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Writing a function to take a number, square each number and return them as a concatenated integer, ie. 3214 => 94116. For some reason, my code appears to occasionally square 2's and 3's twice making a 2 turn into 16 and 3 into 81. I can't figure it out. I'm not a super experienced debugger yet so any help would be appreciated.
function squareDigits(num){
let digits = (""+num).split("");
let intDigits = [];
for (x of digits) {
intDigits.push(parseInt(x));
}
for (x of intDigits) {
intDigits.splice(intDigits.indexOf(x), 1, x * x);
}
return parseInt(intDigits.join(""));
}
console.log(squareDigits(24));
Leaving aside the fact that you can do this more elegantly with something like map() the issue in your code is that it uses indexOf() while changing the values on each iteration. Since indexOf() returns the index of the first occurrence it is going to find digits that you have already replaced.
This is your original code with a few logs so you can understand what I mean:
function squareDigits(num){
let digits = (""+num).split("");
let intDigits = [];
for (x of digits) {
intDigits.push(parseInt(x));
}
for (x of intDigits) {
console.log('intDigits: ' + intDigits);
console.log(` index of ${x} = ${intDigits.indexOf(x)}`);
intDigits.splice(intDigits.indexOf(x), 1, x * x);
}
return parseInt(intDigits.join(""));
}
console.log('RESULT: ' + squareDigits(24));
Notice how in the second pass the index of 4 is 0 (the first position in the array) because you have replaced the original 2 by it's squared value 4.
A simple way to fix this is to not rely on indexOf() and iterate over the array the good old way, like this:
function squareDigits(num){
let digits = (""+num).split("");
let intDigits = [];
for (x of digits) {
intDigits.push(parseInt(x));
}
for (let i = 0; i < intDigits.length; i++) {
const x = intDigits[i];
console.log('intDigits: ' + intDigits);
console.log(` index of ${x} = ${i}`);
intDigits.splice(i, 1, x * x);
}
return parseInt(intDigits.join(""));
}
console.log('RESULT: ' + squareDigits(24));
A simplistic (and bug free) version of your function could be this:
const squareDigits = num => [...num.toString()].map(x => x ** 2).join('');
console.log(squareDigits(24));
Other variant:
const squareDigits = num => num.toString().replaceAll(/\d/g, x => x ** 2);
console.log(squareDigits(24));
Instead of looping twice use array methods .map() , .reduce() etc it will make your code effective .. wrote a simple function
See =>
function squareDigits(num){
let digits = String(num).split("");
digits = digits.reduce((final , digit)=> final += String( parseInt(digit) **2 ), "");
return digits
}
console.log(squareDigits(312));
As #Sebastian mentioned, intDigits.indexOf(x) will find the
first index. So after replacing the first one, there's a change you'll find the number you've just replaced.
We can simplify the function to :
function squareDigits(num){
return num.toString().split('').map(n => n ** 2).join('');
}
Where :
We convert the number to a string using toString()
We split the string into loose numbers using split('')
map() over each number
Return the square by using the Exponentiation (**) operator
join() the numbers to get the result
Example snippet:
function squareDigits(num){
return num.toString().split('').map(n => n ** 2).join('');
}
console.log(squareDigits(3214)); // 94116
I was asked during an interview to implement toString() to convert a number into a string.
toString()
n => s
123 => "123"
Aside from:
converting the number by concatenating an empty string
123+""
using the native toString() function
(123).toString()
creating a new string
String(123)
How else could toString() be implemented in javascript?
You can use it as the property name of an object.
function toString(value) {
// Coerces value to a primitive string (or symbol)
var obj = {};
obj[value] = true;
return Object.getOwnPropertyNames(obj)[0];
}
console.log(toString(123)); // 123 -> "123"
console.log(toString(1.23)); // 1.23 -> "1.23"
console.log(toString(NaN)); // NaN -> "NaN"
console.log(Infinity); // Infinity -> "Infinity"
console.log(toString(-0)); // -0 -> "0"
console.log(toString(1e99)); // 1e99 -> "1e+99"
You can also use DOM attributes:
var obj = document.createElement('div');
obj.setAttribute('data-toString', value);
return obj.getAttribute('data-toString');
Or join an array
return [value].join();
And a big etcetera. There are lots of things which internally use the ToString abstract operation.
This works for integers. It takes the number modulo 10 and divides it by 10 repeatedly, then adds 48 to the digits and uses String.fromCharCode to get a string value of the digits, then joins everything.
function toString(n){
var minus = (n < 0
? "-"
: ""),
result = [];
n = Math.abs(n);
while(n > 0){
result.unshift(n % 10);
n = Math.floor(n / 10);
}
return minus + (result.map(function(d){
return String.fromCharCode(d + 48);
})
.join("") || "0");
}
console.log(toString(123123));
console.log(toString(999));
console.log(toString(0));
console.log(toString(-1));
The trick here is to consider a number as a series of digits. This is not an inherent property of numbers, since the base-10 representation that we use is quite arbitrary. But once a number is represented as a series of digits, it is quite easy to convert each digit individually to a string, and concatenate all such strings.
EDIT: As pointed out, this only takes integers into consideration (which is probably acceptable for an interview question).
var intToDigits = function(n) {
var highestPow = 1;
while (highestPow < n) highestPow *= 10;
var div = highestPow / 10;
// div is now the largest multiple of 10 smaller than n
var digits = [];
do {
var digit = Math.floor(n / div);
n = n - (digit * div);
div /= 10;
digits.push(digit);
} while (n > 0);
return digits;
};
var toString = function(n) {
var digitArr = intToDigits(n);
return digitArr.map(function(n) {
return "0123456789"[n];
}).join('');
};
Usage:
>> toString(678)
"678"
If you're using ES6 you could use template literals.
var a = 5;
console.log(`${a}`);
Below is my source code to reverse (as in a mirror) the given number.
I need to reverse the number using the reverse method of arrays.
<script>
var a = prompt("Enter a value");
var b, sum = 0;
var z = a;
while(a > 0)
{
b = a % 10;
sum = sum * 10 + b;
a = parseInt(a / 10);
}
alert(sum);
</script>
Low-level integer numbers reversing:
function flipInt(n){
var digit, result = 0
while( n ){
digit = n % 10 // Get right-most digit. Ex. 123/10 → 12.3 → 3
result = (result * 10) + digit // Ex. 123 → 1230 + 4 → 1234
n = n/10|0 // Remove right-most digit. Ex. 123 → 12.3 → 12
}
return result
}
// Usage:
alert(
"Reversed number: " + flipInt( +prompt("Enter a value") )
)
The above code uses bitwise operators for quick math
This method is MUCH FASTER than other methods which convert the number to an Array and then reverse it and join it again. This is a low-level blazing-fast solution.
Illustration table:
const delay = (ms = 1000) => new Promise(res => setTimeout(res, ms))
const table = document.querySelector('tbody')
async function printLine(s1, s2, op){
table.innerHTML += `<tr>
<td>${s1}</td>
<td>${s2||''}</td>
</tr>`
}
async function steps(){
printLine(123)
await delay()
printLine('12.3 →')
await delay()
printLine(12, 3)
await delay()
printLine('1.2', '3 × 10')
await delay()
printLine('1.2 →', 30)
await delay()
printLine(1, 32)
await delay()
printLine(1, '32 × 10')
await delay()
printLine('1 →', 320)
await delay()
printLine('', 321)
await delay()
}
steps()
table{ width: 200px; }
td {
border: 1px dotted #999;
}
<table>
<thead>
<tr>
<th>Current</th>
<th>Output</th>
</tr>
</thead>
<tbody>
</tbody>
</table>
Assuming #DominicTobias is correct, you can use this:
console.log(
+prompt("Enter a value").split("").reverse().join("")
)
I was recently asked how to solve this problem and this was my initial solution:
The desired output: 123 => 321, -15 => -51, 500 => 5
function revInt(num) {
// Use toString() to convert it into a String
// Use the split() method to return a new array: -123 => ['-', '1','2','3']
// Use the reverse() method to reverse the new created array: ['-', '1','2','3'] => ['3','2','1','-'];
// Use the join() method to join all elements of the array into a string
let val = num.toString().split('').reverse().join('');
// If the entered number was negative, then that '-' would be the last character in
// our newly created String, but we don't want that, instead what we want is
// for it to be the first one. So, this was the solution from the top of my head.
// The endsWith() method determines whether a string ends with the characters of a specified string
if (val.endsWith('-')) {
val = '-' + val;
return parseInt(val);
}
return parseInt(val);
}
console.log(revInt(-123));
A way better solution:
After I gave it some more thought, I came up with the following:
// Here we're converting the result of the same functions used in the above example to
// an Integer and multiplying it by the value returned from the Math.sign() function.
// NOTE: The Math.sign() function returns either a positive or negative +/- 1,
// indicating the sign of a number passed into the argument.
function reverseInt(n) {
return parseInt(n.toString().split('').reverse().join('')) * Math.sign(n)
}
console.log(reverseInt(-123));
NOTE: The 2nd solution is much more straightforward, IMHO
This is my solution, pure JS without predefined functions.
function reverseNum(number) {
var result = 0,
counter = 0;
for (i = number; i >= 1 - Number.EPSILON; i = i / 10 - (i % 10) * 0.1) {
counter = i % 10;
result = result * 10 + counter;
}
return result;
}
console.log(reverseNum(547793));
Firstly, I don't think you are using an array to store the number. You are using a java script variable.
Try out this code and see if it works.
var a = prompt("Enter a value");
var z = a;
var reverse = 0;
while(z > 0)
{
var digit = z % 10;
reverse = (reverse * 10) + digit;
z = parseInt(z / 10);
}
alert("reverse = " + reverse);
Or, as a one-liner ( x contains the integer number to be inversed):
revX=x.toFixed(0).split('').reverse().join('')-0;
The number will be separated into its individual digits, reversed and then reassembled again into a string. The -0 then converts it into a number again.
Explanation
Using the JavaScript reverse() array method you can reverse the order of the array elements.
Code
var a = prompt("Enter a value");
var arr = [];
for (var i = 0; i < a.length; i++) {
arr[i] = a.charAt(i);
}
arr.reverse();
alert(arr);
Assuming you may want to reverse it as a true number and not a string try the following:
function reverseNumber(num){
num = num + '';
let reversedText = num.split('').reverse().join('');
let reversedNumber = parseInt(reversedText, 10);
console.log("reversed number: ", reversedNumber);
return reversedNumber;
}
Using JavaScript reverse() and Math.sign() you can reverse a number both positive and negative numbers.
var enteredNum = prompt("Enter integer");
function reverseInteger(enteredNum) {
const reveredNumber = enteredNum.toString().split('').reverse().join('');
return parseInt(reveredNumber)*Math.sign(enteredNum);
}
alert(reverseInteger(enteredNum));
function add( num:number){ //159
let d : number;
let a : number =0;
while(num > 0){ //159 15 1
d = num % 10;
a = a * 10 + d; //9 95 951
num = Math.floor(num/10); // 15 1 0
}
return a; //951
}
console.log(add(159));
Reversing a number without converting it into the string using the recursive approach.
const num = 4578;
const by10 = (num) => {
return Math.floor(num / 10);
};
const remBy10 = (num) => {
return Math.floor(num % 10);
};
const reverseNum = (num, str = "") => {
if (num.toString().length == 1) return (str += num);
return reverseNum(by10(num), (str += remBy10(num)));
};
console.log(reverseNum(num, ""));
The simplest solution is to reverse any integer in js. Doesn't work with float.
const i2a = number.toString().split("");
const a2i = parseInt(i2a.reverse().join(""));
console.log(a2i);
Apply logic of reversing number in paper and try, and you have to care about dividing because it gives float values. That's why we have to use parseInt().
function palindrome()
{
var a = document.getElementById('str').value;
var r=0 ,t=0;
while(a>0){
r=a%10;
t=t*10+r;
a=parseInt(a/10);
}
document.write(t);
}
<form>
<input type="text" id="str"/>
<input type="submit" onClick="palindrome()" />
<form>
var reverse = function(x) {
if (x > 2147483647 || x < -2147483648 || x === 0) {
return 0;
}
let isNegative = false;
if(x < 0){
isNegative = true;
x = -x;
}
const length = parseInt(Math.log10(x));
let final = 0;
let digit = x;
let mul = 0;
for(let i = length ; i >= 0; i--){
digit = parseInt(x / (10**i));
mul = 10**(length-i);
final = final + digit * mul;
x = parseInt(x % 10**i);
}
if (final > 2147483647 || final < -2147483648 ) {
return 0;
}
if(isNegative){
return -final;
}
else{
return final;
}
};
console.log(reverse(1534236469));
console.log(reverse(-123));
console.log(reverse(120));
console.log(reverse(0));
console.log(reverse(2,147,483,648));
function reverseInt(n) {
let x = n.toString();
let y = '';
for(let i of x) {
y = i + y
}
return parseInt(y) * Math.sign(n);
}
Sweet and simple:
function reverseNumber(num){
return parseInt(num.toString().split("").reverse().join(""));
}
The above code will not work for negative numbers. Instead, use the following:
/**
* #param {number} x
* #return {boolean}
*/
var isPalindrome = function(x) {
return ((x>=0) ? ((x==(x = parseInt(x.toString().split("").reverse().join("")))) ? true:false) : false);
};
The simplest way is to
Covert it into a string and apply the reverse() method
Change it back to number
Check for the value provided if negative or positive with Math.sign()
Below is my solution to that.
function reverseInt(n) {
const reversed =
n.toString().split('').reverse().join('');
return parseInt(reversed) * Math.sign(n);
}
console.log(reverseInt(12345));
My solution to reverse a string:
var text = ""
var i = 0
var array = ["1", "2", "3"]
var number = array.length
var arrayFinal = []
for (i = 0; i < array.length; i++) {
text = array[number - 1]
arrayFinal.push(text)
text = ""
number = number - 1
}
console.log(arrayFinal)
This question already has answers here:
How can I pad a value with leading zeros?
(76 answers)
Closed 3 years ago.
In JavaScript, I need to have padding.
For example, if I have the number 9, it will be "0009". If I have a number of say 10, it will be "0010". Notice how it will always contain four digits.
One way to do this would be to subtract the number minus 4 to get the number of 0s I need to put.
Is there was a slicker way of doing this?
ES2017 Update
You can use the built-in String.prototype.padStart()
n = 9;
String(n).padStart(4, '0'); // '0009'
n = 10;
String(n).padStart(4, '0'); // '0010'
Not a lot of "slick" going on so far:
function pad(n, width, z) {
z = z || '0';
n = n + '';
return n.length >= width ? n : new Array(width - n.length + 1).join(z) + n;
}
When you initialize an array with a number, it creates an array with the length set to that value so that the array appears to contain that many undefined elements. Though some Array instance methods skip array elements without values, .join() doesn't, or at least not completely; it treats them as if their value is the empty string. Thus you get a copy of the zero character (or whatever "z" is) between each of the array elements; that's why there's a + 1 in there.
Example usage:
pad(10, 4); // 0010
pad(9, 4); // 0009
pad(123, 4); // 0123
pad(10, 4, '-'); // --10
function padToFour(number) {
if (number<=9999) { number = ("000"+number).slice(-4); }
return number;
}
Something like that?
Bonus incomprehensible-but-slicker single-line ES6 version:
let padToFour = number => number <= 9999 ? `000${number}`.slice(-4) : number;
ES6isms:
let is a block-scoped variable (as opposed to var’s functional scoping)
=> is an arrow function that, among other things, replaces function and is prepended by its parameters
If an arrow function takes a single parameter, you can omit the parentheses (hence number =>)
If an arrow function body has a single line that starts with return, you can omit the braces and the return keyword and simply use the expression
To get the function body down to a single line, I cheated and used a ternary expression
Try:
String.prototype.lpad = function(padString, length) {
var str = this;
while (str.length < length)
str = padString + str;
return str;
}
Now test:
var str = "5";
alert(str.lpad("0", 4)); //result "0005"
var str = "10"; // note this is string type
alert(str.lpad("0", 4)); //result "0010"
DEMO
In ECMAScript 2017 , we have new method padStart and padEnd which has below syntax.
"string".padStart(targetLength [,padString]):
So now we can use
const str = "5";
str.padStart(4, "0"); // "0005"
Funny, I recently had to do this.
function padDigits(number, digits) {
return Array(Math.max(digits - String(number).length + 1, 0)).join(0) + number;
}
Use like:
padDigits(9, 4); // "0009"
padDigits(10, 4); // "0010"
padDigits(15000, 4); // "15000"
Not beautiful, but effective.
You did say you had a number-
String.prototype.padZero= function(len, c){
var s= '', c= c || '0', len= (len || 2)-this.length;
while(s.length<len) s+= c;
return s+this;
}
Number.prototype.padZero= function(len, c){
return String(this).padZero(len,c);
}
You could do something like this:
function pad ( num, size ) {
return ( Math.pow( 10, size ) + ~~num ).toString().substring( 1 );
}
Edit: This was just a basic idea for a function, but to add support for larger numbers (as well as invalid input), this would probably be better:
function pad ( num, size ) {
if (num.toString().length >= size) return num;
return ( Math.pow( 10, size ) + Math.floor(num) ).toString().substring( 1 );
}
This does 2 things:
If the number is larger than the specified size, it will simply return the number.
Using Math.floor(num) in place of ~~num will support larger numbers.
This is not really 'slick' but it's faster to do integer operations than to do string concatenations for each padding 0.
function ZeroPadNumber ( nValue )
{
if ( nValue < 10 )
{
return ( '000' + nValue.toString () );
}
else if ( nValue < 100 )
{
return ( '00' + nValue.toString () );
}
else if ( nValue < 1000 )
{
return ( '0' + nValue.toString () );
}
else
{
return ( nValue );
}
}
This function is also hardcoded to your particular need (4 digit padding), so it's not generic.
For fun, instead of using a loop to create the extra zeros:
function zeroPad(n,length){
var s=n+"",needed=length-s.length;
if (needed>0) s=(Math.pow(10,needed)+"").slice(1)+s;
return s;
}
Since you mentioned it's always going to have a length of 4, I won't be doing any error checking to make this slick. ;)
function pad(input) {
var BASE = "0000";
return input ? BASE.substr(0, 4 - Math.ceil(input / 10)) + input : BASE;
}
Idea: Simply replace '0000' with number provided... Issue with that is, if input is 0, I need to hard-code it to return '0000'. LOL.
This should be slick enough.
JSFiddler: http://jsfiddle.net/Up5Cr/
What is the recommended way to zerofill a value in JavaScript? I imagine I could build a custom function to pad zeros on to a typecasted value, but I'm wondering if there is a more direct way to do this?
Note: By "zerofilled" I mean it in the database sense of the word (where a 6-digit zerofilled representation of the number 5 would be "000005").
I can't believe all the complex answers on here... Just use this:
var zerofilled = ('0000'+n).slice(-4);
let n = 1
var zerofilled = ('0000'+n).slice(-4);
console.log(zerofilled)
Simple way. You could add string multiplication for the pad and turn it into a function.
var pad = "000000";
var n = '5';
var result = (pad+n).slice(-pad.length);
As a function,
function paddy(num, padlen, padchar) {
var pad_char = typeof padchar !== 'undefined' ? padchar : '0';
var pad = new Array(1 + padlen).join(pad_char);
return (pad + num).slice(-pad.length);
}
var fu = paddy(14, 5); // 00014
var bar = paddy(2, 4, '#'); // ###2
Since ECMAScript 2017 we have padStart:
const padded = (.1 + "").padStart(6, "0");
console.log(`-${padded}`);
Before ECMAScript 2017
With toLocaleString:
var n=-0.1;
var res = n.toLocaleString('en', {minimumIntegerDigits:4,minimumFractionDigits:2,useGrouping:false});
console.log(res);
I actually had to come up with something like this recently.
I figured there had to be a way to do it without using loops.
This is what I came up with.
function zeroPad(num, numZeros) {
var n = Math.abs(num);
var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );
var zeroString = Math.pow(10,zeros).toString().substr(1);
if( num < 0 ) {
zeroString = '-' + zeroString;
}
return zeroString+n;
}
Then just use it providing a number to zero pad:
> zeroPad(50,4);
"0050"
If the number is larger than the padding, the number will expand beyond the padding:
> zeroPad(51234, 3);
"51234"
Decimals are fine too!
> zeroPad(51.1234, 4);
"0051.1234"
If you don't mind polluting the global namespace you can add it to Number directly:
Number.prototype.leftZeroPad = function(numZeros) {
var n = Math.abs(this);
var zeros = Math.max(0, numZeros - Math.floor(n).toString().length );
var zeroString = Math.pow(10,zeros).toString().substr(1);
if( this < 0 ) {
zeroString = '-' + zeroString;
}
return zeroString+n;
}
And if you'd rather have decimals take up space in the padding:
Number.prototype.leftZeroPad = function(numZeros) {
var n = Math.abs(this);
var zeros = Math.max(0, numZeros - n.toString().length );
var zeroString = Math.pow(10,zeros).toString().substr(1);
if( this < 0 ) {
zeroString = '-' + zeroString;
}
return zeroString+n;
}
Cheers!
XDR came up with a logarithmic variation that seems to perform better.
WARNING: This function fails if num equals zero (e.g. zeropad(0, 2))
function zeroPad (num, numZeros) {
var an = Math.abs (num);
var digitCount = 1 + Math.floor (Math.log (an) / Math.LN10);
if (digitCount >= numZeros) {
return num;
}
var zeroString = Math.pow (10, numZeros - digitCount).toString ().substr (1);
return num < 0 ? '-' + zeroString + an : zeroString + an;
}
Speaking of performance, tomsmeding compared the top 3 answers (4 with the log variation). Guess which one majorly outperformed the other two? :)
Modern browsers now support padStart, you can simply now do:
string.padStart(maxLength, "0");
Example:
string = "14";
maxLength = 5; // maxLength is the max string length, not max # of fills
res = string.padStart(maxLength, "0");
console.log(res); // prints "00014"
number = 14;
maxLength = 5; // maxLength is the max string length, not max # of fills
res = number.toString().padStart(maxLength, "0");
console.log(res); // prints "00014"
Here's what I used to pad a number up to 7 characters.
("0000000" + number).slice(-7)
This approach will probably suffice for most people.
Edit: If you want to make it more generic you can do this:
("0".repeat(padding) + number).slice(-padding)
Edit 2: Note that since ES2017 you can use String.prototype.padStart:
number.toString().padStart(padding, "0")
Unfortunately, there are a lot of needless complicated suggestions for this problem, typically involving writing your own function to do math or string manipulation or calling a third-party utility. However, there is a standard way of doing this in the base JavaScript library with just one line of code. It might be worth wrapping this one line of code in a function to avoid having to specify parameters that you never want to change like the local name or style.
var amount = 5;
var text = amount.toLocaleString('en-US',
{
style: 'decimal',
minimumIntegerDigits: 3,
useGrouping: false
});
This will produce the value of "005" for text. You can also use the toLocaleString function of Number to pad zeros to the right side of the decimal point.
var amount = 5;
var text = amount.toLocaleString('en-US',
{
style: 'decimal',
minimumFractionDigits: 2,
useGrouping: false
});
This will produce the value of "5.00" for text. Change useGrouping to true to use comma separators for thousands.
Note that using toLocaleString() with locales and options arguments is standardized separately in ECMA-402, not in ECMAScript. As of today, some browsers only implement basic support, i.e. toLocaleString() may ignore any arguments.
Complete Example
If the fill number is known in advance not to exceed a certain value, there's another way to do this with no loops:
var fillZeroes = "00000000000000000000"; // max number of zero fill ever asked for in global
function zeroFill(number, width) {
// make sure it's a string
var input = number + "";
var prefix = "";
if (input.charAt(0) === '-') {
prefix = "-";
input = input.slice(1);
--width;
}
var fillAmt = Math.max(width - input.length, 0);
return prefix + fillZeroes.slice(0, fillAmt) + input;
}
Test cases here: http://jsfiddle.net/jfriend00/N87mZ/
The quick and dirty way:
y = (new Array(count + 1 - x.toString().length)).join('0') + x;
For x = 5 and count = 6 you'll have y = "000005"
Here's a quick function I came up with to do the job. If anyone has a simpler approach, feel free to share!
function zerofill(number, length) {
// Setup
var result = number.toString();
var pad = length - result.length;
while(pad > 0) {
result = '0' + result;
pad--;
}
return result;
}
ECMAScript 2017:
use padStart or padEnd
'abc'.padStart(10); // " abc"
'abc'.padStart(10, "foo"); // "foofoofabc"
'abc'.padStart(6,"123465"); // "123abc"
More info:
https://github.com/tc39/proposal-string-pad-start-end
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
I often use this construct for doing ad-hoc padding of some value n, known to be a positive, decimal:
(offset + n + '').substr(1);
Where offset is 10^^digits.
E.g., padding to 5 digits, where n = 123:
(1e5 + 123 + '').substr(1); // => 00123
The hexadecimal version of this is slightly more verbose:
(0x100000 + 0x123).toString(16).substr(1); // => 00123
Note 1: I like #profitehlolz's solution as well, which is the string version of this, using slice()'s nifty negative-index feature.
I really don't know why, but no one did it in the most obvious way. Here it's my implementation.
Function:
/** Pad a number with 0 on the left */
function zeroPad(number, digits) {
var num = number+"";
while(num.length < digits){
num='0'+num;
}
return num;
}
Prototype:
Number.prototype.zeroPad=function(digits){
var num=this+"";
while(num.length < digits){
num='0'+num;
}
return(num);
};
Very straightforward, I can't see any way how this can be any simpler. For some reason I've seem many times here on SO, people just try to avoid 'for' and 'while' loops at any cost. Using regex will probably cost way more cycles for such a trivial 8 digit padding.
In all modern browsers you can use
numberStr.padStart(numberLength, "0");
function zeroFill(num, numLength) {
var numberStr = num.toString();
return numberStr.padStart(numLength, "0");
}
var numbers = [0, 1, 12, 123, 1234, 12345];
numbers.forEach(
function(num) {
var numString = num.toString();
var paddedNum = zeroFill(numString, 5);
console.log(paddedNum);
}
);
Here is the MDN reference https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
I use this snippet to get a five-digits representation:
(value+100000).toString().slice(-5) // "00123" with value=123
The power of Math!
x = integer to pad
y = number of zeroes to pad
function zeroPad(x, y)
{
y = Math.max(y-1,0);
var n = (x / Math.pow(10,y)).toFixed(y);
return n.replace('.','');
}
This is the ES6 solution.
function pad(num, len) {
return '0'.repeat(len - num.toString().length) + num;
}
alert(pad(1234,6));
Not that this question needs more answers, but I thought I would add the simple lodash version of this.
_.padLeft(number, 6, '0')
I didn't see anyone point out the fact that when you use String.prototype.substr() with a negative number it counts from the right.
A one liner solution to the OP's question, a 6-digit zerofilled representation of the number 5, is:
console.log(("00000000" + 5).substr(-6));
Generalizing we'll get:
function pad(num, len) { return ("00000000" + num).substr(-len) };
console.log(pad(5, 6));
console.log(pad(45, 6));
console.log(pad(345, 6));
console.log(pad(2345, 6));
console.log(pad(12345, 6));
Don't reinvent the wheel; use underscore string:
jsFiddle
var numToPad = '5';
alert(_.str.pad(numToPad, 6, '0')); // Yields: '000005'
After a, long, long time of testing 15 different functions/methods found in this questions answers, I now know which is the best (the most versatile and quickest).
I took 15 functions/methods from the answers to this question and made a script to measure the time taken to execute 100 pads. Each pad would pad the number 9 with 2000 zeros. This may seem excessive, and it is, but it gives you a good idea about the scaling of the functions.
The code I used can be found here:
https://gist.github.com/NextToNothing/6325915
Feel free to modify and test the code yourself.
In order to get the most versatile method, you have to use a loop. This is because with very large numbers others are likely to fail, whereas, this will succeed.
So, which loop to use? Well, that would be a while loop. A for loop is still fast, but a while loop is just slightly quicker(a couple of ms) - and cleaner.
Answers like those by Wilco, Aleksandar Toplek or Vitim.us will do the job perfectly.
Personally, I tried a different approach. I tried to use a recursive function to pad the string/number. It worked out better than methods joining an array but, still, didn't work as quick as a for loop.
My function is:
function pad(str, max, padder) {
padder = typeof padder === "undefined" ? "0" : padder;
return str.toString().length < max ? pad(padder.toString() + str, max, padder) : str;
}
You can use my function with, or without, setting the padding variable. So like this:
pad(1, 3); // Returns '001'
// - Or -
pad(1, 3, "x"); // Returns 'xx1'
Personally, after my tests, I would use a method with a while loop, like Aleksandar Toplek or Vitim.us. However, I would modify it slightly so that you are able to set the padding string.
So, I would use this code:
function padLeft(str, len, pad) {
pad = typeof pad === "undefined" ? "0" : pad + "";
str = str + "";
while(str.length < len) {
str = pad + str;
}
return str;
}
// Usage
padLeft(1, 3); // Returns '001'
// - Or -
padLeft(1, 3, "x"); // Returns 'xx1'
You could also use it as a prototype function, by using this code:
Number.prototype.padLeft = function(len, pad) {
pad = typeof pad === "undefined" ? "0" : pad + "";
var str = this + "";
while(str.length < len) {
str = pad + str;
}
return str;
}
// Usage
var num = 1;
num.padLeft(3); // Returns '001'
// - Or -
num.padLeft(3, "x"); // Returns 'xx1'
First parameter is any real number, second parameter is a positive integer specifying the minimum number of digits to the left of the decimal point and third parameter is an optional positive integer specifying the number if digits to the right of the decimal point.
function zPad(n, l, r){
return(a=String(n).match(/(^-?)(\d*)\.?(\d*)/))?a[1]+(Array(l).join(0)+a[2]).slice(-Math.max(l,a[2].length))+('undefined'!==typeof r?(0<r?'.':'')+(a[3]+Array(r+1).join(0)).slice(0,r):a[3]?'.'+a[3]:''):0
}
so
zPad(6, 2) === '06'
zPad(-6, 2) === '-06'
zPad(600.2, 2) === '600.2'
zPad(-600, 2) === '-600'
zPad(6.2, 3) === '006.2'
zPad(-6.2, 3) === '-006.2'
zPad(6.2, 3, 0) === '006'
zPad(6, 2, 3) === '06.000'
zPad(600.2, 2, 3) === '600.200'
zPad(-600.1499, 2, 3) === '-600.149'
The latest way to do this is much simpler:
var number = 2
number.toLocaleString(undefined, {minimumIntegerDigits:2})
output: "02"
Just another solution, but I think it's more legible.
function zeroFill(text, size)
{
while (text.length < size){
text = "0" + text;
}
return text;
}
This one is less native, but may be the fastest...
zeroPad = function (num, count) {
var pad = (num + '').length - count;
while(--pad > -1) {
num = '0' + num;
}
return num;
};
My solution
Number.prototype.PadLeft = function (length, digit) {
var str = '' + this;
while (str.length < length) {
str = (digit || '0') + str;
}
return str;
};
Usage
var a = 567.25;
a.PadLeft(10); // 0000567.25
var b = 567.25;
b.PadLeft(20, '2'); // 22222222222222567.25
With ES6+ JavaScript:
You can "zerofill a number" with something like the following function:
/**
* #param number The number
* #param minLength Minimal length for your string with leading zeroes
* #return Your formatted string
*/
function zerofill(nb, minLength) {
// Convert your number to string.
let nb2Str = nb.toString()
// Guess the number of zeroes you will have to write.
let nbZeroes = Math.max(0, minLength - nb2Str.length)
// Compute your result.
return `${ '0'.repeat(nbZeroes) }${ nb2Str }`
}
console.log(zerofill(5, 6)) // Displays "000005"
With ES2017+:
/**
* #param number The number
* #param minLength Minimal length for your string with leading zeroes
* #return Your formatted string
*/
const zerofill = (nb, minLength) => nb.toString().padStart(minLength, '0')
console.log(zerofill(5, 6)) // Displays "000005"
Use recursion:
function padZero(s, n) {
s = s.toString(); // In case someone passes a number
return s.length >= n ? s : padZero('0' + s, n);
}
Some monkeypatching also works
String.prototype.padLeft = function (n, c) {
if (isNaN(n))
return null;
c = c || "0";
return (new Array(n).join(c).substring(0, this.length-n)) + this;
};
var paddedValue = "123".padLeft(6); // returns "000123"
var otherPadded = "TEXT".padLeft(8, " "); // returns " TEXT"
function pad(toPad, padChar, length){
return (String(toPad).length < length)
? new Array(length - String(toPad).length + 1).join(padChar) + String(toPad)
: toPad;
}
pad(5, 0, 6) = 000005
pad('10', 0, 2) = 10 // don't pad if not necessary
pad('S', 'O', 2) = SO
...etc.
Cheers