问题描述
zio-streams提供throttleShape
,
/**
* Delays the chunks of this stream according to the given bandwidth parameters using the token bucket
* algorithm. Allows for burst in the processing of elements by allowing the token bucket to accumulate
* tokens up to a `units + burst` threshold. The weight of each chunk is determined by the `costFn`
* function.
*/
final def throttleShape(units: Long,duration: Duration,burst: Long = 0)(
costFn: Chunk[O] => Long
): ZStream[R with Clock,E,O]
我正在努力了解如何使用参数unit
,duration
burst
和costFun
。根据我对token bucket
throttleShape(1,1.second)(_ => 1)
处理一个元素的成本为一个令牌(costFun = _ => 1
),一秒钟后会补充一个令牌(unit = 1
)(duration = 1.second
)。但是,除了
throttleShape(1,1.second)(_ => 2)
这使其挂起。例如,如何解释以下使用无穷持续时间的片段(来自PR)的节流
Stream(1,2,3,4)
.throttleShape(1,Duration.Infinity)(_ => 0)
.runCollect
Stream(1,4)
.throttleShape(2,Duration.Infinity)(_ => 1)
.take(2)
.runCollect
具体来说,假设我想每分钟最多处理100个元素,那么应该如何指定throttleShape
?
解决方法
问题在于您的初始流是单个Chunk[Int]
,并且在throttleShape
中是注释中所说的-您按块而不是按元素来限制。
单个块是根据Stream(1,2,3,4)
构造的,因为它对应于
/**
* Creates a pure stream from a variable list of values
*/
def apply[A](as: A*): ZStream[Any,Nothing,A] = fromIterable(as)
其中
/**
* Creates a stream from an iterable collection of values
*/
def fromIterable[O](as: => Iterable[O]): ZStream[Any,O] =
fromChunk(Chunk.fromIterable(as))
因此,如果要限制元素,则应按.chunkN(1)
将块重新缩放为1个元素。您应该在节流之前这样做。
如果是
说我想每分钟最多处理100个元素
如果您不需要块的优化(以批量/块方式处理项目),则可以将块缩放到1,然后throttleShape(100,1.minute)(_ => 1)
stream.Stream.fromIterable(1 to 1000)
.chunkN(1)
.throttleShape(100,1.minute)(_ => 1)
.foreachChunk(chunk => console.putStrLn(s"processed '${chunk.foldLeft("")(_ + _)}'"))
或者,如果您要分块处理并保持相同的处理速率-您可以将costFn
写为_.size
:
stream.Stream.fromIterable(1 to 1000)
.chunkN(5)
.throttleShape(100,1.minute)(_.size)
.foreachChunk(chunk => console.putStrLn(s"processed '${chunk.foldLeft("")(_ + _)}'"))