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.length;
for(var i = 0; i<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?
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:
[1,2,3] start = 1,` len = -3` => [2,1,3].
let val = arr[Math.abs(i % n)]; ?i. The result -- remainder of the division -- will be different from the result expected from modular arithmetic. For example in JS (-1 % 10 + 10) % 10 evaluates to 9, but Math.abs(-1 % 10) evaluates to 1.
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('****')
}how about this one-liner I made ?
var nextItem = (list.indexOf(currentItem) < list.length - 1)
? list[list.indexOf(currentItem) + 1] : list[0];
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(','));
}
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());});
array.cirSlice(start, len). [1,3,4].cirSlice(3,5) results [4,1,3,4,1]?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));
}You could get the sliced part from index and the rest of slicing from start, if necessary.
This appraoch does not mutate the array.
const
getCircular = (array, size) => array.map((_, i, a) => [
...a.slice(i, i + size),
...a.slice(0, i + size < a.length ? 0 : i + size - array.length)
]);
console.log(getCircular(['A', 'B', 'C', 'D'], 3).map(a => a.join('')));
console.log(getCircular(['A', 'B', 'C', 'D'], 5).map(a => a.join('')));.as-console-wrapper { max-height: 100% !important; top: 0; }We can simply achieve this by using Array.splice() method along with the Destructuring assignment
Live Demo :
// Input array.
let arr = ['A', 'B', 'C', 'D'];
// Finding the length of an array.
const len = arr.length;
// Iterattion based on array length.
for(let i = 0; i < len; i++) {
const splittedArr = arr.splice(0, 3);
arr.push(splittedArr[0]);
arr = [splittedArr[1], splittedArr[2], ...arr]
console.log(splittedArr);
}