Swift - 如何获取数组的过滤项的索引

Question

let items: [String] = ["A", "B", "A", "C", "A", "D"]

items.whatFunction("A") // -> [0, 2, 4]
items.whatFunction("B") // -> [1]

Swift 3是否支持这样的功能whatFunction(_: Element)？

如果没有,最有效的逻辑是什么？

Answer 1

您可以indices直接过滤数组,避免额外的mapping.

let items = ["A", "B", "A", "C", "A", "D"]
let filteredIndices = items.indices.filter {items[$0] == "A"}

或作为Array扩展名:

extension Array where Element: Equatable {

    func whatFunction(_ value :  Element) -> [Int] {
        return self.indices.filter {self[$0] == value}
    }

}

items.whatFunction("A") // -> [0, 2, 4]
items.whatFunction("B") // -> [1]

或者更通用

extension Collection where Element: Equatable {

    func whatFunction(_ value :  Element) -> [Index] {
        return self.indices.filter {self[$0] == value}
    }

}

Answer 2

您可以为阵列创建自己的扩展.

extension Array where Element: Equatable {
    func indexes(of element: Element) -> [Int] {
        return self.enumerated().filter({ element == $0.element }).map({ $0.offset })
    }
}

你可以简单地这样称呼它

items.indexes(of: "A") // [0, 2, 4]
items.indexes(of: "B") // [1]

Answer 3

您可以通过以下链条实现此目的:

1. enumerated() - 添加索引;
2. filter() 出不必要的物品;
3. map() 我们的索引.

示例(适用于Swift 3 - Swift 4.x):

let items: [String] = ["A", "B", "A", "C", "A", "D"]  
print(items.enumerated().filter({ $0.element == "A" }).map({ $0.offset })) // -> [0, 2, 4]

另一种方法是使用flatMap,它允许您在一个闭包中检查元素并返回索引(如果需要).

示例(适用于Swift 3 - Swift 4.0):

print(items.enumerated().flatMap { $0.element == "A" ? $0.offset : nil }) // -> [0, 2, 4]

但是,由于flatMap可以返回非零对象的Swift 4.1 变得弃用,而是应该使用compactMap.

示例(自Swift 4.1起作用):

print(items.enumerated().compactMap { $0.element == "A" ? $0.offset : nil }) // -> [0, 2, 4]

最干净,最便宜的内存方式是迭代数组索引并检查当前索引处的数组元素是否等于所需元素.

示例(适用于Swift 3 - Swift 4.x):

print(items.indices.filter({ items[$0] == "A" })) // -> [0, 2, 4]

Answer 4

在Swift 3和Swift 4中你可以这样做:

let items: [String] = ["A", "B", "A", "C", "A", "D"]

extension Array where Element: Equatable {

    func indexes(of item: Element) -> [Int]  {
        return enumerated().compactMap { $0.element == item ? $0.offset : nil }
    }
}

items.indexes(of: "A")

我希望我的回答很有帮助