rxjs map operator evaluated for each subscriber

Question

Why is the map operator evaluated for each subscriber instead of once?

const obs1 = Rx.Observable.interval(1000).take(1).map((x, i) => {
  console.log(i+1 + ':1 map')
  return 'obs1';
})

const obs2 = Rx.Observable.interval(1300).take(1).map((x, i) => {
  console.log(i+1 + ':2 map')
  return 'obs2';
})

const obs3 = Rx.Observable.interval(1700).take(2).map((x, i) => {
  console.log(i+1 + ':3 map')
  return 'obs3';
})

const view = obs1.combineLatest(obs2, obs3, (obs1, obs2, obs3) => {                 return obs1 + ', ' + obs2 + ', ' + obs3; });

// Every subscriber adds more calls to map - why is it called multiple     times at the same time ?

view.subscribe((value) => {
  console.log('sub1: ' + value)
});

view.subscribe((value) => {
  console.log('sub2: ' + value)
});

view.subscribe((value) => {
  console.log('sub3: ' + value)
});

I created a testcase here: http://jsbin.com/jubinuf/3/edit?js,console

Can I write this testcase differently to avoid this behaviour?

Answer 1

Every subscriber will run through the Observable sequence. If you want everyone to get the resulting stream instead, use .publish().refCount().

http://jsbin.com/ditikonopi/edit?js,console

The .publish() will return an observable sequence that shares a single subscription to the underlying sequence. refCount() will stay connected to the source so long as there is at least one subscription.