Question

struct Human{
    let name: String
    let id: String
    let travelled: String
}

struct Animal {
    let name: String
    let id: String
    let travelled: String
}

// Consider no human and animal object can have same name
let human1 = Human(name: "vikas", id: "12", travelled: "123")
let human2 = Human(name: "jacky", id: "15", travelled: "343")
let human3 = Human(name: "lucy", id: "32", travelled: "132")
let animal1 = Animal(name: "jacky", id: "56", travelled: "8979")
let animal2 = Animal(name: "lucy", id: "78", travelled: "678")
let animal3 = Animal(name: "jimmy", id: "98", travelled: "690")
let humans = [human1, human2, human3]
let animals = [animal1, animal2, animal3]

var list = [[String: String]]()
// can we eleminate this for loop with filter or something else
for human in humans {
    if let animal = animals.first(where: {$0.name == human.name}){
        let data = ["humanId": human.id, "animalId": animal.id]
        list.append(data)
    }
}
print(list)

输出：

[["humanId": "15", "animalId": "56"], ["humanId": "32", "animalId": "78"]]

有没有一种方法我们可以一次应用多个过滤器以获得所需的输出找不到或创建一个

Answer 1

您可以尝试改用 compactMap(_:) 。

let list = humans.compactMap { (human) -> [String:String]? in
    if let animal = animals.first(where: { $0.name == human.name }) {
        return ["humanId": human.id, "animalId": animal.id]
    }
    return nil
}

Answer 2

此解决方案仍使用循环，但如果存在则将找到多个匹配项

animals.forEach {animal in
    let filtered = humans.filter { $0.name == animal.name }
        .map { human -> [String: String] in ["humanId": human.id, "animalId": animal.id] }

    list.append(contentsOf: filtered)
}

Answer 3

在这里

for index in 0..<humans.count {
    list[index]["humanId"] = humans[index]
    list[index]["animalId"] = animals[index]
}

从两种不同类型的数组获取公共数据

3 个答案: