ios - 使用条件排名数组进行快速归约和求和

Question

我想计算胜利的次数并将所有分数相加。

我愿意使用平面 map 和减少，以降低复杂性。为了安全起见，我可以使用 for 循环来获取它

问题是在 allMatch 中，我可以有不同的 teamID(团队可以单人或多人比赛)并且与顺序无关(例如 teamID 1,2 等于 teamID 2,1

struct InfoMatch: Hashable {
    let teamId: Set<Int>
    let rank: Int
    let mPoints: Int
    let vPoints: Int
}

所有比赛都是数组的数组，因为它返回每场比赛每支球队的信息，在本例中，第一场比赛有 2 支球队，最后一场比赛有 3 支球队

let allMatch: [[InfoMatch]] = [
    [InfoMatch(teamId:[1,2],rank:1,mPoints: 200,vPoints: 0),InfoMatch(teamId:[3,4],rank:2,mPoints: 100,vPoints: 0)],
    [InfoMatch(teamId:[1,2],rank:2,mPoints: 0,vPoints: 10),InfoMatch(teamId:[4,3],rank:1,mPoints: 0,vPoints: 20)],
    [InfoMatch(teamId:[2,1],rank:1,mPoints: 40,vPoints: 0),InfoMatch(teamId:[3,4],rank:2,mPoints: 30,vPoints: 0),InfoMatch(teamId:[5,6],rank:3,mPoints: 5,vPoints: 0)]
]
let winners = allMatch.map{$0.filter{$0.rank == 1}}
let losers = allMatch.map{$0.filter{$0.rank != 1}}

print(winners)

打印:

[[__lldb_expr_7.InfoMatch(teamId: Set([1, 2]), rank: 1, mPoints: 200, vPoints: 0)], [__lldb_expr_7.InfoMatch(teamId: Set([3, 4]), rank: 1, mPoints: 0, vPoints: 20)], [__lldb_expr_7.InfoMatch(teamId: Set([1, 2]), rank: 1, mPoints: 40, vPoints: 0)]]

我想要一个返回的数组

teamID: victoryCount: loserCount: mPoints: vPoints:

举个例子:

teamID:1,2 victoryCount:2 loserCount:1 mPoints:240 vPoints:10 teamID:3,4 victoryCount:1 loserCount:2 mPoints:130 vPoints:20 teamID:5,6 victoryCount:0 loserCount:1 mPoints:5 vPoints:0

Answer 1

我真的想知道使用flatmap映射和reduce是否可以降低复杂性，但是你可以这样使用reduce。

首先，你说你想要teamID:victoryCount:loserCount:mPoints:vPoints:，那么你应该更好地定义代表它的类型:

struct MatchStats {
    var teamID: Set<Int>
    var victoryCount: Int
    var loserCount: Int
    var mPoints: Int
    var vPoints: Int
}

//For convenience...
extension MatchStats {
    init(teamID: Set<Int>) {
        self.teamID = teamID
        victoryCount = 0
        loserCount = 0
        mPoints = 0
        vPoints = 0
    }
}

//For debugging...
extension MatchStats: CustomStringConvertible {
    public var description: String {
        let teamIDStr = teamID.sorted().map(String.init).joined(separator: ",")
        return "teamID:\(teamIDStr) victoryCount:\(victoryCount) loserCount:\(loserCount) mPoints:\(mPoints) vPoints:\(vPoints)"

    }
}

并在 reduce 和 flatMap 中按如下方式使用它:

let teamStats: [Set<Int>: MatchStats] = allMatch.flatMap{$0}.reduce(into: [:]) {result, info in
    result[info.teamId, default: MatchStats(teamID: info.teamId)].victoryCount += info.rank == 1 ? 1 : 0
    result[info.teamId]!.loserCount += info.rank == 1 ? 0 : 1
    result[info.teamId]!.mPoints += info.mPoints
    result[info.teamId]!.vPoints += info.vPoints
}

print(teamStats.values)

输出:

[teamID:1,2 victoryCount:2 loserCount:1 mPoints:240 vPoints:10, teamID:3,4 victoryCount:1 loserCount:2 mPoints:130 vPoints:20, teamID:5,6 victoryCount:0 loserCount:1 mPoints:5 vPoints:0]

(每次调用时输出可能是随机顺序的。但排序是另一个问题。)

<小时/>

你认为使用reduce比这更好吗？

var teamStats: [Set<Int>: MatchStats] = [:]
for match in allMatch {
    for info in match {
        teamStats[info.teamId, default: MatchStats(teamID: info.teamId)].victoryCount += info.rank == 1 ? 1 : 0
        teamStats[info.teamId]!.loserCount += info.rank == 1 ? 0 : 1
        teamStats[info.teamId]!.mPoints += info.mPoints
        teamStats[info.teamId]!.vPoints += info.vPoints
    }
}

print(teamStats.values)