Question

假设我有2个类的实例数组＆＃34; Person＆＃34;。 Class Person具有属性＆＃34; name＆＃34;，＆＃34; age＆＃34;，＆＃34; id＆＃34;。＆＃34; ID＆＃34;是一个唯一的标识符，因此可用于比较人。所以，假设我有两个人：

[ {Name: Bob,Age: 18,ID: 142353}, {Name: Rob, Age: 40,ID: 142350}, {Name: Andy, Age: 30,ID: 142359}, {Name: Andy, Age: 21,ID: 142340}  ]

[ {Name: Jack,Age: 18,ID: 142362}, {Name: James, Age: 40,ID: 142311}, {Name: Rob, Age: 40,ID: 142350}, {Name: Andy, Age: 21,ID: 142340}  ]

我想比较人的ID并获得那些在数组1中与数组2中的ID不同的ID，因此结果应如下所示：

[{Name: Bob,Age: 18,ID: 142353}, {Name: Andy, Age: 30,ID: 142359}]

我现在在Swift中这样做的方式是：

   let new = newPeople.filter({ p1 in
                        previousPeople.contains(where: { p2 in
                            p1.id != p2.id
         })
   })

但我认为这不起作用。无法弄清楚我的实施有什么问题。

Answer 1

您希望将Person类符合Equatable协议。

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

然后你可以对这两个数组进行简单的过滤：

let person1 = Person(name: "J", age: 20, id: UUID())
let person2 = Person(name: "K", age: 40, id: UUID())
let person3 = Person(name: "L", age: 30, id: UUID())
let person4 = Person(name: "M", age: 25, id: UUID())
let person5 = Person(name: "N", age: 39, id: UUID())

let personArray1 = [person1, person2, person3]
let personArray2 = [person4, person2, person3]

let filteredPersonArray = personArray1.filter { !personArray2.contains($0) }
//filteredPersonArray.count == 1 which is person4

Answer 2

创建您不想保留的Set个ID，并通过针对此contains检查Set来过滤您的用户。您可以使用Array从Array个人处获得people.map{ $0.id }个ID。

struct Person {
    let name: String
    let age: Int
    let id: Int
}

let people1 = [
    Person(name: "Bob", age: 18, id: 142353),
    Person(name: "Rob", age: 40, id: 142350),
    Person(name: "Andy", age: 30, id: 142359),
    Person(name: "Andy", age: 21, id: 142340)
]

let people2 = [
    Person(name: "Jack", age: 18, id: 142362),
    Person(name: "James", age: 40, id: 142311),
    Person(name: "Rob", age: 40, id: 142350),
    Person(name: "Andy", age: 21, id: 142340)
]


let undesiredIDs = Set(people2.map{ $0.id })
print("Keeping all people that don't have one of these IDs: \(undesiredIDs)\r\n\r\n")

let filteredPeople = people1.filter{ !undesiredIDs.contains($0.id) }

print("Original list: \(people1).\r\n\r\n")
print("Filtered: \(filteredPeople)")

根据对象的属性减去两个对象数组

2 个答案: