Question

我有一个array个对象中的一个MyData（MyData是一个struct）：

[
MyData(id: 3, locale: "en", title: "p1", date: "10/15/2019"), 
MyData(id: 3, locale: "de", title: "p2", date: "11/12/2019"), 
MyData(id: 32, locale: "fr", title: "free", date: "10/11/2019"), 
MyData(id: 15, locale: "de", title: "free", date: "10/11/2019"), 
MyData(id: 19, locale: "de", title: "p1", date: "11/10/2019"),
MyData(id: 19, locale: "de", title: "p2", date: "11/10/2019"),
MyData(id: 19, locale: "de", title: "p3", date: "11/10/2019"),
]

我想根据id键将此数组分组（甚至有意创建一个新数组）。

结果应该是这样的：

[
[MyData(id: 3, locale: "en", title: "p1", date: "10/15/2019"), MyData(id: 3, locale: "de", title: "p2", date: "11/12/2019")], 
MyData(id: 32, locale: "fr", title: "free", date: "10/11/2019"), 
MyData(id: 15, locale: "de", title: "free", date: "10/11/2019"), 
[MyData(id: 19, locale: "de", title: "p1", date: "11/10/2019"),MyData(id: 19, locale: "de", title: "p2", date: "11/10/2019"),MyData(id: 19, locale: "de", title: "p3", date: "11/10/2019")]
]

即：具有相同ID的数组应形成一个新数组。

当然，我可以简单地循环遍历第一个数组并创建第二个数组，但是我想知道Swift是否可以对其过滤器进行处理。任何帮助表示赞赏。

Answer 1

您可以肯定地使用高阶函数，但不能完全使用 100 ％来产生所需的结果，但要获得很大的一部分，因为所需的数组类型是:[Any] 。

查看下面的代码：

var myGroup = Dictionary(grouping: arrayOne, by: { $0.id }) // group each element by id -type of: [Int:[Data]]
let resultArray = myGroup.map { $0.value } //map out the elements without the id key. -type of: [[Data]]

//Create hetro Array so we can use it later to append the results
var myHetroArray: [Any] = []
// loop each array in the result array and check if it only contains 1 element if so append that one element to the hetro array otherwise just append the whole thing.
for array in resultArray {
    if array.count ==  1 {
    myHetroArray.append(array.first!)
    } else {
        myHetroArray.append(array)
    }
}

print(myHetroArray) // produce the desired result.

输出： [
  [数据（id：19，区域设置：“ de”，标题：“ p1”，日期：“ 11/10/2019”），数据（id：19，区域设置：“ de”，标题：“ p2”，日期： “” 11/10/2019“），数据（id：19，语言环境：” de“，标题：” p3“，日期：” 11/10/2019“））]，
  数据（id：15，语言环境：“ de”，标题：“ free”，日期：“ 10/11/2019”），
  数据（id：32，语言环境：“ fr”，标题：“ free”，日期：“ 10/11/2019”），
  [数据（id：3，区域设置：“ en”，标题：“ p1”，日期：“ 10/15/2019”），数据（id：3，区域设置：“ de”，标题：“ p2”，日期： “ 11/12/2019”）]
  ]

Answer 2

还有两种方法。就像这里的注释。

    struct MyData{
        let id : Int
        let locale : String
        let title : String
        let date : String
    }

    let data = [
        MyData(id: 3, locale: "en", title: "p1", date: "10/15/2019"),
        MyData(id: 3, locale: "de", title: "p2", date: "11/12/2019"),
        MyData(id: 32, locale: "fr", title: "free", date: "10/11/2019"),
        MyData(id: 15, locale: "de", title: "free", date: "10/11/2019"),
        MyData(id: 19, locale: "de", title: "p1", date: "11/10/2019"),
        MyData(id: 19, locale: "de", title: "p2", date: "11/10/2019"),
        MyData(id: 19, locale: "de", title: "p3", date: "11/10/2019"),
    ]


    data.reduce(into: [:]) { ( result, next) in
    if result.keys.contains(next.id){
        (result[next.id] as? [MyData]).map{result[next.id] =  $0 + [next] }
        ([result[next.id]] as? [MyData]).map{result[next.id] = $0 + [next]}
    }
    else{ result[next.id] = next }
}.values


    Set(data.map{$0.id}).map{id -> Any in
        let result = data.filter{$0.id == id}
        return result.count == 1 ? result.first! : result
    }

根据键值创建多维数组

2 个答案: