Iterate an array as a pair (current, next) in JavaScript

Question

In the question Iterate a list as pair (current, next) in Python, the OP is interested in iterating a Python list as a series of current, next pairs. I have th

Answer 1

Here's a generic functional solution without any dependencies:

const nWise = (n, array) => {
  iterators = Array(n).fill()
    .map(() => array[Symbol.iterator]());
  iterators
    .forEach((it, index) => Array(index).fill()
      .forEach(() => it.next()));
  return Array(array.length - n + 1).fill()
    .map(() => (iterators
      .map(it => it.next().value);
};

const pairWise = (array) => nWise(2, array);

I know doesn't look nice at all but by introducing some generic utility functions we can make it look a lot nicer:

const sizedArray = (n) => Array(n).fill();

I could use sizedArray combined with forEach for times implementation, but that'd be an inefficient implementation. IMHO it's ok to use imperative code for such a self-explanatory function:

const times = (n, cb) => {
  while (0 < n--) {
    cb();
  }
}

If you're interested in more hardcore solutions, please check this answer.

Unfortunately Array.fill only accepts a single value, not a callback. So Array(n).fill(array[Symbol.iterator]()) would put the same value in every position. We can get around this the following way:

const fillWithCb = (n, cb) => sizedArray(n).map(cb);

The final implementation:

const nWise = (n, array) => {
  iterators = fillWithCb(n, () => array[Symbol.iterator]());
  iterators.forEach((it, index) => times(index, () => it.next()));
  return fillWithCb(
    array.length - n + 1,
    () => (iterators.map(it => it.next().value),
  );
};

By changing the parameter style to currying, the definition of pairwise would look a lot nicer:

const nWise = n => array => {
  iterators = fillWithCb(n, () => array[Symbol.iterator]());
  iterators.forEach((it, index) => times(index, () => it.next()));
  return fillWithCb(
    array.length - n + 1,
    () => iterators.map(it => it.next().value),
  );
};

const pairWise = nWise(2);

And if you run this you get:

> pairWise([1, 2, 3, 4, 5]);
// [ [ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 5 ] ]