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在Swift中按不同属性值对自定义对象排序

nssortdescriptor sorting swift arrays

UKDataGeek · 技术社区 · 8 年前

我试图通过不同的属性值轻松地对自定义结构数组进行排序。

struct Customer: Comparable, Equatable {
    var name: String
    var isActive: Bool
    var outstandingAmount: Int
    var customerGroup: String
}

var customerlist: [Customer] // This is downloaded from our backend.

我希望能够在用户选择各种图标时,根据所有字段值对UI中的customerlist数组进行排序。

我尝试了几种使用switch语句对其进行排序的方法,但我被告知,正确的方法是使用排序描述符(这似乎是基于Objective-C的,意味着我需要将数组转换为NSArray)当我使用本机Swift结构尝试这种方法时,我不断出错。

允许用户使用Swift对上述数组进行排序的最佳方式是什么?

下面似乎很冗长!

func sortCustomers(sortField:ColumnOrder, targetArray:[Customer]) -> [Customer] { //Column Order is the enum where I have specified the different possible sort orders
        var result = [Customer]()
    switch sortField {
        case .name:
             result = targetArray.sorted(by: { (cust0: Customer, cust1: Customer) -> Bool in
                return cust0.name > cust1.name
            })
        case .isActive:
             result = targetArray.sorted(by: { (cust0: Customer, cust1: Customer) -> Bool in
                return cust0.isActive > cust1.isActive
            })
        case .outstandingAmount:
            result = targetArray.sorted(by: { (cust0: Customer, cust1: Customer) -> Bool in
                return cust0.outstandingAmount > cust1.outstandingAmount
            })
        case .customerGroup:
            result = targetArray.sorted(by: { (cust0: Customer, cust1: Customer) -> Bool in
                return cust0.customerGroup > cust1.customerGroup
            })
    }
    return result
}

2 回复 | 直到 8 年前

GetSwifty 8 年前

我将使用关键路径:

func sortCustomers<T: Comparable>(customers: [Customer], with itemPath: KeyPath<Customer, T>) -> [Customer] {
    return customers.sorted() {
       $0[keyPath: itemPath] < $1[keyPath: itemPath]
    }
}

这种方法完全避免了对枚举的需要,并允许您只需执行以下操作

let testData = [Customer(name: "aaaa", isActive: false, outstandingAmount: 1, customerGroup: "aaa"),
                Customer(name: "bbbb", isActive: true, outstandingAmount: 2, customerGroup: "bbb")];

let testResultsWithName = sortCustomers(customers: testData, with: \Customer.name)
let testResultsWithActive = sortCustomers(customers: testData, with: \Customer.isActive) 
// etc

注意,我切换了 > 到a < . 这是默认预期,将导致“b”之前的“a”,“2”之前的“1”等。

此外,您需要为Bool添加一个可比较的扩展:

extension Bool: Comparable {
    public static func <(lhs: Bool, rhs: Bool) -> Bool {
        return lhs == rhs || (lhs == false && rhs == true)
    }
}

为了完善该方法,您还可以传入一个比较函数:

func sortCustomers<T: Comparable>(customers: [Customer], comparing itemPath: KeyPath<Customer, T>, using comparitor: (T, T) -> Bool) -> [Customer] {
    return customers.sorted() {
        comparitor($0[keyPath: itemPath], $1[keyPath: itemPath])
    }
}
let testResults = sortCustomers(customers: testData, comparing: \Customer.name, using: <)

这样,如果需要自定义排序,您可以使用普通比较运算符:(<、<=、>、>)以及闭包。

Jeffery Thomas 8 年前

我重新打包了冗长的解决方案,使其变得更好。我将属性添加到 ColumnOrder 返回一个有序闭包。

struct Customer {
    var name: String
    var isActive: Bool
    var outstandingAmount: Int
    var customerGroup: String
}

enum ColumnOrder {
    case name
    case isActive
    case outstandingAmount
    case customerGroup

    var ordering: (Customer, Customer) -> Bool {
        switch self {
        case .name:              return { $0.name > $1.name }
        case .isActive:          return { $0.isActive && !$1.isActive }
        case .outstandingAmount: return { $0.outstandingAmount > $1.outstandingAmount}
        case .customerGroup:     return { $0.customerGroup > $1.customerGroup }
        }
    }
}

以下是它的使用方法:

let sortedCustomers = customers.sorted(by: ColumnOrder.name.ordering)

接下来,我扩展了 Sequence

extension Sequence where Element == Customer {
    func sorted(by columnOrder: ColumnOrder) -> [Element] {
        return sorted(by: columnOrder.ordering)
    }
}

最终结果:

let sortedCustomers = customers.sorted(by: .name)