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
}
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()
reduce的另一个更通用的实现,
let numbers = [[1,2,3],[4],[5,6,7,8,9]]
let reduced = reduce(numbers,[],+)
这完成了同样的事情,但可能会对reduce中发生的事情有更多的了解。
从苹果公司的文件来看,
func reduce<S : SequenceType, U>(sequence: S, initial: U, combine: (U, S.Generator.Element) -> U) -> U
描述
返回反复调用combine的结果,并依次返回初始化为initial的累积值和sequence的每个元素。
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]