无初始延迟去抖 [英] Debounce without initial delay

问题描述

RxJS 中是否有一个操作符可以在不延迟突发中的第一个事件"的情况下进行去抖动，但会延迟(并始终发出)突发中的最后一个事件"?

Is there an operator in RxJS that debounces without delaying the "first event in a burst", but delaying (and always emitting) the "last event in a burst"?

像这样:

---a----b-c-d-----e-f---

在 awesome-debounce(2 dashes) 之后变成:

---a----b------d--e----f

而正常的去抖动是:

-----a---------d-------f

这是油门和去抖动的混合......

It's kind of a mix between throttle and debounce...

推荐答案

嗯，这是我能想到的最简单的解决方案.对您来说有趣的部分是创建子链的 awesomeDebounce() 函数.

Hmmm, this is the easiest solution I can think of. The interesting part for you is the awesomeDebounce() function that creates the sub-chain.

它基本上只是结合了 throttle() 和 debounceTime() 操作符:

It basically just combines throttle() and debounceTime() operators:

const Rx = require('rxjs');
const chai = require('chai');

let scheduler = new Rx.TestScheduler((actual, expected) => {
  chai.assert.deepEqual(actual, expected);
  console.log(actual);
});

function awesomeDebounce(source, timeWindow = 1000, scheduler = Rx.Scheduler.async) {
  let shared = source.share();
  let notification = shared
      .switchMap(val => Rx.Observable.of(val).delay(timeWindow, scheduler))
      .publish();

  notification.connect();

  return shared
    .throttle(() => notification)
    .merge(shared.debounceTime(timeWindow, scheduler))
    .distinctUntilChanged();
}

let sourceMarbles =   '---a----b-c-d-----e-f---';
let expectedMarbles = '---a----b------d--e----f';

// Create the test Observable
let observable = scheduler
  .createHotObservable(sourceMarbles)
  .let(source => awesomeDebounce(source, 30, scheduler));

scheduler.expectObservable(observable).toBe(expectedMarbles);
scheduler.flush();

内部 notification Observable 仅用于 throttle() 运算符，因此我可以在需要时手动重置其计时器.我还必须将这个 Observable 变成 "hot" 以独立于来自 throttle() 的内部订阅.

The inner notification Observable is used only for the throttle() operator so I can reset its timer manually when I need. I also had to turn this Observable into "hot" to be independent on the internal subscriptions from throttle().

这篇关于无初始延迟去抖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！