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
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([], +)
}
}
其他回答
Apple Swift 5.1.2版本(swiflang -1100.0.278) 目标:x86_64-apple-darwin19.2.0
let optionalNumbers = [[1, 2, 3, nil], nil, [4], [5, 6, 7, 8, 9]]
print(optionalNumbers.compactMap { $0 }) // [[Optional(1), Optional(2), Optional(3), nil], [Optional(4)], [Optional(5), Optional(6), Optional(7), Optional(8), Optional(9)]]
print(optionalNumbers.compactMap { $0 }.reduce([], +).map { $0 as? Int ?? nil }.compactMap{ $0 }) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(optionalNumbers.compactMap { $0 }.flatMap { $0 }.map { $0 as? Int ?? nil }.compactMap{ $0 }) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(Array(optionalNumbers.compactMap { $0 }.joined()).map { $0 as? Int ?? nil }.compactMap{ $0 }) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
let nonOptionalNumbers = [[1, 2, 3], [4], [5, 6, 7, 8, 9]]
print(nonOptionalNumbers.compactMap { $0 }) // [[1, 2, 3], [4], [5, 6, 7, 8, 9]]
print(nonOptionalNumbers.reduce([], +)) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(nonOptionalNumbers.flatMap { $0 }) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(Array(nonOptionalNumbers.joined())) // [1, 2, 3, 4, 5, 6, 7, 8, 9]
迅速的4.倍和5.倍
为了在数组中增加一点复杂性，如果有一个数组包含数组的数组，那么flatMap实际上会失败。
假设数组是
var array:[Any] = [1,2,[[3,4],[5,6,[7]]],8]
flatMap或compactMap返回的是:
array.compactMap({$0})
//Output
[1, 2, [[3, 4], [5, 6, [7]]], 8]
为了解决这个问题，我们可以使用简单的for循环逻辑+递归
func flattenedArray(array:[Any]) -> [Int] {
var myArray = [Int]()
for element in array {
if let element = element as? Int {
myArray.append(element)
}
if let element = element as? [Any] {
let result = flattenedArray(array: element)
for i in result {
myArray.append(i)
}
}
}
return myArray
}
用给定的数组调用这个函数
flattenedArray(array: array)
结果是:
[1, 2, 3, 4, 5, 6, 7, 8]
考虑到这里Int的情况，这个函数将有助于将任何类型的数组平化
操场上的输出:
flatten()在Swift 3中根据SE-0133重命名为joined():
https://github.com/apple/swift-evolution/blob/master/proposals/0133-rename-flatten-to-joined.md
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]
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()
