arrays - 在 Swift 中对数组进行分组和排序

Question

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得到一个输出如下的数组:

- [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 你能检查一下吗？？

Answer 1

这似乎有点矫枉过正，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，这可能比其他解决方案的性能低。好的部分是您不必依赖并行数组来获取与特定日期关联的小时数。