如何在事件回调之间保持python生成器的状态

Question

所以我在尝试学习leap-motion SDK并在4-5年没有接触过python之后重新学习它，我在python2.7中遇到了一个生成器问题

基本上，我有一个单词列表，每次跳跃动作拿起一个新的“圆圈”手势时，我想打印列表中的下一个单词。我看到的是每次on_frame回调函数只打印列表中的第一个单词。我相信发生的是python运行时在事件之间忘记了生成器的状态。有没有办法在手势事件之间保持生成器状态？在

if not frame.hands.is_empty:
        for gesture in frame.gestures():
            if gesture.type == Leap.Gesture.TYPE_CIRCLE:
                circle = CircleGesture(gesture)
                # Determine clock direction using the angle between the pointable and the circle normal
                action = None
                if circle.pointable.direction.angle_to(circle.normal) <= Leap.PI/4:
                    action = GestureActions.clockwiseCircleGesture()
                else:
                    action = GestureActions.counterClockwiseCircleGesture()

                print action.next()

  def clockwiseCircleGesture():
       words = ["You", "spin", "me", "right", "round", "baby", "right", "round", "like", "a", "record", "baby",                          "Right", "round", "round", "round", "You", "spin", "me", "right", "round", "baby", "Right", "round", "like", "a",
         "record", "baby", "Right", "round", "round", "round"]
      for word in words:
          yield word

任何对这方面的洞察都会很好。谢谢

Answer 1

对python不熟悉，但当人们学习生成器/迭代器/枚举器/任何东西时，这是一个常见的问题。每次遍历循环时都会重新创建迭代器并丢失状态。在

而是最多创建一次

clockwise = GestureActions.clockwiseCircleGesture()
counter_clockwise = GestureActions.counterClockwiseCircleGesture()
# then

action = clockwise if foo else counter_clockwise

action.next()

Answer 2

我怀疑每次触发事件时，action变量都会被重置。在

在事件处理函数之外初始化生成器。看起来您可能想要两个，clockwise_action和{}。在