RxJS_Q-001.js

//1. 正常 情況
var people = [
    {name: 'Anna', score: 100, subject: 'English'},
    {name: 'Anna', score: 90, subject: 'Math'},
    {name: 'Anna', score: 96, subject: 'Chinese' }, 
    {name: 'Jerry', score: 80, subject: 'English'},
    {name: 'Jerry', score: 100, subject: 'Math'},
    {name: 'Jerry', score: 90, subject: 'Chinese' }, 
];
var source = Rx.Observable
	.from(people)
	.zip(Rx.Observable.interval(100),(x, y) => x);

var example = source
	.groupBy(person => person.name)
	.map(group => group.reduce((acc, curr) => ({ 
	    name: curr.name,
	    score: curr.score + acc.score 
	})))
	.mergeAll();
	
example.subscribe(console.log);

RxJS_Q-002.js

//2. 加上.take(3) 變更 people 順序，為什麼輸出多出subject欄位
var people = [
    {name: 'Anna', score: 100, subject: 'English'},
    {name: 'Jerry', score: 80, subject: 'English'},
    {name: 'Anna', score: 90, subject: 'Math'},
    {name: 'Anna', score: 96, subject: 'Chinese' }, 
    {name: 'Jerry', score: 100, subject: 'Math'},
    {name: 'Jerry', score: 90, subject: 'Chinese' }, 
];
var source = Rx.Observable
	.from(people)
	.zip(Rx.Observable.interval(100).take(3),(x, y) => x);

var example = source
	.groupBy(person => person.name)
	.map(group => group.reduce((acc, curr) => ({ 
	    name: curr.name,
	    score: curr.score + acc.score 
	})))
	.mergeAll();
	
example.subscribe(console.log);

RxJS_Q-003.js

//3. 把 mergeAll 換成 concatAll，為什麼只輸出一個
var people = [
    {name: 'Anna', score: 100, subject: 'English'},
    {name: 'Anna', score: 90, subject: 'Math'},
    {name: 'Anna', score: 96, subject: 'Chinese' }, 
    {name: 'Jerry', score: 80, subject: 'English'},
    {name: 'Jerry', score: 100, subject: 'Math'},
    {name: 'Jerry', score: 90, subject: 'Chinese' }, 
];
var source = Rx.Observable
	.from(people)
	.zip(Rx.Observable.interval(100),(x, y) => x);

var example = source
	.groupBy(person => person.name)
	.map(group => group.reduce((acc, curr) => ({ 
	    name: curr.name,
	    score: curr.score + acc.score 
	})))
	.concatAll();
	
example.subscribe(console.log);