
case even, odd
init(_ value: Int) {
self = value % 2 == 0 ? .even : .odd
}
}
let parity = Dictionary(grouping: 0 ..< 10 , by: Parity.init )
let parity2 = Dictionary(grouping: 0 ..< 10) { $0 % 2 }
struct Person : CustomStringConvertible {
let dateOfBirth : Date
let name :String
var description: String {
return "\(name)"
}
}
extension Date {
init(dateString:String) {
let formatter = DateFormatter()
formatter.timeZone = NSTimeZone.default
formatter.dateFormat = "MM/dd/yyyy"
self = formatter.date(from: dateString)!
}
}
let people = [Person(dateOfBirth:Date(dateString:"01/01/2017"),name:"Foo"),
Person(dateOfBirth:Date(dateString:"01/01/2017"),name:"Bar"),
Person(dateOfBirth:Date(dateString:"02/01/2017"),name:"FooBar")]
let parityFields = Dictionary(grouping: people) {$0.dateOfBirth}
输出:
[2017-01-01: [Foo, Bar], 2017-02-01: [FooBar] ]
这是一种清晰的分组方法:
let grouped = allRows.group(by: {$0.groupId}) // Dictionary with the key groupId
class ContactPerson {
var groupId:String?
var name:String?
var contactRecords:[PhoneBookEntry] = []
}
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
}
}