我有大量的对象列表,我需要将它们分成一组两个元素用于UI propouse.
例：
[0,1,2,3,4,5,6]
成为这四个数组的数组
[[0,1],[2,3],[4,5],[6]]
分割阵列有很多种方法.但是,如果阵列很大,那么效率最高(成本最低)的是什么.
如果你正在寻找效率,你可以有一个方法,可以懒惰地生成每个2个元素的数组,所以你一次只能在内存中存储2个元素：
public struct ChunkGen<G : GeneratorType> : GeneratorType {
private var g: G
private let n: Int
private var c: [G.Element]
public mutating func next() -> [G.Element]? {
var i = n
return g.next().map {
c = [$0]
while --i > 0,let next = g.next() { c.append(next) }
return c
}
}
private init(g: G,n: Int) {
self.g = g
self.n = n
self.c = []
self.c.reserveCapacity(n)
}
}
public struct ChunkSeq<S : SequenceType> : SequenceType {
private let seq: S
private let n: Int
public func generate() -> ChunkGen<S.Generator> {
return ChunkGen(g: seq.generate(),n: n)
}
}
public extension SequenceType {
func chunk(n: Int) -> ChunkSeq<Self> {
return ChunkSeq(seq: self,n: n)
}
}
var g = [1,5].chunk(2).generate()
g.next() // [1,2]
g.next() // [3,4]
g.next() // [5]
g.next() // nil
此方法适用于任何SequenceType,而不仅仅是Arrays.
对于Swift 1,没有协议扩展,你有：
public struct ChunkGen<T> : GeneratorType {
private var (st,en): (Int,Int)
private let n: Int
private let c: [T]
public mutating func next() -> ArraySlice<T>? {
(st,en) = (en,en + n)
return st < c.endIndex ? c[st..<min(en,c.endIndex)] : nil
}
private init(c: [T],n: Int) {
self.c = c
self.n = n
self.st = 0 - n
self.en = 0
}
}
public struct ChunkSeq<T> : SequenceType {
private let c: [T]
private let n: Int
public func generate() -> ChunkGen<T> {
return ChunkGen(c: c,n: n)
}
}
func chunk<T>(ar: [T],#n: Int) -> ChunkSeq<T> {
return ChunkSeq(c: ar,n: n)
}
对于Swift 3：
public struct ChunkIterator<I: IteratorProtocol> : IteratorProtocol {
fileprivate var i: I
fileprivate let n: Int
public mutating func next() -> [I.Element]? {
guard let head = i.next() else { return nil }
var build = [head]
build.reserveCapacity(n)
for _ in (1..<n) {
guard let x = i.next() else { break }
build.append(x)
}
return build
}
}
public struct ChunkSeq<S: Sequence> : Sequence {
fileprivate let seq: S
fileprivate let n: Int
public func makeIterator() -> ChunkIterator<S.Iterator> {
return ChunkIterator(i: seq.makeIterator(),n: n)
}
}
public extension Sequence {
func chunk(_ n: Int) -> ChunkSeq<Self> {
return ChunkSeq(seq: self,5].chunk(2).makeIterator()
g.next() // [1,4]
g.next() // [5]
g.next() // nil
