JavaScript Number Split into individual digits

You can also do it in the "mathematical" way without treating the number as a string:

var num = 278; var digits = []; while (num != 0) { digits.push(num % 10); num = Math.trunc(num / 10); } digits.reverse(); console.log(digits);

One upside I can see is that you won't have to run parseInt() on every digit, you're dealing with the actual digits as numeric values.

This is the shortest I've found, though it does return the digits as strings:

let num = 12345; [...num+''] //["1", "2", "3", "4", "5"] 

Or use this to get back integers:

[...num+''].map(n=>+n) //[1, 2, 3, 4, 5] 

I will provide a variation on an answer already given so you can see a different approach that preserves the numeric type all along:

var number = 12354987, output = []; while (number) { output.push(number % 10); number = Math.floor(number/10); } console.log(output.reverse().join(',')); // 1,2,3,5,4,9,8,7 

I've used a technique such as the above to good effect when converting a number to Roman numerals, which is one of my favorite ways to begin to learn a programming language I'm not familiar with. For instance here is how I devised a way to convert numbers to Roman numerals with Tcl slightly after the turn of the century: http://code.activestate.com/recipes/68379-conversion-to-roman-numerals/

The comparable lines in my Tcl script being:

 while {$arabic} { set digit [expr {$arabic%10}] set arabic [expr {$arabic/10}] 

Use String, split and map :

String(number).split("").map(Number); 

function splitNum(num) { return String(num).split("").map(Number); } console.log(splitNum(1523)); // [1, 5, 2, 3] console.log(splitNum(2341)); // [2, 3, 4, 1] console.log(splitNum(325)); // [3, 2, 5]

You can work on strings instead of numbers to achieve this. You can do it like this

(111 + '').split('') 

This will return an array of strings ['1','1','1'] on which you can iterate upon and call parseInt method.

parseInt('1') === 1 

If you want the sum of individual digits, you can use the reduce function (implemented from Javascript 1.8) like this

(111 + '').split('').reduce(function(previousValue, currentValue){ return parseInt(previousValue,10) + parseInt(currentValue,10); }) 

Without converting to string:

function toDigits(number) { var left; var results = []; while (true) { left = number % 10; results.unshift(left); number = (number - left) / 10; if (number === 0) { break; } } return results; } 

Using String, ... and map

const num = 7890; const digits = [...String(num)].map(Number); console.log(digits)

Alternatively, using ... and reduce to get digits and their sum.

const sumOfDigits = num => [...""+num].reduce((acc, dig) => acc + +dig, 0); console.log('Sum of digits: ', sumOfDigits(7890));

Separate each 2 parametr.

function separator(str,sep) { var output = ''; for (var i = str.length; i > 0; i-=2) { var ii = i-1; if(output) { output = str.charAt(ii-1)+str.charAt(ii)+sep+output; } else { output = str.charAt(ii-1)+str.charAt(ii); } } return output; } console.log(separator('123456',':')); //Will return 12:34:56 

A fun introduction to recursion. This answer takes a Number and returns an array of Number digits. It does not convert the number to a string as an intermediate step.

Given n = 1234,

• n % 10 will return first (right-moist) digit, 4
• n / 10 will return 123 with some remainder
• Using Math.floor we can chop the remainder off
• Repeating these steps, we can form the entire result

Now we just have to build the recursion condition,

• If the number is already a single digit (n < 10), return an array singleton of the digit
• otherwise (inductive) the number is 10 or greater; recur and prepend to the first digit

const digits = (n = 0) => n < 10 ? [ n ] : [ ... digits (Math.floor (n / 10)), n % 10 ] console.log (digits ()) // [ 0 ] console.log (digits (1)) // [ 1 ] console.log (digits (12)) // [ 1, 2 ] console.log (digits (123)) // [ 1, 2, 3 ] console.log (digits (11234)) // [ 1, 2, 3, 4 ] console.log (digits (123456789012)) // [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2 ]

With ES6, you could use Array.from with a stringed number as iterables and Number as mapping function.

const getDigits = n => Array.from(n.toString(), Number); console.log(getDigits(12345));

This also works:

var number = 12354987; console.log(String(number).split('').map(Number));

Shadow Wizard , extended version by Orien

var num:Number = 1523; var digits:Array = []; var cnt:int = 0; while (num > 0) { var mod:int = num % 10; digits.push(mod * Math.pow(10, cnt)) num = Math.floor(num / 10); cnt++; } digits.reverse(); trace(digits); 

output:1000,500,20,3

I used this simple way of doing it.

To split digits

var N = 69; var arr = N.toString().split('').map(Number) // outputs [6,9] console.log( arr ); 

To add them together

console.log(arr.reduce( (a,b) => a+b )); // 15 

And the easiest.... num_string.split('').map(Number)

Try below:

console.log((''+123).split('').map(Number))

To just split an integer into its individual digits in the same order, Regular Expression is what I used and prefer since it prevents the chance of loosing the identity of the numbers even after they have been converted into string.

The following line of code convert the integer into a string, uses regex to match any individual digit inside the string and return an array of those, after which that array is mapped to be converted back to numbers.

const digitize = n => String(n).match(/\d/g).map(Number);

I am posting this answer to introduce the use of unshift which is a modern solution. With push, you add to the end of an array while unshift adds to the beginning. This makes the mathematical approach more powerful as you won't need to reverse anymore.

let num = 278; let digits = []; while (num > 0) { digits.unshift(num % 10); num = parseInt(num / 10); } console.log(digits);

And as a one liner.

console.log(BigInt(1e200).toString().split('').reduce((pv, v) => Number(v) + pv, 0));

var num = 111, separateDigits = num.toString().split(""), i, l = separateDigits.length; for( i = 0; i < l; ++i ) { someObject.someMethod( +separateDigits[i] ); } 

A functional approach in order to get digits from a number would be to get a string from your number, split it into an array (of characters) and map each element back into a number.

For example:

var number = 123456; var array = number.toString() .split('') .map(function(item, index) { return parseInt(item); }); console.log(array); // returns [1, 2, 3, 4, 5, 6] 

If you also need to sum all digits, you can append the reduce() method to the previous code:

var num = 123456; var array = num.toString() .split('') .map(function(item, index) { return parseInt(item); }) .reduce(function(previousValue, currentValue, index, array) { return previousValue + currentValue; }, 0); console.log(array); // returns 21 

As an alternative, with ECMAScript 2015 (6th Edition), you can use arrow functions:

var number = 123456; var array = number.toString().split('').map((item, index) => parseInt(item)); console.log(array); // returns [1, 2, 3, 4, 5, 6] 

If you need to sum all digits, you can append the reduce() method to the previous code:

var num = 123456; var result = num.toString() .split('') .map((item, index) => parseInt(item)) .reduce((previousValue, currentValue) => previousValue + currentValue, 0); console.log(result); // returns 21 

You can try this.

 var num = 99; num=num.toString().split("").map(value=>parseInt(value,10)); //output [9,9] 

Hope this helped!

function iterateNumber(N, f) { let n = N; var length = Math.log(n) * Math.LOG10E + 1 | 0; for (let i = 0; i < length; i++) { const pow = Math.pow(10, length - i - 1) let c = (n - (n % pow)) / pow f(c, i) n %= pow } } 

('' + 123456789).split('').map( x => +x ).reduce( (a,b) => a+b ) === 45

true

or without map

('' + 123456789).split('').reduce( (a,b) => (+a)+(+b) ) === 45

true

You can do it in single line, seperate each digits than add them together :

var may = 12987; var sep = (""+may).split("").map(n=>+n).reduce((a,b)=>a+b); 

This is my short solution.. with sum of number

 function sum (num) { let sNumber = num .toString() .split('') .reduce((el1, el2) => { return Number(el1) + Number(el2) }, 0) return sNumber } 

console.log(sum(123))
console.log(sum(456))

javascript has a function for it and you can use it easily.

console.log(new Intl.NumberFormat().format(number)); 

for example :

console.log(new Intl.NumberFormat().format(2334325443534)); ==> 2,334,325,443,534 

I might be wrong, but a solution picking up bits and pieces. Perhaps, as I still learning, is that the functions does many things in the same one. Do not hesitate to correct me, please.

const totalSum = (num) => [...num + ' '].map(Number).reduce((a, b) => a + b); 

So we take the parameter and convert it to and arr, adding empty spaces. We do such operation in every single element and push it into a new array with the map method. Once splited, we use reduce to sum all the elements and get the total.

As I said, don't hesitate to correct me or improve the function if you see something that I don't.

Almost forgot, just in case:

const totalSum = (num) => ( num === 0 || num < 0) ? 'I need a positive number' : [...num + ' '].map(Number).reduce((a, b) => a + b); 

If negatives numbers or just plain zero go down as parameters. Happy coding to us all.