Hello, If you have events:
A@10m, B@14m, Watermark@15m C@16m D@25m, Watermark@30m then the result will be: [A,B]@14.59 [C,D]@29.59 This means that for the next windowing, you will have the elements [A,B]@14.59 and [C,D]@29.59 to window in windows of 5min. Given that they are 15min apart, this means that you will have: [A,B]@14.59 in one window and [C,D]@29.59 This is what I meant by at most one element per window, because window 0 to 5 min, 5 to 10 and 15 to 20 and 20 to 25 will be empty. Does this make it clearer? Kostas > On Nov 30, 2016, at 11:48 AM, Janardhan Reddy <janardhan.re...@olacabs.com> > wrote: > > HI > i didn't get it , can you please clarify with an example in case each of > operation A and B emit multiple elements. > > On Wed, Nov 30, 2016 at 3:34 PM, Kostas Kloudas <k.klou...@data-artisans.com > <mailto:k.klou...@data-artisans.com>> wrote: > Hi Janardhan, > > After the first windowing operation, the timestamp of the emitted element for > each window > will be the (endOfWindow - 1). So in your case, in the second windowing > operation (window by 5) > there will be at most one element per window. > > I hope this answers your question. > > Kostas > > > On Nov 29, 2016, at 7:25 PM, Janardhan Reddy <janardhan.re...@olacabs.com > > <mailto:janardhan.re...@olacabs.com>> wrote: > > > > Hi, > > > > Suppose we have a stream like this. > > > > someStream.timeWindow(Time.minutes(15)).apply { > > operation A > > > > }.keyby("....").window(TumblingEventTimeWindows.ofseconds(5)).apply { > > > > operation B > > > > }.keyby("....").window(TumblingEventTimeWindows.ofseconds(5)).apply { > > > > operation C > > > > } > > > > Say operation A emits some elements => it would be emitted every 15 minutes. > > > > How would be the window behaviour of where operation B takes place if > > operation A takes more than 5 seconds with ingestion Time characteristic. > > Similarly how would windows behave near operation C if operation B takes > > more than 5 seconds. > > > > > > Thanks > > Janardhan > >
