问题描述
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] = []
我可以多次手动调用下一个函数,以便将 2 个数组按同名"分组.
I could call as many times manually the next function in order to have 2 arrays grouped by "same name".
filteredArray1 = currentStat.statEvents.filter({$0.name == "dinner"})
filteredArray2 = currentStat.statEvents.filter({$0.name == "lunch"})
问题是我不知道变量值,在这种情况下是晚餐"和午餐",所以我想按名称自动对这个statEvents数组进行分组,所以我得到的数组与名称一样多变得不一样了.
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.
我怎么能这样做?
推荐答案
Swift 4:
自 Swift 4 起,此功能已添加到标准库中.你可以这样使用它:
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)
]
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
}
}
不幸的是,上面的 append
函数复制了底层数组,而不是就地改变它,这会更好.这会导致相当大的放缓.您可以通过使用引用类型包装器来解决这个问题:
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:
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
}
}
即使你两次遍历最终字典,这个版本在大多数情况下仍然比原始版本快.
Even though you traverse the final dictionary twice, this version is still faster than the original in most cases.
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
}
}
在您的情况下,您可以将 keyFunc
返回的键"作为名称:
In your case, you could have the "keys" returned by keyFunc
be the names:
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)
]
]
所以你会得到一个字典,其中每个键都是一个名称,每个值都是具有该名称的 StatEvents 数组.
So you'll get a dictionary, where every key is a name, and every value is an array of the StatEvents with that name.
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 }
给出输出:
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
]
]
(swiftstub 在这里)
(The swiftstub is here)
这篇关于如何在 Swift 中按数组的元素分组的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!