Generate all consecutive element subarrays from a JavaScript array

Getting a single element from an array is easy, but what if you want to get all possible n-tuples of consecutive elements from an array? This is a fairly uncommon problem, yet it can be solved with a simple function.

Arrays of consecutive elements

All you need to do is use Array.prototype.map() in order to map each element to an array of n consecutive elements, using Array.prototype.slice() to extract the subarray. In order to make sure no subarrays with fewer than n elements are returned, you can use Array.prototype.slice() to remove the first n - 1 elements.

Finally, you can use Array.prototype.length to check if the length of the array is greater than n and return an empty array if it is not.

const aperture = (n, arr) =>
  n > arr.length
    ? []
    : arr.slice(n - 1).map((v, i) => arr.slice(i, i + n));

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

Mapped arrays of consecutive elements

Subsequently, if you want to apply a function to each of the subarrays, you can simply apply the function to each subarray during the mapping process.

const apertureMap = (arr, n, fn) =>
  n > arr.length
    ? []
    arr.slice(n - 1).map((v, i) => fn(arr.slice(i, i + n)));

apertureMap([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 3, x => x.join('-'));
// ['1-2-3', '2-3-4', '3-4-5', '4-5-6', '5-6-7', '6-7-8', '7-8-9', '8-9-10'];