RXJS:聚合去抖动 [英] RXJS: Aggregated debounce

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本文介绍了RXJS:聚合去抖动的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我的用例如下:我得到事件,这些事件有时会突然发生.如果发生突发,我只需要处理一次.去抖就是这样做的.

然而,去抖动只给我一个突发的最后一个元素,但我需要知道突发中的所有元素才能聚合它们(使用平面图).

这可以通过定时窗口或缓冲区来完成,但是,这些是固定间隔,因此缓冲区/窗口超时可能发生在突发的中间,因此将突发分成 2 部分而不是 1 部分来处理.

所以我想要的是像

<预><代码>..事件:一个..->一个......事件:b.事件:c.事件:d..->乙、丙、丁.....事件:e..->电子.

解决方案

这可以通过 buffer 通过将去抖动流作为关闭选择器传入,例如:

var s = Rx.Observable.of('a').merge(Rx.Observable.of('b').delay(100)).merge(Rx.Observable.of('c').delay(150)).merge(Rx.Observable.of('d').delay(200)).merge(Rx.Observable.of('e').delay(300)).分享();s.buffer(s.debounce(75)).subscribe(x => console.log(x));

这是一个可运行的版本:https://jsbin.com/wusasi/edit?js,控制台,输出

My use case is as following: I get events, which sometimes happen in bursts. If a burst occurs, I only need to handle it once though. Debounce does this.

However, debounce only gives me the last element of a burst, but I need to know about all elements in a burst to aggregate on them (using flatmap).

This could be done by a timed window or buffer, however, these are fixed intervals, so a buffer/window timeout could occur in the middle of a burst, therefore splitting the burst in 2 parts to handle instead of 1.

So what I'd want is something like

.
.
event: a
.
. -> a
.
.
.
.
.
.event: b
.event: c
.event: d
.
.-> b,c,d
. 
.
.
.
.event : e
.
. -> e
.

解决方案

This can be achieved with buffer by passing a debounced stream in as a closing selector, e.g.:

var s = Rx.Observable.of('a')
  .merge(Rx.Observable.of('b').delay(100))
  .merge(Rx.Observable.of('c').delay(150))
  .merge(Rx.Observable.of('d').delay(200))
  .merge(Rx.Observable.of('e').delay(300))
  .share()
;

s.buffer(s.debounce(75)).subscribe(x => console.log(x));

Here's a runnable version: https://jsbin.com/wusasi/edit?js,console,output

这篇关于RXJS:聚合去抖动的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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