How to store last N values in an array with auto loop to first item when pass last item?

Question

Is there a standard method to have special Array which can store last N elements (probably with an internal index to keep current item) such that when it reaches to and pass the last index, it goes to beginning and over-write the most old items. This way we have always the last recent N items. This is useful for example to store recent 10 price values of a product.

I can write a custom function may be like as below but i guess there may be some built-in method for it as it has many uses.

var n = 10; //max number of elements var a = new Array(n); //init the array var i = 0; //current position function setNext(value) { i++; if(i >= n) i=0; a[i] = value; } function getLast() { return a[i]; } function getAll() { //return concat(a[i+1...n], a[0...i]); //all items from old to new ones }

Sample data:

// consider n = 10 // assume we set 10 numbers in array, setNext(101); setNext(102); //now a is: a = [101, 102, 103, 104, 105, 106, 107, 108, 109, 110]; // cur index: 9 //now, if we add another value like as 111, it should take // position of oldest item of list which is currently 101: a = [ *111*, 102, 103, 104, 105, 106, 107, 108, 109, 110]; //cur index: 1

do you have some data as well and the wanted result?

Nina Scholz
– Nina Scholz

2020-12-07 18:05:32 +00:00
Commented Dec 7, 2020 at 18:05 — Nina Scholz
– Nina Scholz, Commented Dec 7, 2020 at 18:05
@NinaScholz yes, it is included to the question

S.Serpooshan
– S.Serpooshan

2020-12-07 18:14:08 +00:00
Commented Dec 7, 2020 at 18:14 — S.Serpooshan
– S.Serpooshan, Commented Dec 7, 2020 at 18:14

Nina Scholz · Accepted Answer · 2020-12-07 18:31:33Z

You could splice the array and add the last item to the end.

function lastN(limit, array = []) { if (array.length > limit) array.splice(0, array.length - limit); const fluent = { add: value => { if (array.length >= limit) array.splice(0, array.length - limit + 1); array.push(value); return fluent; }, getAll: () => array }; return fluent; } const five = lastN(5), two = lastN(2, ['a', 'b', 'c', 'd', 'e']); console.log(...five.getAll()); five.add(0); console.log(...five.getAll()); five.add(1); console.log(...five.getAll()); five.add(2); console.log(...five.getAll()); five.add(3); console.log(...five.getAll()); five.add(4); console.log(...five.getAll()); five.add(5); console.log(...five.getAll()); five.add(6); console.log(...five.getAll()); five.add(7); console.log(...five.getAll()); console.log('---') console.log(...two.getAll()); two.add('f'); console.log(...two.getAll()); two.add('g'); console.log(...two.getAll()); two.add('h'); console.log(...two.getAll());

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Thanks, it seems good (+1). I'm thinking about its performance. Is it efficient for large arrays?
in this case your round robin with a storage of the last or next index is better.
best answer up to now, your code is simple but professional!

Kousha · Accepted Answer · 2020-12-07 18:03:20Z

Not directly, but you could create a function that reads the array and uses the modulo % to perform this operation, something like

const n = 10; const array = new Array(n); const get = (a, index) => a(index % a.length); // all three are the same item const elem1 = get(array, 1); const elem2 = get(array, 11); const elem3 = get(array, 21);

Basically, we are using % to calculate the remainder which would act like a "loop" as you put it:

0 % 10 = 0 // first element 1 % 10 = 1 // second element ... 10 % 10 = 0 // Looped once, gives first element again 11 % 10 = 1 // Looped once, gives the second element ... 20 % 10 = 0 // Looped twice, gives first element again

Thanks, but the main goal is how to get the mixed parts of array. This is just a shortcut for calculating the index

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