在Swift中有没有对应的Scala、Xtend、Groovy、Ruby等等?

var aofa = [[1,2,3],[4],[5,6,7,8,9]]
aofa.flatten() // shall deliver [1,2,3,4,5,6,7,8,9] 

当然我可以用reduce来做,但那有点糟糕

var flattened = aofa.reduce(Int[]()){
    a,i in var b : Int[] = a
    b.extend(i)
    return b
}

当前回答

func convert(){
    let arr = [[1,2,3],[4],[5,6,7,8,9]]
    print("Old Arr = ",arr)
    var newArr = [Int]()
    for i in arr{
        for j in i{
            newArr.append(j)
        }
    }
    print("New Arr = ",newArr)
}

其他回答

Swift >= 3.0

减少:

let numbers = [[1,2,3],[4],[5,6,7,8,9]]
let reduced = numbers.reduce([], +)

flatMap:

let numbers = [[1,2,3],[4],[5,6,7,8,9]]
let flattened = numbers.flatMap { $0 }

加入:

let numbers = [[1,2,3],[4],[5,6,7,8,9]]
let joined = Array(numbers.joined())

在Swift标准库中,为所有符合Sequence协议的类型实现了join函数(或在Swift 3之前在SequenceType上flatten),其中包括Array:

let numbers = [[1,2,3],[4],[5,6,7,8,9]]
let flattened = Array(numbers.joined())

在某些情况下,使用joined()可能是有益的,因为它返回一个惰性集合而不是一个新数组,但当传递给array()初始化器时,总是可以转换为一个数组,就像上面的例子一样。

斯威夫特5.1

public extension Array where Element: Collection {

    func flatten() -> [Element.Element] {
        return reduce([], +)
    }
}

如果你也想在Dictionary值中使用它:

public extension Dictionary.Values where Value : Collection {
    func flatten() -> [Value.Element]{
         return self.reduce([], +)
    }
}
struct Group {
    var members: [String]?
}

let groups = [Group]()
let outputMembers: [String] = Array(groups.compactMap({ $0.members }).joined())

描述

如果要制作单数组的数组对象模型。示例:我们从所有组中获取outputMembers单个数组。

你可以用下面的方法来平嵌套数组:

var arrays = [1, 2, 3, 4, 5, [12, 22, 32], [[1, 2, 3], 1, 3, 4, [[[777, 888, 8999]]]]] as [Any]

func flatten(_ array: [Any]) -> [Any] {

    return array.reduce([Any]()) { result, current in
        switch current {
        case(let arrayOfAny as [Any]):
            return result + flatten(arrayOfAny)
        default:
            return result + [current]
        }
    }
}

let result = flatten(arrays)

print(result)

/// [1, 2, 3, 4, 5, 12, 22, 32, 1, 2, 3, 1, 3, 4, 777, 888, 8999]