Creating a sequential publisher in Swift Combine

Question

I want to debounce a batch of events and process them after a delay of ~1.5 seconds. Here's what I've done.

class ViewModel: ObservableObject { @Published var pending: [TaskInfo] private var cancellable: AnyCancellable? = nil init() { processPendingTasks() } func queueTask(task: TaskInfo) { pending.append(task) } private func processPendingTasks() { cancellable = $pendingTasks .debounce(for: 1.5, scheduler: RunLoop.main) .sink(receiveValue: { batch in // Iterate though elements and process events. }) } }

Issue: This works fine, but the issue that I've is that it performs unnecessary view updates since the array is tagged @Published.

What I'm looking for: The ideal approach would be a streaming setup where I get all events (in a batched fashion) but the sink should wait exactly for 1.5 seconds after the last event was added.

I tried PassthroughSubject, but it seems like it only gets me the last event that happened in the last 1.5 seconds.

vadian · Accepted Answer · 2021-12-25 18:28:33Z

1

A possible solution is a combination of a PassthroughSubject and the collect operator. In queueTask send the tasks to the subject.

func queueTask(task: TaskInfo) { subject.send(task) }

1.5 seconds after receiving the last item send

subject.send(completion: .finished)

and subscribe

subject .collect() .sink { [weak self] tasks in self?.pending = tasks }

If the interval of the incoming tasks is < 1.5 seconds you could also use the .timeout(1.5) operator which terminates the pipeline after the timeout interval.

answered Dec 25, 2021 at 18:28

vadian

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4 Comments

Siddharth Kamaria Over a year ago

This is a nice solution. I tried it and I've only one issue now. Since I'm calling .finished, the sink is only called once. What I want is to continue the stream even after the sink is called, i.e. after one batch is finished processing - it should start over again listening for a new batch.

vadian Over a year ago

As you wrote after the last event was added I assumed that the transmission was finished after the last item. Alternatively you could use collect(n) which emits the items in chunks of n.

Siddharth Kamaria Over a year ago

Yeah I ended up using the .collect(.byTime) pattern with a delay of 1.5. It works for my use case although not exactly what I asked in the question.

Siddharth Kamaria Over a year ago

collect() was the missing piece that I was looking for. Thank you for that!

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