Nope, we're setting retainDuplicates to false.
On Tue, Jul 3, 2018 at 6:55 AM, Damian Guy <damian....@gmail.com> wrote:
> Hi,
>
> When you create your window store do you have `retainDuplicates` set to
> `true`? i.e., assuming you use `Stores.persistentWindowStore(...)` is the
> last param `true`?
>
> Thanks,
> Damian
>
> On Mon, 2 Jul 2018 at 17:29 Christian Henry <christian.henr...@gmail.com>
> wrote:
>
> > We're using the latest Kafka (1.1.0). I'd like to note that when we
> > encounter duplicates, the window is the same as well.
> >
> > My original code was a bit simplifier -- we also insert into the store if
> > iterator.hasNext() as well, before returning null. We're using a window
> > store because we have a punctuator that runs every few minutes to count
> > GUIDs with similar metadata, and reports that in a healthcheck. Since our
> > healthcheck window is less than the retention period of the store
> > (retention period might be 1 hour, healthcheck window is ~5 min), the
> > window store seemed like a good way to efficiently query all of the most
> > recent data. Note that since the healthcheck punctuator needs to
> aggregate
> > on all the recent values, it has to do a *fetchAll(start, end) *which is
> > how these duplicates are affecting us.
> >
> > On Fri, Jun 29, 2018 at 7:32 PM, Guozhang Wang <wangg...@gmail.com>
> wrote:
> >
> > > Hello Christian,
> > >
> > > Since you are calling fetch(key, start, end) I'm assuming that
> > > duplicateStore
> > > is a WindowedStore. With a windowed store, it is possible that a single
> > key
> > > can fall into multiple windows, and hence be returned from the
> > > WindowStoreIterator,
> > > note its type is <Windowed<K>, V>
> > >
> > > So I'd first want to know
> > >
> > > 1) which Kafka version are you using.
> > > 2) why you'd need a window store, and if yes, could you consider using
> > the
> > > single point fetch (added in KAFKA-6560) other than the range query
> > (which
> > > is more expensive as well).
> > >
> > >
> > >
> > > Guozhang
> > >
> > >
> > > On Fri, Jun 29, 2018 at 11:38 AM, Christian Henry <
> > > christian.henr...@gmail.com> wrote:
> > >
> > > > Hi all,
> > > >
> > > > I'll first describe a simplified view of relevant parts of our setup
> > > (which
> > > > should be enough to repro), describe the behavior we're seeing, and
> > then
> > > > note some information I've come across after digging in a bit.
> > > >
> > > > We have a kafka stream application, and one of our transform steps
> > keeps
> > > a
> > > > state store to filter out messages with a previously seen GUID. That
> > is,
> > > > our transform looks like:
> > > >
> > > > public KeyValue<byte[], String> transform(byte[] key, String guid) {
> > > > try (WindowStoreIterator<DuplicateMessageMetadata> iterator =
> > > > duplicateStore.fetch(correlationId, start, now)) {
> > > > if (iterator.hasNext()) {
> > > > return null;
> > > > } else {
> > > > duplicateStore.put(correlationId, some metadata);
> > > > return new KeyValue<>(key, message);
> > > > }
> > > > }}
> > > >
> > > > where the duplicateStore is a persistent windowed store with caching
> > > > enabled.
> > > >
> > > > I was debugging some tests and found that sometimes when calling
> > > > *all()* or *fetchAll()
> > > > *on the duplicate store and stepping through the iterator, it would
> > > return
> > > > the same guid more than once, even if it was only inserted into the
> > store
> > > > once. More specifically, if I had the following guids sent to the
> > stream:
> > > > [11111, 22222, ... 99999] (for 9 values total), sometimes it would
> > return
> > > > 10 values, with one (or more) of the values being returned twice by
> the
> > > > iterator. However, this would not show up with a *fetch(guid)* on
> that
> > > > specific guid. For instance, if 11111 was being returned twice by
> > > > *fetchAll()*, calling *duplicateStore.fetch("11111", start, end)*
> will
> > > > still return an iterator with size of 1.
> > > >
> > > > I dug into this a bit more by setting a breakpoint in
> > > > *SegmentedCacheFunction#compareSegmentedKeys(cacheKey,
> > > > storeKey)* and watching the two input values as I looped through the
> > > > iterator using "*while(iterator.hasNext()) { print(iterator.next())
> > }*".
> > > In
> > > > one test, the duplicate value was 66666, and saw the following
> behavior
> > > > (trimming off the segment values from the byte input):
> > > > -- compareSegmentedKeys(cacheKey = 66666, storeKey = 22222)
> > > > -- next() returns 66666
> > > > and
> > > > -- compareSegmentedKeys(cacheKey = 77777, storeKey = 66666)
> > > > -- next() returns 66666
> > > > Besides those, the input values are the same and the output is as
> > > expected.
> > > > Additionally, a coworker noted that the number of duplicates always
> > > matches
> > > > the number of times *Long.compare(cacheSegmentId, storeSegmentId)
> > > *returns
> > > > a non-zero value, indicating that duplicates are likely arising due
> to
> > > the
> > > > segment comparison.
> > > >
> > >
> > >
> > >
> > > --
> > > -- Guozhang
> > >
> >
>
