Convert first N item in iterable to Array

Question

Something similar to question Convert ES6 Iterable to Array. But I only want first N items. Is there any built-in for me to do so? Or how can I achieve this more elegantly?

let N = 100;
function *Z() { for (let i = 0; ; i++) yield i; }

// This wont work
// Array.from(Z()).slice(0, N);
// [...Z()].slice(0, N)

// This works, but a built-in may be preferred
let a = [], t = Z(); for (let i = 0; i < N; i++) a.push(t.next().value);

Answer 1

To get the first n values of an iterator, you could use one of:

Array.from({length: n}, function(){ return this.next().value; }, iterator);
Array.from({length: n}, (i => () => i.next().value)(iterator));

To get the iterator of an arbitrary iterable, use:

const iterator = iterable[Symbol.iterator]();

In your case, given a generator function Z:

Array.from({length: 3}, function(){ return this.next().value; }, Z());

If you need this functionality more often, you could create a generator function:

function* take(iterable, length) {
  const iterator = iterable[Symbol.iterator]();
  while (length-- > 0) yield iterator.next().value;
}

// Example:
const set = new Set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]);
console.log(...take(set, 3));

Answer 2

There is no built in method to take only a certain number of items from an iterable (ala something like take()). Although your snippet can be somewhat improved with a for of loop, which is specifically meant to work with iterables, eg:

let a = []; let i = 0; for (let x of Z()) { a.push(x); if (++i === N) break; }

Which might be better since your original snippet would continue looping even if there are not N items in the iterable.