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
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。
其他回答
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]
编辑:使用joined()代替:
https://developer.apple.com/documentation/swift/sequence/2431985-joined
最初的回答:
let numbers = [[1, 2, 3], [4, 5, 6]]
let flattenNumbers = numbers.reduce([], combine: +)
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([], +)
}
}
斯威夫特4.2
我在下面写了一个简单的数组扩展。可用于将包含另一个数组或元素的数组平展。不像joined()方法。
public extension Array {
public func flatten() -> [Element] {
return Array.flatten(0, self)
}
public static func flatten<Element>(_ index: Int, _ toFlat: [Element]) -> [Element] {
guard index < toFlat.count else { return [] }
var flatten: [Element] = []
if let itemArr = toFlat[index] as? [Element] {
flatten = flatten + itemArr.flatten()
} else {
flatten.append(toFlat[index])
}
return flatten + Array.flatten(index + 1, toFlat)
}
}
用法:
let numbers: [Any] = [1, [2, "3"], 4, ["5", 6, 7], "8", [9, 10]]
numbers.flatten()
修改了@RahmiBozdag的回答， 1. 公共扩展中的方法是公共的。 2. 删除了额外的方法，因为开始索引将始终为零。 3.我没有找到一种方法把compactMap内部为nil和可选的，因为内部方法T总是[Any?]，欢迎提出任何建议。
let array = [[[1, 2, 3], 4], 5, [6, [9], 10], 11, nil] as [Any?]
public extension Array {
func flatten<T>(_ index: Int = 0) -> [T] {
guard index < self.count else {
return []
}
var flatten: [T] = []
if let itemArr = self[index] as? [T] {
flatten += itemArr.flatten()
} else if let element = self[index] as? T {
flatten.append(element)
}
return flatten + self.flatten(index + 1)
}
}
let result: [Any] = array.flatten().compactMap { $0 }
print(result)
//[1, 2, 3, 4, 5, 6, 9, 10, 11]
