合并为flatMap

Question

Theoretically it should be possible to implement any RxJS operator (except just() and flatMap() ) through flatMap() . For instance map() can be implemented as

function map(source, selector) {
  return source.flatMap(x => Rx.Observable.just(selector(x)));
}

How to implement merge() through flatMap() ? (avoiding mergeAll() too, of course)

Answer 1

It looks possible if you take advantage of the fact that flatMap can also take array return values.

Rx.Observable.prototype.merge = function(other) {
  var source = this;
  return Rx.Observable.just([source, other])
           //Flattens the array into observable of observables
           .flatMap(function(arr) { return arr; })
           //Flatten out the observables
           .flatMap(function(x) { return x; });
}

EDIT 1

Using RxJS 6 and the pipe syntax

import {of} from 'rxjs'
import {flatMap} from 'rxjs/operators'

function merge (other) {
  return source => of([source, other]).pipe(
           //Flattens the array into observable of observables
           flatMap(arr => arr)
           //Flatten out the observables
           flatMap(x => x)
         );
}

 const {timestamp, map, flatMap, take} = rxjs.operators; const {interval, of: just} = rxjs; const source1 = interval(2000).pipe( timestamp(), map(x => "Interval 1 at " + x.timestamp + " w/ " + x.value) ) const source2 = interval(3000).pipe( timestamp(), map(x => "Interval 2 at " + x.timestamp + " w/ " + x.value) ) function mergeFromFlatMap (other) { return source => just([source, other]).pipe( flatMap(arr => arr), flatMap(seq => seq) ) } source1.pipe( mergeFromFlatMap(source2), take(20) ).subscribe(console.log.bind(console)); 

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