如果没有匹配的累加器，RxJS如何减少？

问题描述

我正在浏览Reactive Programming in JavaScript上的一篇文章，不确定其中列出的以下示例如何产生输出27

import {Observable} from 'rxjs-es';

let output = Observable.interval(500)
             .map(i => [1,2,3,4,5,6][i]);

let result = output.map(num1 => num1)
    .filter(num1 => num1 > 4)
    .reduce((num1,num2) => num1 + num2);

result.subscribe(number => console.log(number));

Output --> 27

据我了解，每隔500毫秒，流[1,6]中的每个数字都会被一个一个地过滤。因此，只有5和6可以通过过滤器。

但是，由于在处理过程中的任何给定点上只有一个元素可用，所以我想知道reduce如何将结果累加为27？

解决方法

我已将其翻译为RxJs 6，但不输出27

const { interval } = rxjs;
const { map,filter,reduce,take } = rxjs.operators;

let output = interval(500).pipe(
 map(i => [1,2,3,4,5,6][i]),// very strange way of adding one to the interval
 take(6) // had to add a take so observable would complete else reduce would never emit
);

let result = output.pipe(
  map(num1 => num1),// Does nothing
  filter(num1 => num1 > 4),// filters out those less than 5
  reduce((num1,num2) => num1 + num2) // add the leftovers 5 and 6
);

result.subscribe(number => console.log(number)); // 11 is the output

<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.6.2/rxjs.umd.min.js"></script>