Generate a stream with delays for each emit in dart

Question

I want to emit values in a stream from a list with a specific delay in dart.

So, from [1,2,3], which is a regular List I want to emit values like 1...2...3 in separated events.

I tried something like this

List<int> myList = [1,2,3];
Subject _current = BehaviorSubject<int>();
Stream<int> get current$ => _current.stream.delay(Duration(seconds:1));

myList.forEach(current.add);

But I get ...123 instead, so this is delaying the whole stream 1 seconds, rather than every value in the list.

Any ideas? Thanks

score 4 · Answer 1 · answered Feb 22 '22 at 20:51

4

If you'd like to avoid rxdart, you can use simple transformation function like:

Stream<T> streamDelayer<T>(Stream<T> inputStream, Duration delay) async* {
  await for (final val in inputStream) {
    yield val;
    await Future.delayed(delay);
  }
}

void main() {
  var s = Stream.fromIterable([1, 2, 3]);
  streamDelayer(s, Duration(seconds: 1)).forEach(print);
}

answered Feb 22 '22 at 20:51

user2424794

209
1
8

2

moving yield below the await will delay the first value also. – Polymath Jun 27 '22 at 13:41

score 1 · Answer 2 · answered Jan 06 '21 at 06:32

1

One solution using rxdart 0.25.0:

final Stream<int> stream = 
    Rx.range(1, 3)
      .asyncExpand((event) => 
        event == 1
        ? Stream.value(event) // return first value immediately
        : Rx.timer(event, Duration(seconds: 1))
    );

answered Jan 06 '21 at 06:32

David Lawson

7,802
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score 1 · Answer 3 · answered Feb 05 '21 at 20:25

1

using interval from rxdart package

Stream.fromIterable(List<int>.generate(100, (i) => i))
        .interval(Duration(milliseconds: 10))
        .listen((v) {
      print(v);
});

answered Feb 05 '21 at 20:25

Ali80

6,333
2
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It's emitting values like: `...1...2...3` But it should emit like: `1...2...3` – David Araujo Aug 19 '22 at 16:07
@DavidAraujo if you mean it starts from 0 and should start from 1, change `=> i` to `=> i+1` – Ali80 Aug 19 '22 at 18:27
@Ali80 I think the original commenter meant that they don't want an initial delay before the first emitted event. – masterwok Jan 05 '23 at 19:38

score 0 · Answer 4 · answered Jan 03 '19 at 15:22

0

The rxdart

package provides a delay transformer or method that does that

new Observable.fromIterable([1, 2, 3, 4])
  .delay(new Duration(seconds: 1))
  .listen(print); // [after one second delay] prints 1, 2, 3, 4 immediately

There are others like debounce

answered Jan 03 '19 at 15:22

Günter Zöchbauer

623,577
216
2,003
1,567

Thanks, but that's doing the same that I have, it waits for 1 second and then prints immediately; what I want is to print 1, wait 1 second, print 2, wait 1 second, print 3, wait 1 second, print 4, wait 1 second. – David Jan 03 '19 at 15:26
1

I actually got the answer from rxdart itself, using concatMap, thanks. – David Jan 03 '19 at 15:35

score 0 · Answer 5 · answered Jan 03 '19 at 15:39

0

I managed to get it working

Observable.range(myList)
  .concatMap((i) =>
    new Observable.timer(i, new Duration(seconds: 1))
  .listen(print);

From the rxdart documentation itself. Thanks Günter for the good tip.

answered Jan 03 '19 at 15:39

David

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score 0 · Answer 6 · answered Aug 27 '23 at 21:04

Best solution using StreamController, which gives you the ability to control the stream using pause and resume as follows:

List<int> myList = [1, 2, 3];
StreamController current = StreamController();
late StreamSubscription listenController;

void waitFunction() async {
  listenController.pause();
  await Future.delayed(const Duration(seconds: 2));
  listenController.resume();
}

myList.forEach(current.add);

listenController = current.stream.listen((event) async {
  print(event);
  waitFunction();
});

Generate a stream with delays for each emit in dart

6 Answers6