默认情况下,.windowedBy(SessionWindows.with(Duration.ofSeconds(60))
为每个传入记录返回一条记录。
结合.count()
和.filter()
,可以轻松检索第一条记录。
用 .suppress(Suppressed.untilWindowCloses(unbounded()))
检索上一条记录也很容易。
所以。。。我做了两次处理,如你所看到的适应字数示例:
final KStream<String, String> streamsBranches = builder.<String,String>stream("streams-plaintext-input");
streamsBranches
.flatMapValues(value -> Arrays.asList(value.toLowerCase(Locale.getDefault()).split("\W+")))
.groupBy((key, value) -> ""+value)
.windowedBy(SessionWindows.with(Duration.ofSeconds(60)).grace(Duration.ofSeconds(2)))
.count(Materialized.with(Serdes.String(), Serdes.Long()))
.toStream()
.map((wk, v) -> new KeyValue<>(wk.key(), v == null ? -1l : v))
.filter((wk, v) -> v == 1)
.to("streams-wordcount-output", Produced.with(Serdes.String(), Serdes.Long()));
streamsBranches
.flatMapValues(value -> Arrays.asList(value.toLowerCase(Locale.getDefault()).split("\W+")))
.groupBy((key, value) -> ""+value)
.windowedBy(SessionWindows.with(Duration.ofSeconds(60)).grace(Duration.ofSeconds(2)))
.count(Materialized.with(Serdes.String(), Serdes.Long()))
.suppress(Suppressed.untilWindowCloses(unbounded()))
.toStream()
.map((wk, v) -> new KeyValue<>(wk.key(), v))
.filter((wk, v) -> v != null)
.to("streams-wordcount-output", Produced.with(Serdes.String(), Serdes.Long()));
但我想知道是否有一种更简单、更美丽的方法来做同样的事情。
我认为您应该使用SessionWindowedKStream::aggregate(...)
并根据您的逻辑在聚合器中累积结果(第一个和最后一个值(
示例代码可能如下所示:
streamsBranches.groupByKey()
.windowedBy(SessionWindows.with(Duration.ofSeconds(60)).grace(Duration.ofSeconds(2)))
.aggregate(
AggClass::new,
(key, value, oldAgg) -> oldAgg.update(value),
(key, agg1, agg2) -> agg1.merge(agg2),
Materialized.with(Serdes.String(), new AggClassSerdes())
).suppress(Suppressed.untilWindowCloses(unbounded()))
.toStream().map((wk, v) -> new KeyValue<>(wk.key(), v))
.to("streams-wordcount-output", Produced.with(Serdes.String(), new AggClassSerdes()));
其中AggClass
是累加器,AggClassSerdes
是该累加器的Serdes
public class AggClass {
private String first;
private String last;
public AggClass() {}
public AggClass(String first, String last) {
this.first = first;
this.last = last;
}
public AggClass update(String value) {
if (first == null)
first = value;
last = value;
return this;
}
public AggClass merge(AggClass other) {
if (this.first == null)
return other;
else return new AggClass(this.first, other.last);
}
}