Matt的语法非常好，如果你需要使块失效，你可能想使用这个:

typealias dispatch_cancelable_closure = (cancel : Bool) -> Void

func delay(time:NSTimeInterval, closure:()->Void) ->  dispatch_cancelable_closure? {

    func dispatch_later(clsr:()->Void) {
        dispatch_after(
            dispatch_time(
                DISPATCH_TIME_NOW,
                Int64(time * Double(NSEC_PER_SEC))
            ),
            dispatch_get_main_queue(), clsr)
    }

    var closure:dispatch_block_t? = closure
    var cancelableClosure:dispatch_cancelable_closure?

    let delayedClosure:dispatch_cancelable_closure = { cancel in
        if closure != nil {
            if (cancel == false) {
                dispatch_async(dispatch_get_main_queue(), closure!);
            }
        }
        closure = nil
        cancelableClosure = nil
    }

    cancelableClosure = delayedClosure

    dispatch_later {
        if let delayedClosure = cancelableClosure {
            delayedClosure(cancel: false)
        }
    }

    return cancelableClosure;
}

func cancel_delay(closure:dispatch_cancelable_closure?) {

    if closure != nil {
        closure!(cancel: true)
    }
}

使用方法如下

let retVal = delay(2.0) {
    println("Later")
}
delay(1.0) {
    cancel_delay(retVal)
}

学分

上面的链接似乎坏了。来自Github的原始Objc代码

我经常使用dispatch_after，所以我写了一个顶级实用函数来简化语法:

func delay(delay:Double, closure:()->()) {
    dispatch_after(
        dispatch_time(
            DISPATCH_TIME_NOW,
            Int64(delay * Double(NSEC_PER_SEC))
        ),
        dispatch_get_main_queue(), closure)
}

现在你可以这样说:

delay(0.4) {
    // do stuff
}

哇，一种你可以改进的语言。还有什么比这更好的呢?

更新Swift 3, Xcode 8种子6

看起来几乎不值得费心，现在他们已经改进了调用语法:

func delay(_ delay:Double, closure:@escaping ()->()) {
    let when = DispatchTime.now() + delay
    DispatchQueue.main.asyncAfter(deadline: when, execute: closure)
}

使用这段代码在2.0秒后执行一些UI相关的任务。

            let delay = 2.0
            let delayInNanoSeconds = dispatch_time(DISPATCH_TIME_NOW, Int64(delay * Double(NSEC_PER_SEC)))
            let mainQueue = dispatch_get_main_queue()

            dispatch_after(delayInNanoSeconds, mainQueue, {

                print("Some UI related task after delay")
            })

Swift 3.0版本

以下闭包函数在主线程上执行一些延迟后的任务。

func performAfterDelay(delay : Double, onCompletion: @escaping() -> Void){

    DispatchQueue.main.asyncAfter(deadline: DispatchTime.now() + delay, execute: {
       onCompletion()
    })
}

像这样调用这个函数:

performAfterDelay(delay: 4.0) {
  print("test")
}

Swift 4有一个很短的方法来做到这一点:

Timer.scheduledTimer(withTimeInterval: 2, repeats: false) { (timer) in
    // Your stuff here
    print("hello")
}

我总是喜欢使用扩展而不是自由函数。

斯威夫特4

public extension DispatchQueue {

  private class func delay(delay: TimeInterval, closure: @escaping () -> Void) {
    let when = DispatchTime.now() + delay
    DispatchQueue.main.asyncAfter(deadline: when, execute: closure)
  }

  class func performAction(after seconds: TimeInterval, callBack: @escaping (() -> Void) ) {
    DispatchQueue.delay(delay: seconds) {
      callBack()
    }
  }

}

按以下方法使用。

DispatchQueue.performAction(after: 0.3) {
  // Code Here
}

斯威夫特 3+

这在Swift 3+中是超级简单和优雅的:

DispatchQueue.main.asyncAfter(deadline: .now() + 4.5) {
    // ...
}

年长的回答:

为了扩展Cezary的答案，它将在1纳秒后执行，我必须执行以下操作以在4秒半后执行。

let delay = 4.5 * Double(NSEC_PER_SEC)
let time = dispatch_time(DISPATCH_TIME_NOW, Int64(delay))
dispatch_after(time, dispatch_get_main_queue(), block)

编辑:我发现我原来的代码有一点错误。如果不将NSEC_PER_SEC转换为Double类型，隐式类型将导致编译错误。

如果有人能提出一个更优的解决方案，我很乐意听听。