由某些属性组合数组的元素

文档具有此方法的简单示例. 我只是发布以下示例:

let students = ["Kofi", "Abena", "Efua", "Kweku", "Akosua"]
let studentsByLetter = Dictionary(grouping: students, by: { $0.first! })

结果将是:

["E": ["Efua"], "K": ["Kofi", "Kweku"], "A": ["Abena", "Akosua"]]

改进oriyentel解决方案以允许对任何事物进行有序分组:

extension Sequence {
    func group(by key: (Iterator.Element) -> GroupingType) -> [[Iterator.Element]] {
        var groups: [GroupingType: [Iterator.Element]] = [:]
        var groupsOrder: [GroupingType] = []
        forEach { element in
            let key = key(element)
            if case nil = groups[key]?.append(element) {
                groups[key] = [element]
                groupsOrder.append(key)
            }
        }
        return groupsOrder.map { groups[$0]! }
    }
}

然后它将适用于任何元组,结构或类以及任何属性:

let a = [(grouping: 10, content: "a"),
         (grouping: 20, content: "b"),
         (grouping: 10, content: "c")]
print(a.group { $0.grouping })

struct GroupInt {
    var grouping: Int
    var content: String
}
let b = [GroupInt(grouping: 10, content: "a"),
         GroupInt(grouping: 20, content: "b"),
         GroupInt(grouping: 10, content: "c")]
print(b.group { $0.grouping })