Javascript lazy enumerable

Question

Is there any way to convert regular array into lazy enumerable? I want the code below produces only one log, not ten ones:

const items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
items.map((i) => {
    console.log('expensive process');
    return i;
  })
  .findIndex((i) => true);

To answer your question more accurately, please describe your use cases... — Nur
– Nur, Commented May 19, 2021 at 10:37
i want doing something like this const goodChild = Array.from(parent.children).filter(calculateCondition).map(c => { child: c, area: calcArea(c) })[0] — Heyyy Marco
– Heyyy Marco, Commented May 19, 2021 at 10:51

Matvii Hodovaniuk · Accepted Answer · 2021-05-19 10:42:24Z

1

You can use the Array.find method to stop right after you find the right one element.

const items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];


items.find((i) => {
    if(i){
        console.log('expensive process');
    }
    
    return i;
});

answered May 19, 2021 at 10:42

Matvii Hodovaniuk

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2 Comments

Heyyy Marco Over a year ago

i can't do that, there are severals expensive .map & .filter before the last .find

Matvii Hodovaniuk Over a year ago

@HeyyyMarco You can't stop .map. You can use yeild to pause and resume a generator function. You can use async/await: const arr = [1, 2, 3]; const asyncRes = await Promise.all(arr.map(async (i) => { await sleep(10); return i + 1; })); console.log(asyncRes); // 2,3,4

trincot · Accepted Answer · 2024-11-14 21:45:01Z

1

Since ECMAScript 2025 we have iterator helpers, including map and find methods, such that we get the "lazy" behaviour. For your example you can consume the map-returned iterator with find, and if its callback returns a truthy value, nothing more will be consumed from the map-returned iterator:

const items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];

const item = items.values().map((i) => {
    console.log('expensive process');
    return i;
})
.find((i) => true);

console.log("item is", item);

answered Nov 14, 2024 at 21:45

trincot

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