How to access array in circular manner in JavaScript

I have an array like [A,B,C,D]. I want to access that array within a for loop like as

var arr = [A,B,C,D]; var len = arr.len; for(var i = 0;i<arr.len;i++){ 0 - A,B,C 1 - B,C,D 2 - C,D,A 3 - D,A,B } 

I want to access that like in JavaScript, any ideas?

7 Answers

Answering to the main question, someone can access an array in a circular manner using modular arithmetic. That can be achieved in JavaScript with the modulus operator (%) and a workaround.

Given an array arr of a length n and a value val stored in it that will be obtained through an access index i, the circular manner, and safer way, to access the array, disregarding the value and sign of i, would be:

let val = arr[(i % n + n) % n]; 

This little trick is necessary -- someone can not use the modulus result straightforwardly -- because JavaScript always evaluates a modulus operation as the remainder of the division between dividend (the first operand) and divisor (the second operand) disconsidering their signs but assigning to the remainder the sign of the dividend. That behavior does not always result in the desired "wrap around" effect of the modular arithmetic and could result in a wrong access of a negative position of the array.

References for more information:

Try this:

var arr = ["A","B","C","D"]; for (var i=0, len=arr.length; i<len; i++) { alert(arr.slice(0, 3).join(",")); arr.push(arr.shift()); } 

Without mutating the array, it would be

for (var i=0, len=arr.length; i<len; i++) { var str = arr[i]; for (var j=1; j<3; j++) str += ","+arr[(i+j)%len]; // you could push to an array as well alert(str); } // or for (var i=0, len=arr.length; i<len; i++) alert(arr.slice(i, i+3).concat(arr.slice(0, Math.max(i+3-len, 0)).join(",")); 

Simply using modulus operator you can access array in circular manner.

var arr = ['A', 'B', 'C', 'D']; for (var i = 0, len = arr.length; i < len; i++) { for (var j = 0; j < 3; j++) { console.log(arr[(i + j) % len]) } console.log('****') }

for (var i = 0; i < arr.length; i++) { var subarr = []; for (var j = 0; j < 3; j++) { subarr.push(arr[(i+j) % arr.length]); } console.log(i + " - " + subarr.join(',')); } 

how about this one-liner I made ?

var nextItem = (list.indexOf(currentItem) < list.length - 1) ? list[list.indexOf(currentItem) + 1] : list[0]; 

One line solution for "in place" circular shift:

const arr = ["A","B","C","D"]; arr.forEach((x,i,t) => {console.log(i,t); t.push(t.shift());}); console.log("end of cycle", arr); // control: cycled back to the original 

logs:

0 Array ["A", "B", "C", "D"] 1 Array ["B", "C", "D", "A"] 2 Array ["C", "D", "A", "B"] 3 Array ["D", "A", "B", "C"] "end of cycle" Array ["A", "B", "C", "D"] 

If you want only the first 3 items, use:

arr.forEach((x,i,t) => {console.log(i,t.slice(0, 3)); t.push(t.shift());}); 

Another solutions:

 var arr = ['A','B','C','D']; var nextVal = function (arr) { return arr[( ( ( nextVal.counter < ( arr.length - 1 ) ) ? ++nextVal.counter : nextVal.counter=0 ) )]; }; for(var i=0;i<arr.length;i++){ console.log(nextVal(arr)+','+nextVal(arr)+','+nextVal(arr)); }

And based on Modulo :

var arr = ['A','B','C','D']; var len = arr.length; var nextVal = function (arr, dir = 1) { if ( dir < 0 ) { nextVal.counter--;} let i = (nextVal.counter % len + len) % len; if ( dir > 0 ) { nextVal.counter++; } return arr[i]; }; nextVal.counter=0; for(var i=0;i<arr.length;i++){ console.log(nextVal(arr)+','+nextVal(arr)+','+nextVal(arr)); } // in reverse console.log('-------------------'); nextVal.counter=0; for(var i=0; i<10; i++) { console.log(nextVal(arr, -1)+','+nextVal(arr, -1)+','+nextVal(arr, -1)); }