Promise All with Limit of Concurrent N

README.org

The Promise All Problem

in case of processing a very large array e.g. Promise.all(A_VERY_LARGE_ARRAY_OF_XHR_PROMISE)

which would probably blow you browser memory by trying to send all requests at the same time

solution is limit the concurrent of requests, and wrap promise in thunk

Promise.allConcurrent(2)([()=>fetch('BLAH1'), ()=>fetch('BLAH2'),...()=>fetch('BLAHN')])

if set concurrent to 2, it will send request BLAH1 and BLAH2 at the same time

if BLAH1 return response and resolved, will immediatly send request to BLAH3

in this way promise sending at the same time always keep the limit 2 which we’ve just configed before

prelude-test.js

describe('promiseAllStepN', function(){
  describe('3 tasks, and cucurrent is 2', function(){
    let tasks;
    beforeEach(function(){
      tasks = range(3).map(x=>sinon.stub())
    })
    describe('1 is finish', function(){
      it('will kickoff the third task', function(done){
        tasks[0].returns(Promise.resolve(0))
        let task2 = tasks[2]
        tasks[1].returns(new Promise(resolve=>setTimeout(()=>{
          expect(task2.called).to.be.equal(true)
          resolve(1)
          done()
        }, 1000)))
        tasks[2].returns(Promise.resolve(2))

        return Promise.allConcurrent(2)(tasks).then(x=>console.log(x))
      })
    })
  })

  describe('10 tasks, and cucurrent is 3', function(){
    let tasks;
    beforeEach(function(){
      tasks = range(10).map(x=>sinon.stub())
    })
    describe('1st is finish but 2nd stuck', function(){
      it.only('final task will run before 2nd', function(done){
        tasks.forEach((task,index) => task.returns(Promise.resolve(index)))
        let task10 = tasks[9]
        tasks[1].returns(new Promise(resolve=>setTimeout(()=>{
          expect(task10.called).to.be.equal(true)
          resolve(1)
          done()
        }, 1000)))
        return Promise.allConcurrent(2)(tasks).then(x=>console.log(x))
      })
    })
  })

})

prelude.js

function promiseAllStepN(n, list) {
  let tail = list.splice(n)
  let head = list
  let resolved = []
  let processed = 0
  return new Promise(resolve=>{
    head.forEach(x=>{
      let res = x()
      resolved.push(res)
      res.then(y=>{
        runNext()
        return y
      })
    })
    function runNext(){
      if(processed == tail.length){
        resolve(Promise.all(resolved))
      }else{
        resolved.push(tail[processed]().then(x=>{
          runNext()
          return x
        }))
        processed++
      }
    }
  })
}
Promise.allConcurrent = n => list =>  promiseAllStepN(n, list)

This is a great tool, thanks!

Not sure if it is intentional but I have noticed that if all the promises in a thread are rejected, the next thread is not executed. To prevent this I am catching the error and returning it.

results[curIndex] = await queue[curIndex]().catch(err => err);

@jacksonv1lle it happens because Promise.all rejects instantly upon any of promises reject. You can also wrap the call into try/catch:

try {
  results[curIndex] = await queue[curIndex]();
} catch(err) {
  console.log("ERROR", err);
  results[curIndex] = err;
}

The pure JS implementation without the callback hell :)

const pAll = async (queue, concurrency) => {
  let index = 0;
  const results = [];

  // Run a pseudo-thread
  const execThread = async () => {
    while (index < queue.length) {
      const curIndex = index++;
      // Use of `curIndex` is important because `index` may change after await is resolved
      results[curIndex] = await queue[curIndex]();
    }
  };

  // Start threads
  const threads = [];
  for (let thread = 0; thread < concurrency; thread++) {
    threads.push(execThread());
  }
  await Promise.all(threads);
  return results;
};

and a simple use case:

const test = async () => {
  const urls = ["url1", "url2", "url3"];
  const res = await pAll(
    urls.map(url => () => makeHTTPRequest(url)),
    5
  );
  console.log(res);
};

Thank you! This works! But can you explain how this code works? I don't understand the magic of how the splitting into a limited number of requests occurs?

I know this is four years ago, but given it's still a top google result, I wanted to share my approach / style too. I made a couple of decisions in my implementation:

• Typescript
• Readability is more important than Size
• Using Promise.race instead of Promise.all internally in order to maximize "worker" usage
• Using const generic types to support promise tuple types in returned Promise.all
• Guardless approach (loops just check queues, operations are only performed when needed without additional guards, eg. splice works with deleteCount=0 naturally and empty array is not processed)
• Mix of data types (Array for remainingTasks, Map for executingTasks and array again for finishedTasks) to have native solutions for performance optimizations

If you prefer some of those decisions for your own code then hopefully this solution is useful to you. I did not yet test the function very extensively, so there might be bugs I did not account for yet.

export type AsyncFunction<T = unknown> = () => Promise<T>;
export type TaskResult<T = unknown> = {
  index: number;
  value: T;
  task: AsyncFunction<T>;
};

export async function promiseAllMaxConcurrency<const Tasks extends AsyncFunction<unknown>[]>(
  tasks: Tasks,
  maxConcurrency = 4
): Promise<{ [K in keyof Tasks]: Awaited<ReturnType<Tasks[K]>> }> {
  const remainingTasks = [...tasks];
  const finishedTasks: Promise<unknown>[] = [];
  const executingTasks = new Map<number, Promise<TaskResult>>();
  do {
    const tasksToAdd = maxConcurrency - executingTasks.size;
    for (const task of remainingTasks.splice(0, tasksToAdd)) {
      const index = tasks.indexOf(task);
      const taskPromise = task().then((value) => ({
        index: tasks.indexOf(task),
        value,
        task,
      } satisfies TaskResult));
      executingTasks.set(index, taskPromise);
    }

    const finishedPromise = Promise.race(executingTasks.values());
    const result = await finishedPromise;
    finishedTasks[result.index] = finishedPromise.then(result => result.value);
    executingTasks.delete(result.index);
  } while (remainingTasks.length > 0 || executingTasks.size > 0);

  return Promise.all(finishedTasks) as Promise<{ [K in keyof Tasks]: Awaited<ReturnType<Tasks[K]>> }>;
}

promiseAllMaxConcurrency([
  () => new Promise<number>((resolve) => resolve(1)),
  () => new Promise<string>((resolve) => setTimeout(() => resolve('two'), 1000)),
  () => new Promise<string>((resolve) => setTimeout(() => resolve('three'), 1000)),
  () => new Promise<string>((resolve) => setTimeout(() => resolve('four'), 1000)),
  () => new Promise<string>((resolve) => setTimeout(() => resolve('long running...'), 5000)),
  () => new Promise<boolean>((resolve) => setTimeout(() => resolve(true), 1000)),
]).then((results) => {
  console.log(results);
}).catch(e => console.error(e));