假设我有这段代码:
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: "02-01-2015", hours: 2),
StatEvents(name: "dinner", date: "03-01-2015", hours: 3),
StatEvents(name: "lunch", date: "04-01-2015", hours: 4),
StatEvents(name: "dinner", date: "05-01-2015", hours: 5),
StatEvents(name: "breakfast", date: "06-01-2015", hours: 6),
StatEvents(name: "lunch", date: "07-01-2015", hours: 7),
StatEvents(name: "breakfast", date: "08-01-2015", hours: 8)
]
我想知道是否有办法获得一个像这样的输出数组:
- [0]
- name : "lunch"
- date
- [0] : "01-01-2015"
- [1] : "04-01-2015"
- [2] : "07-01-2015"
- hours
- [0] : 1
- [1] : 4
- [2] : 7
- [1]
- name : "dinner"
- date
- [0] : "02-01-2015"
- [1] : "03-01-2015"
- [2] : "05-01-2015"
- hours
- [0] : 2
- [1] : 3
- [2] : 5
- [2]
- name : "breakfast"
- date
- [0] : "06-01-2015"
- [1] : "08-01-2015"
- hours
- [0] : 6
- [1] : 8
如您所见,最终数组应按“名称”后代分组。 @oisdk你可以看一下吗?
答案 0 :(得分:2)
这看起来有点矫枉过正,但这是我想到的解决方案。
extension Array {
/**
Indicates whether there are any elements in self that satisfy the predicate.
If no predicate is supplied, indicates whether there are any elements in self.
*/
func any(predicate: T -> Bool = { t in true }) -> Bool {
for element in self {
if predicate(element) {
return true
}
}
return false
}
/**
Takes an equality comparer and returns a new array containing all the distinct elements.
*/
func distinct(comparer: (T, T) -> Bool) -> [T] {
var result = [T]()
for t in self {
// if there are no elements in the result set equal to this element, add it
if !result.any(predicate: { comparer($0, t) }) {
result.append(t)
}
}
return result
}
}
let result = currentStat.statEvents
.map({ $0.name })
.distinct(==)
.sorted(>)
.map({ name in currentStat.statEvents.filter({ $0.name == name }) })
现在您有一个列表列表,其中第一个列表包含晚餐类型的所有statEvents,下一个列表包含午餐类型的事件等。
明显的缺点是,这可能不如其他解决方案那么高效。好的部分是你不必依赖并行数组来获得与特定日期相关的小时数。
答案 1 :(得分:2)
已经有一些答案,但到底是什么,这很有趣。我的回答并没有在Swift中使用很多高阶函数,但它完成了工作:
// Get the list of unique event names
var eventNames = [String]()
for event in currentStat.statEvents {
if !eventNames.contains(event.name) {
eventNames.append(event.name)
}
}
// The type of the result
struct ResultType {
var name : String
var date : [String]
var hours : [Int]
}
var result = [ResultType]()
for name in eventNames {
let matchingEvents = currentStat.statEvents.filter { $0.name == name }
let dates = matchingEvents.map { $0.date }
let hours = matchingEvents.map { $0.hours }
result.append(ResultType(name: name, date: dates, hours: hours))
}
答案 2 :(得分:1)
最终结果是[[String:AnyObject]]类型,或者您创建一个包含这些值的新结构类型,结果类型为[String:NewStructType]:
struct NewStructType
{
var dates: [String]?
var hours: [Int]?
}
所以你必须决定那个,然后你必须编写自己的函数来对StatEvents对象进行排序和分组。也许你可以优化它的性能,但这是第一个想法如何实现第二个版本(使用NewStructType):
var result = [String : NewStructType]()
for statEvent in currentStat.statEvents
{
if (result[statEvent.name] != nil)
{
var newStructType = result[statEvent.name]!
newStructType.dates.append(statEvent.date)
newStructType.hours.append(statEvent.hours)
}
else
{
result[statEvent.name] = NewStructType(dates: [statEvent.date], hours: [statEvent.hours])
}
}
答案 3 :(得分:1)
我的看法:
/proc/self/status
测试:
extension StatEvents : Comparable {}
func < (lhs:StatEvents, rhs:StatEvents) -> Bool {
if lhs.name != rhs.name {
return lhs.name > rhs.name
} else if lhs.date != rhs.date {
return lhs.date < rhs.date
} else {
return lhs.hours < rhs.hours
}
}
func == (lhs:StatEvents, rhs:StatEvents) -> Bool {
return lhs.name == rhs.name
&& lhs.date == rhs.date
&& lhs.hours == rhs.hours
}
struct ResultRow {
var name: String
var dates: [String]
var hours: [Int]
}
var result : [ResultRow] = []
let sorted = currentStat.statEvents.sort()
for event in sorted {
if result.last?.name != event.name {
result.append(ResultRow(name: event.name, dates: [], hours: []))
}
result[result.endIndex - 1].dates.append(event.date)
result[result.endIndex - 1].hours.append(event.hours)
}
打印:
for r in result { print(r) }