Sorry I mistyped the code, it should be
*timeWindow**(Time.minutes(10))* instead
of *window**(Time.minutes(10)).*
On Wed, Aug 24, 2016 at 9:29 PM, Yassine Marzougui
wrote:
> Hi subash,
>
> A stream is infinite, hence it has no notion of "final" count. To get
> distinct counts you need to define
Hi subash,
A stream is infinite, hence it has no notion of "final" count. To get
distinct counts you need to define a period (= a window [1] ) over which
you count elements and emit a result, by adding a winow operator before the
reduce.
For example the following will emit distinct counts every 10
Hello Kostas,
Sorry for late reply. But I couldn't understand how to apply split in
datastream, such as in below to get the distinct output stream element with
the count after applying group by and reduce.
DataStream> gridWithDensity =
pointsWithGridCoordinates.map(new AddCountAppender())
.keyBy(
Hi Subash,
You should also split your elements in windows.
If not, Flink emits an element for each incoming record.
That is why you have:
(1,1)
(1,2)
(1,3)
…
Kostas
> On Aug 22, 2016, at 5:58 PM, subash basnet wrote:
>
> Hello all,
>
> I grouped by the input based on it's id to count the n
Hello all,
I grouped by the input based on it's id to count the number of elements in
each group.
DataStream> gridWithCount;
Upon printing the above datastream it shows with duplicate rows:
Output:
(1, 1)
(1,2)
(2,1)
(1,3)
(2,2)...
Whereas I wanted the distinct rows with final count:
Needed O
