我有一个名为person的结构
struct person{
var name : String
var score: Int
}
然后我创建一个名为
let people : [person] = [person("a", 1), person("a", 3), person("b", 5)]
如您所见，这里有两个同名“a”的对象。
现在我想把这个变成一个累积字典，显示每个人的总分。在这种情况下
dict = {"a": 4 (3+1), "b": 5}
我知道我违反了 OO 设计规则。 谢谢
【问题讨论】：
UpperCamelCase。
let people = [("a",1), ("a",3), ("b", 5)].map(Person.init)
标签： arrays swift dictionary
这里有一个基本的建模问题。您的结构 person 实际上并没有为人建模。它模拟RoundResult 之类的东西。
我将通过创建一个真正模拟人的Player 来重构它（只有name: String 之类的字段），并创建一个包含winner: Player 和score: Score 的RoundResult。
struct Player: Hashable { // Perhaps should be a class, if names aren't unique.
let name: String
}
struct RoundResult {
let winner: Player
let score: Int
}
let playerA = Player(name: "a")
let playerB = Player(name: "b")
let roundResults = [
RoundResult(winner: playerA, score: 1),
RoundResult(winner: playerA, score: 3),
RoundResult(winner: playerB, score: 5),
]
let scoresByPlayer = Dictionary(grouping: roundResults, by: \.winner)
.mapValues { roundResults -> Int in
let scores = roundResults.lazy.map(\.score)
return scores.reduce(0, +)
}
print(scoresByPlayer)
从这里，您可以在玩家上添加一个 score 变量，实际上模拟玩家得分，而不仅仅是单轮/游戏/比赛/任何事情中的一小部分 p>
【讨论】：
您可以通过两步来实现：1) 分组到字典中，2) 对分组的分数求和：
// grouping = { "a": [person("a", 1), person("a", 2)], "b": [person("b": 3)]
let grouping = Dictionary.init(grouping: people, by: { person in person.name })
let dict = grouping.mapValues { group in
group.reduce(0, { sum, person in sum + person.score })
}
或者更短但更神秘的形式：
let d = Dictionary.init(grouping: people, by: { $0.name })
.mapValues{ $0.reduce(0, $0 + $1.score ) }
【讨论】：
let dict: [String: Int] = .init(people.map{($0.name, $0.score)}, uniquingKeysWith: +)
我假设您是 Swift 新手。让我带你看一下这个的几个迭代。
最基本的是直截了当的方法。
var dict = [String: Int]()
for person in people {
if let score = dict[person.name] {
// If the person already has a score in dict, then add this new score to it.
dict[person.name] = score + person.score
} else {
// If this is the first score for this person, then add the person to dict.
dict[person.name] = person.score
}
}
接下来，我们将使用subscript(_:default:) 运算符将if 条件的两部分结合起来。
var dict = [String: Int]()
for person in people {
// Add the new score to the person's existing score. If there is no existing
// score, then default to 0 and add the new score.
dict[person.name, default: 0] += person.score
}
最后，使用 reduce(into:_:) 摆脱 for 循环（参见 no raw loops segment of C++ Seasoning）
// Reduce people into a dictionary of names and cumulative scores.
let dict = people.reduce(into: [String: Int]()) { result, person in
result[person.name, default: 0] += person.score
}
【讨论】：