Question

我编写了一个方法，将变量rank添加到结构数组中。数组friutsArray的创建方式与函数makeFriuts()类似。之后，数据被排序，根据这一点，每个项目分别获得rank索引。

最后我需要FriutsWithRank结构。但我想知道是否有更好，更有效的方法。甚至可以跳过整个Friuts结构：

struct Friuts {
    var name: String
    var price: Double
}

struct FriutsWithRank {
    var name: String
    var price: Double
    var rank: Int
}

var friutsArray = [Friuts]()

func makeFriuts() {
    friutsArray.append(Friuts(name: "mango", price: 1.2))
    friutsArray.append(Friuts(name: "banana", price: 0.79))
    friutsArray.append(Friuts(name: "orange", price: 2.2))
}

func makeFriutsWithRank(data: [Friuts]) -> [FriutsWithRank] {

    let dataSorted = data.sorted { $1.price < $0.price }

    var datatoappend = [FriutsWithRank]()

    var i = 0
    dataSorted.forEach { fruit in
        i += 1
        let name = fruit.name
        let price = fruit.price
        let rank = i

        let result = FriutsWithRank(name: name, price: price, rank: rank)

        datatoappend.append(result)
    }
    return datatoappend
}

let friutsArrayWithRank = makeFriutsWithRank(data: friutsArray)

更有效，我的意思不一定是更少的代码。我认为现在使用两次迭代创建了两个数组。是否可以跳过整个Fruits结构并只使用一个结构和一个迭代？

Answer 1

我已对您的代码进行了一些修改，请阅读内联评论。没有更多优化，但肯定更具可读性。

// Typo fixed + it is a single Fruit, not Fruits
struct Fruit {
    var name: String
    var price: Double
}


struct FruitsWithRank {
    // You already have a variable holding name and price, Friut
    // Lets just reuse Fruit object
    var fruit: Fruits
    var rank: Int
}

var fruits = [Fruit]()

func makeFruits() {
    fruits.append(Fruit(name: "mango", price: 1.2))
    fruits.append(Fruit(name: "banana", price: 0.79))
    fruits.append(Fruit(name: "orange", price: 2.2))
}

func makeFruitsWithRank(data: [Fruits]) -> [FruitsWithRank] {

    let dataSorted = data.sorted { $1.price < $0.price }
    var datatoappend = [FruitsWithRank]()
    // Use `enumerated` to get index and the object straight away
    for (index, fruit) in dataSorted.enumerated() {
        // Just init your `FruitsWithRank` with the fruit and the index
        let rankedFruit = FruitsWithRank(fruit: fruit, rank: index)
        // Add to your array
        datatoappend.append(rankedFruit)
    } 

    return datatoappend
}

let fruitsWithRank = makeFruitsWithRank(data: fruitsArray)

修改在您的问题编辑后，我已经应用了一些更改。如果您需要FruitsWithRank具有名称和价格，您只需创建名称和价格tuples，然后立即创建一个数组，无需任何循环或追加。您可以省略makeFruitsWithRank函数，并在元组数组上直接排序，枚举和创建类型。

struct FruitsWithRank { var rank: Int var name: String var price: Double } let rankedFruits: [FruitsWithRank] = [ (name: "mango", price: 1.2), (name: "banana", price: 0.79), (name: "orange", price: 2.2)] .sorted { $0.price < $1.price } .enumerated() .map({ return FruitsWithRank(rank: $0 + 1, name: $1.name, price: $1.price) })

Answer 2

最后，这不比你的代码更有效，但它更紧凑：

func makeFriutsWithRank(data: [Friuts]) -> [FriutsWithRank] {

    let dataMapped = data.sorted { $1.price < $0.price }
        .enumerated()
        .map { FriutsWithRank(name: $1.name, price: $1.price, rank: $0 + 1) }

    return dataMapped
}

Answer 3

拥有Struct真的很难吗？因为您已经排序，所以索引可以作为排名。

有没有更好的方法来创建数组的索引？

3 个答案: