Paul Murray Paul Murray - 1 year ago 85
Javascript Question

Group Observables and then filter by count

I'm having trouble filtering an Observable that is based on a


Here is an example based on the

var codes = [
{ keyCode: 38}, // up
{ keyCode: 38}, // up
{ keyCode: 40}, // down
{ keyCode: 40}, // down
{ keyCode: 37}, // left
{ keyCode: 39}, // right
{ keyCode: 37}, // left
{ keyCode: 39}, // right
{ keyCode: 66}, // b
{ keyCode: 65} // a

var source = Rx.Observable.from(codes)
function (x) { return x.keyCode; },
function (x) { return x.keyCode; }
.flatMap(obs =>
.filter(x => x > 1)


This returns:


So, the Observables are being filtered, but their values are gone.

Answer Source

It is tricky to reason about observables as a single object (say, an array) instead of as a asynchronous sequence of values. What happens here is :

  • you take the obs observable and then you apply count to it

    count will wait for the observable to finish and will output one value, the number of elements and then finish.

  • you filter that value, but then you are applying toArray to obs which is already finished. That's why you get an empty array.

So the problem here is the same obs variable is a different thing at a different time, so you cannot avoid having to reason about time, when you reuse the same variable at different points of your program.

Now, if you want to get the count number out to the subscribe, you should be able to directly write obs.count().filter(x => x > 1).toArray() instead of what you have right now. flatMap accepts promise and arrays in addition to observables as the return value of the selector function :


UPDATE: with the comments taken into account, this is what I came up with (to be tested) :

var source = Rx.Observable
    .scan(function ( acc, code ) {
            acc[code] = acc[code] + 1;
            acc.latest = code;
            return acc;
          }, {})
    .filter(function ( acc ) {return acc[acc.latest] > 1;})
    .flatMap(function ( acc ) {return acc[acc.latest] == 2? Rx.Observable.repeat(acc.latest, 2) : Rx.just(acc.latest);})

If you can wait to have all the values processed, you can stop the above after the filter and add .last(). This will get you the last accumulator with the count of all values. From there you can do what you want.

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