如何泛化参数是具有不同属性的不同结构的函数？

Question

请参考以下代码：

import UIKit

struct Item {
    var brandId = 1
    var name: String = ""
}
struct Store {
    var areaName = ""
    var name: String = ""
}

let itemArray = [Item(brandId: 1, name: "item1"), Item(brandId: 2, name: "item2"), Item(brandId: 1, name: "item3") ]

let storeArray = [Store(areaName: "hk", name: "store1"), Store(areaName: "bj", name: "store2"), Store(areaName: "hk", name: "store3")]


var intKeys = [Int]()
var groupedItems = [[Item]]()

var stringKeys = [String]()
var groupedStores = [[Store]]()

extension Array {

    func transTo2d() -> [[Element]] {
        let grouped = [[Element]]()

        return grouped
    }

}

itemArray.forEach { (item) in
    let brandId = item.brandId
    if !intKeys.contains(brandId) {
        intKeys.append(brandId)
        var newArray = [Item]()
        newArray.append(item)
        groupedItems.append(newArray)
    } else {
        let index = intKeys.index(of: brandId)!
        groupedItems[index].append(item)
    }
}

我的最终目标是可以使用 itemArray.transTo2d() 获得基于商品 brandId 的二维数组，使用 storeArray.transTo2d() 获得基于商店 areaName 的二维数组。我不知道如何根据键泛化将一维数组转换为二维数组的函数？

Answer 1

我认为您不能为数组编写通用扩展，其中元素的类型为 Item 或 Store，因为它们对您来说没有任何关系编写一个通用的泛型方法。您可以为 Array 编写扩展，其中元素属于上述类型。您只需要使您的两个结构都符合 equatable 协议。

struct Item {
    var brandId = 1
    var name: String = ""
}

extension Item : Equatable{
    static func ==(lhs: Item, rhs: Item) -> Bool{
        return lhs.brandId == rhs.brandId
    }
}

struct Store {
    var areaName = ""
    var name: String = ""
}

extension Store : Equatable{
    static func ==(lhs: Store, rhs: Store) -> Bool{
        return lhs.areaName == rhs.areaName
    }
}

extension Array where Element == Store{
    func transform()->[[Store]]{
        var storeArray = self
        var groupedArray = [[Store]]()
        while storeArray.count > 0{
            if let firstElement = storeArray.first{
                groupedArray.append(storeArray.filter{[=10=].areaName == firstElement.areaName})
                storeArray = storeArray.filter{[=10=].areaName != firstElement.areaName}
            }
        }
        return groupedArray
    }
}

extension Array where Element == Item{
    func transform()->[[Item]]{
        var itemArray = self
        var groupedArray = [[Item]]()
        while itemArray.count > 0{
            if let firstElement = itemArray.first{
                groupedArray.append(itemArray.filter{[=10=].brandId == firstElement.brandId})
                itemArray = itemArray.filter{[=10=].brandId != firstElement.brandId}
            }
        }
        return groupedArray
    }
}

使用变换函数

let storeArray = [Store(areaName: "hk", name: "store1"), Store(areaName: "bj", name: "store2"), Store(areaName: "hk", name: "store3")]
let itemArray = [Item(brandId: 1, name: "item1"), Item(brandId: 2, name: "item2"), Item(brandId: 1, name: "item3") ]
print(storeArray.transform())
print(itemArray.transform())

这将打印我相信您想要的输出。

[[Store(areaName: "hk", name: "store1"), Store(areaName: "hk", name: "store3")], [Store(areaName: "bj", name: "store2")]]
[[Item(brandId: 1, name: "item1"), Item(brandId: 1, name: "item3")], [Item(brandId: 2, name: "item2")]]

如何泛化参数是具有不同属性的不同结构的函数？

How to generic a function where params are different structs with different properties?

arrays

generics

generic-programming

swift

swift-protocols