Swift - How to get indexes of filtered items of ar

Question

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

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

Does Swift 3 support a function like whatFunction(_: Element)?

If not, what is the most efficient logic?

Answer 1

you can use it like that :

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

        let indexes = items.enumerated().filter {
            $0.element == "A"
            }.map{$0.offset}

        print(indexes)

Answer 2

For example finding the indices of p_last values that are in inds1 array: (swift 4+)

let p_last = [51,42]
let inds1 = [1,3,51,42,4]
let idx1 = Array(inds1.filter{ p_last.contains($0) }.indices)

idx1 = [0,1]

Answer 3

You can filter the indices of the array directly, it avoids the extra mapping.

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

or as Array extension:

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]

or still more generic

extension Collection where Element: Equatable {

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

}

Answer 4

In Swift 3 and Swift 4 you can do that:

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")

I hope my answer was helpful

查看更多

0人赞 添加讨论(0) 举报

Answer 5

You can create your own extension for arrays.

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

You can simply call it like this

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

Answer 6

You can achieve this by chain of:

1. enumerated() - add indexes;
2. filter() out unnecessary items;
3. map() our indexes.

Example (works in 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]

---

Another way is using flatMap, which allows you to check the element and return index if needed in one closure.

Example (works in Swift 3 - Swift 4.0):

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

But since Swift 4.1 flatMap that can return non-nil objects become deprecated and instead you should use compactMap.

Example (works since Swift 4.1):

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

---

And the cleanest and the most memory-cheap way is to iterate through array indices and check if element of array at current index equals to required element.

Example (works in Swift 3 - Swift 4.x):

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