Array reduce function with async await

Question

I'm trying to skip one object from an array objects based on a async operator. I've tried following cases, but getting a Type error.

Tried Method 1

newObjectArray = await Promise.all(objectAray.reduce(async (result, el) => {
  const asyncResult = await someAsyncTask(el);
  if (asyncResult) {
      result.push(newSavedFile);
  }
  return result;
}, []));

Tried Method 2

newObjectArray = await Promise.all(objectAray.reduce(async (prevPromise, el) => {
  const collection = await prevPromise;
  const asyncResult = await someAsyncTask(el);
  if (asyncResult) {
      prevPromise.push(newSavedFile);
  }

  collection.push(newSavedFile);
  return collection;
}, Promise.resolve([])));

Error

'TypeError: #<Promise> is not iterable',
'    at Function.all (<anonymous>)',

Could you please explain what is the desired output?

gr4viton
– gr4viton

2018-10-05 22:39:38 +00:00
Commented Oct 5, 2018 at 22:39 — gr4viton
– gr4viton, Commented Oct 5, 2018 at 22:39

Jonas Wilms · Accepted Answer · 2018-10-05 22:40:39Z

15

In your first try, result is a promise as all async functions evaluate to a promise when called, so you have to await result before you can push to the array, and then you don't need the Promise.all:

 newObjectArray = await objectAray.reduce(async (result, el) => {
  const asyncResult = await someAsyncTask(el);
  if (asyncResult) {
    (await result).push(newSavedFile);
  }
  return result;
}, []);

But I'd guess that it is way faster to just filter afterwards:

 newObjectArray = (await Promise.all(objArray.map(someAsyncTask))).filter(el => el);

answered Oct 5, 2018 at 22:40

Jonas Wilms

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3 Comments

Ankit Balyan Over a year ago

Isn't filter will cost on double iteration over the array.

Jonas Wilms Over a year ago

@ankit yup bat that might cost you a few nanoseconds, while executing one someAsyncTask after another will take you much more. Just put a Date.now() before and after this, and see how both perform ...

To Kra Over a year ago

thanks for your answer. Its not possible to use Promise.all() when I reduce to Object. So I love your answer which helped that purpose :) thank you!

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