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Replace / mask alternative N chars in a string

I want to replace some characters in a string with a "*", in the following way:
Given N, leave the first N characters as-is, but mask the next N characters with "*", then leave the next N characters unchanged, ...etc, alternating every N characters in the string.
I am able to mask every alternating character with "*" (the case where N is 1):
let str = "abcdefghijklmnopqrstuvwxyz"
for (let i =0; i<str.length; i +=2){
str = str.substring(0, i) + '*' + str.substring(i + 1);
}
console.log(str)
Output:
"*b*d*f*h*j*l*n*p*r*t*v*x*z"
But I don't know how to perform the mask with different values for N.
Example:
let string = "9876543210"
N = 1; Output: 9*7*5*3*1*
N = 2; Output: 98**54**10
N = 3; Output: 987***321*
What is the best way to achieve this without regular expressions?

You could use Array.from to map each character to either "*" or the unchanged character, depending on the index. If the integer division of the index by n is odd, it should be "*". Finally turn that array back to string with join:
function mask(s, n) {
return Array.from(s, (ch, i) => Math.floor(i / n) % 2 ? "*" : ch).join("");
}
let string = "9876543210";
console.log(mask(string, 1));
console.log(mask(string, 2));
console.log(mask(string, 3));

This code should work:
function stars(str, n = 1) {
const parts = str.split('')
let num = n
let printStars = false
return parts.map((letter) => {
if (num > 0 && !printStars) {
num -= 1
return letter
}
printStars = true
num += 1
if (num === n) {
printStars = false
}
return '*'
}).join('')
}
console.log(stars('14124123123'), 1)
console.log(stars('14124123123', 2), 2)
console.log(stars('14124123123', 3), 3)
console.log(stars('14124123123', 5), 5)
console.log(stars(''))
Cheers

This requires you to use the current mask as an argument and build your code upon it.
I also edited the function to allow other characters than the "*"
const number = 'abcdefghijklmnopqrstuvwxyzzz';
for(let mask = 1; mask <= 9; mask++){
console.log("Current mask:", mask, " Value: ",applyMask(number, mask));
}
function applyMask(data, mask, defaultMask = '*'){
// start i with the value of mask as we allow the first "n" characters to appear
let i;
let str = data;
for(i = mask; i < data.length ; i+=mask*2){
// I used the same substring method you used the diff is that i used the mask to get the next shown values
// HERE
str = str.substring(0, i) + defaultMask.repeat(mask) + str.substring(i + mask);
}
// this is to trim the string if any extra "defaultMask" were found
// this is to handle the not full size of the mask input at the end of the string
return str.slice(0, data.length)
}

Adding numeric values in a string together

I have a string returned from an endpoint in which I need to add certain parts of the string together in order to produce two different values.
Example response:
149,226;147,226;146,224
Now I know I can use the unary plus operator to force the string to be treated as a number like so.
var num1 = '20',
num2 = '22';
var total = (+num1) + (+num2);
or I could do some conversion like so
var number1 = parseInt(num1);
var number2 = parseInt(num2);
var total = number1 + number2;
either of these work fine however this is not what I am looking for exactly.
I want to take this result
149,226;147,226;146,224
Then add all the numbers before the first comer together so that would be (149, 147, 146) to produce one result and then add all the number after the second comer together (226, 226, 224).
I know I probably need some sort of reg expression for this I just dont know what.

You can just use string.split, twice, one for the ; and then again for the ,. And put this through array.reduce.
eg.
var str = '149,226;147,226;146,224';
var result = str.split(';')
.reduce((a,v) => {
var vv = v.split(',');
a[0] += vv[0] | 0;
a[1] += vv[1] | 0;
return a;
}, [0, 0]);
console.log(result);
For a more generic solution, that could handle any number of sub strings, eg. 1,2,3;4,5,6, and also handle alternative split types, and cope with extra , or ;.
function sumStrings(str, outerSplit, innerSplit) {
return str.split(outerSplit || ';')
.filter(Boolean)
.reduce((a,v) => {
v.split(innerSplit || ',')
.filter(Boolean)
.forEach((v,ix) => {
a[ix] = (a[ix] | 0) + (v | 0);
});
return a;
}, []);
}
console.log(sumStrings(
'149,226;147,226;146,224'
));
console.log(sumStrings(
'149.226.201|147.226.112.|146.224.300|',
'|','.'));
//how about total of totals?
console.log(sumStrings(
'149,226;147,226;146,224'
).reduce((a,v) => a + v));
.as-console-wrapper {
min-height: 100%
}

You could do:
const myString = '149,226;147,226;146,224';
/*
* 1. you split the string by ';' to obtain an array of string couples
* then you split each couple by ','. In this way you end up with an array like this:
* [['149', '266'], ['147', '266'], ['146', '264']]
*/
const myNumbers = myString.split(';').map(numCouple => numCouple.split(','));
/*
* 2. you use Array.prototype.reduce() to calculate the sums
*/
const sum1 = myNumbers.reduce((sum, item) => {
return sum += parseInt(item[0]);
}, 0);
const sum2 = myNumbers.reduce((sum, item) => {
return sum += parseInt(item[1]);
}, 0);
// or, as an alternative:
const sumsObj = myNumbers.reduce((obj, item) => {
obj.sum1 += parseInt(item[0]);
obj.sum2 += parseInt(item[1]);
return obj;
}, { sum1: 0, sum2: 0 });
// or also:
const sumsArr = myNumbers.reduce((acc, item) => {
acc[0] += parseInt(item[0]);
acc[1] += parseInt(item[1]);
return acc;
}, [0, 0]);
// test
console.log('sum1:', sum1);
console.log('sum2:', sum2);
console.log('--------------');
console.log('sum1:', sumsObj.sum1);
console.log('sum2:', sumsObj.sum2);
console.log('--------------');
console.log('sum1:', sumsArr[0]);
console.log('sum2:', sumsArr[1]);

without using regex, one possible solution:
var c = '149,226;147,226;146,224'
var d = c.split(";")
var first = d.map(x=>Number(x.split(",")[0]))
var second= d.map(x=>Number(x.split(",")[1]))
console.log(first)
console.log(second)
let resultFirst = first.reduce((a,b) => a + b, 0);
let resultSecond = second.reduce((a,b) => a + b, 0);
console.log(resultFirst)
console.log(resultSecond)

Below supplies regex to String's .split to get the numbers by themselves. Then you could add every other number but I don't see why not just add them all in order.
const str = '149,226;147,226;146,224'
const total = str.split(/[;,]/).map(Number).reduce((a, b) => a + b)
console.log('total', total)

I got this one running:
const nums = ('149,226;147,226;146,224');
var firstNums = nums.match(/(?<=[0-9]*)[0-9]+(?=,)/gs);
var secondNums = nums.match(/(?<=[0-9]*)[0-9]+(?=;|$)/gs);
console.log(firstNums, secondNums);
let sumFirstNums = 0,
sumSecondNums = 0;
firstNums.map(x => {
sumFirstNums += +x;
})
console.log(sumFirstNums)
secondNums.map(x => {
sumSecondNums += +x;
})
console.log(sumSecondNums)
//If you want the result in the same format:
const finalResult = `${sumFirstNums}, ${sumSecondNums};`
console.log(finalResult)
;)

For that string format, you could use a single pattern with 2 capturing groups matching 1+ more digits between a comma, and asserting a ; or the end of the string at the right.
You refer to the group values by indexing into the match of every iteration.
(\d+),(\d+)(?=;|$)
The pattern matches
(\d+) Capture group 1, match 1+ digits
, Match a comma
(\d+) Capture group 2, match 1+ digits
(?=;|$) Positive lookahead, assert directly to the right a ; or end of the string
See a regex demo.
let result1 = 0;
let result2 = 0
for (const match of "149,226;147,226;146,224".matchAll(/(\d+),(\d+)(?=;|$)/g)) {
result1 += +match[1]
result2 += +match[2]
}
console.log(result1, result2);

Alternative way to padStart [duplicate]

I am in need of a JavaScript function which can take a value and pad it to a given length (I need spaces, but anything would do). I found this, but I have no idea what the heck it is doing and it doesn't seem to work for me.
String.prototype.pad = function(l, s, t) {
return s || (s = " "),
(l -= this.length) > 0 ?
(s = new Array(Math.ceil(l / s.length) + 1).join(s))
.substr(0, t = !t ? l : t == 1 ?
0 :
Math.ceil(l / 2)) + this + s.substr(0, l - t) :
this;
};
var s = "Jonas";
document.write(
'<h2>S = '.bold(), s, "</h2>",
'S.pad(20, "[]", 0) = '.bold(), s.pad(20, "[]", 0), "<br />",
'S.pad(20, "[====]", 1) = '.bold(), s.pad(20, "[====]", 1), "<br />",
'S.pad(20, "~", 2) = '.bold(), s.pad(20, "~", 2)
);

ECMAScript 2017 (ES8) added String.padStart (along with String.padEnd) for just this purpose:
"Jonas".padStart(10); // Default pad string is a space
"42".padStart(6, "0"); // Pad with "0"
"*".padStart(8, "-/|\\"); // produces '-/|\\-/|*'
If not present in the JavaScript host, String.padStart can be added as a polyfill.
Pre ES8
I found this solution here and this is for me much much simpler:
var n = 123
String("00000" + n).slice(-5); // returns 00123
("00000" + n).slice(-5); // returns 00123
(" " + n).slice(-5); // returns " 123" (with two spaces)
And here I made an extension to the string object:
String.prototype.paddingLeft = function (paddingValue) {
return String(paddingValue + this).slice(-paddingValue.length);
};
An example to use it:
function getFormattedTime(date) {
var hours = date.getHours();
var minutes = date.getMinutes();
hours = hours.toString().paddingLeft("00");
minutes = minutes.toString().paddingLeft("00");
return "{0}:{1}".format(hours, minutes);
};
String.prototype.format = function () {
var args = arguments;
return this.replace(/{(\d+)}/g, function (match, number) {
return typeof args[number] != 'undefined' ? args[number] : match;
});
};
This will return a time in the format "15:30".

A faster method
If you are doing this repeatedly, for example to pad values in an array, and performance is a factor, the following approach can give you nearly a 100x advantage in speed (jsPerf) over other solution that are currently discussed on the inter webs. The basic idea is that you are providing the pad function with a fully padded empty string to use as a buffer. The pad function just appends to string to be added to this pre-padded string (one string concat) and then slices or trims the result to the desired length.
function pad(pad, str, padLeft) {
if (typeof str === 'undefined')
return pad;
if (padLeft) {
return (pad + str).slice(-pad.length);
} else {
return (str + pad).substring(0, pad.length);
}
}
For example, to zero pad a number to a length of 10 digits,
pad('0000000000',123,true);
To pad a string with whitespace, so the entire string is 255 characters,
var padding = Array(256).join(' '), // make a string of 255 spaces
pad(padding,123,true);
Performance Test
See the jsPerf test here.
And this is faster than ES6 string.repeat by 2x as well, as shown by the revised JsPerf here
Please note that jsPerf is no longer online
Please note that the jsPerf site that we originally used to benchmark the various methods is no longer online. Unfortunately, this means we can't get to those test results. Sad but true.

String.prototype.padStart() and String.prototype.padEnd() are currently TC39 candidate proposals: see github.com/tc39/proposal-string-pad-start-end (only available in Firefox as of April 2016; a polyfill is available).

http://www.webtoolkit.info/javascript_pad.html
/**
*
* JavaScript string pad
* http://www.webtoolkit.info/
*
**/
var STR_PAD_LEFT = 1;
var STR_PAD_RIGHT = 2;
var STR_PAD_BOTH = 3;
function pad(str, len, pad, dir) {
if (typeof(len) == "undefined") { var len = 0; }
if (typeof(pad) == "undefined") { var pad = ' '; }
if (typeof(dir) == "undefined") { var dir = STR_PAD_RIGHT; }
if (len + 1 >= str.length) {
switch (dir){
case STR_PAD_LEFT:
str = Array(len + 1 - str.length).join(pad) + str;
break;
case STR_PAD_BOTH:
var padlen = len - str.length;
var right = Math.ceil( padlen / 2 );
var left = padlen - right;
str = Array(left+1).join(pad) + str + Array(right+1).join(pad);
break;
default:
str = str + Array(len + 1 - str.length).join(pad);
break;
} // switch
}
return str;
}
It's a lot more readable.

Here's a recursive approach to it.
function pad(width, string, padding) {
return (width <= string.length) ? string : pad(width, padding + string, padding)
}
An example...
pad(5, 'hi', '0')
=> "000hi"

ECMAScript 2017 adds a padStart method to the String prototype. This method will pad a string with spaces to a given length. This method also takes an optional string that will be used instead of spaces for padding.
'abc'.padStart(10); // " abc"
'abc'.padStart(10, "foo"); // "foofoofabc"
'abc'.padStart(6,"123465"); // "123abc"
'abc'.padStart(8, "0"); // "00000abc"
'abc'.padStart(1); // "abc"
A padEnd method was also added that works in the same manner.
For browser compatibility (and a useful polyfill) see this link.

Using the ECMAScript 6 method String#repeat, a pad function is as simple as:
String.prototype.padLeft = function(char, length) {
return char.repeat(Math.max(0, length - this.length)) + this;
}
String#repeat is currently supported in Firefox and Chrome only. for other implementation, one might consider the following simple polyfill:
String.prototype.repeat = String.prototype.repeat || function(n){
return n<=1 ? this : (this + this.repeat(n-1));
}

Using the ECMAScript 6 method String#repeat and Arrow functions, a pad function is as simple as:
var leftPad = (s, c, n) => c.repeat(n - s.length) + s;
leftPad("foo", "0", 5); //returns "00foo"
jsfiddle
edit:
suggestion from the comments:
const leftPad = (s, c, n) => n - s.length > 0 ? c.repeat(n - s.length) + s : s;
this way, it wont throw an error when s.lengthis greater than n
edit2:
suggestion from the comments:
const leftPad = (s, c, n) =>{ s = s.toString(); c = c.toString(); return s.length > n ? s : c.repeat(n - s.length) + s; }
this way, you can use the function for strings and non-strings alike.

The key trick in both those solutions is to create an array instance with a given size (one more than the desired length), and then to immediately call the join() method to make a string. The join() method is passed the padding string (spaces probably). Since the array is empty, the empty cells will be rendered as empty strings during the process of joining the array into one result string, and only the padding will remain. It's a really nice technique.

With ES8, there are two options for padding.
You can check them in the documentation.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padEnd
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

Taking up Samuel's ideas, upward here. And remember an old SQL script, I tried with this:
a=1234;
'0000'.slice(a.toString().length)+a;
It works in all the cases I could imagine:
a= 1 result 0001
a= 12 result 0012
a= 123 result 0123
a= 1234 result 1234
a= 12345 result 12345
a= '12' result 0012

Pad with default values
I noticed that I mostly need the padLeft for time conversion / number padding.
So I wrote this function:
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array(b || 2).join(c || 0) + a).slice(-b)
}
This simple function supports Number or String as input.
The default pad is two characters.
The default char is 0.
So I can simply write:
padL(1);
// 01
If I add the second argument (pad width):
padL(1, 3);
// 001
The third parameter (pad character)
padL('zzz', 10, 'x');
// xxxxxxxzzz
#BananaAcid: If you pass a undefined value or a 0 length string, you get 0undefined, so:
As suggested
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array((b || 1) + 1).join(c || 0) + (a || '')).slice(-(b || 2))
}
But this can also be achieved in a shorter way.
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array(b || 2).join(c || 0) + (a || c || 0)).slice(-b)
}
It also works with:
padL(0)
padL(NaN)
padL('')
padL(undefined)
padL(false)
And if you want to be able to pad in both ways:
function pad(a, b, c, d) { // string/number, length=2, char=0, 0/false=Left-1/true=Right
return a = (a || c || 0), c = new Array(b || 2).join(c || 0), d ? (a + c).slice(0, b) : (c + a).slice(-b)
}
which can be written in a shorter way without using slice.
function pad(a, b, c, d) {
return a = (a || c || 0) + '', b = new Array((++b || 3) - a.length).join(c || 0), d ? a+b : b+a
}
/*
Usage:
pad(
input // (int or string) or undefined, NaN, false, empty string
// default:0 or PadCharacter
// Optional
,PadLength // (int) default:2
,PadCharacter // (string or int) default:'0'
,PadDirection // (bolean) default:0 (padLeft) - (true or 1) is padRight
)
*/
Now if you try to pad 'averylongword' with 2... that’s not my problem.
I said that I would give you a tip.
Most of the time, if you pad, you do it for the same value N times.
Using any type of function inside a loop slows down the loop!!!
So if you just want to pad left some numbers inside a long list, don't use functions to do this simple thing.
Use something like this:
var arrayOfNumbers = [1, 2, 3, 4, 5, 6, 7],
paddedArray = [],
len = arrayOfNumbers.length;
while(len--) {
paddedArray[len] = ('0000' + arrayOfNumbers[len]).slice(-4);
}
If you don't know how the maximum padding size based on the numbers inside the array.
var arrayOfNumbers = [1, 2, 3, 4, 5, 6, 7, 49095],
paddedArray = [],
len = arrayOfNumbers.length;
// Search the highest number
var arrayMax = Function.prototype.apply.bind(Math.max, null),
// Get that string length
padSize = (arrayMax(arrayOfNumbers) + '').length,
// Create a Padding string
padStr = new Array(padSize).join(0);
// And after you have all this static values cached start the loop.
while(len--) {
paddedArray[len] = (padStr + arrayOfNumbers[len]).slice(-padSize); // substr(-padSize)
}
console.log(paddedArray);
/*
0: "00001"
1: "00002"
2: "00003"
3: "00004"
4: "00005"
5: "00006"
6: "00007"
7: "49095"
*/

padding string has been inplemented in new javascript version.
str.padStart(targetLength [, padString])
https://developer.mozilla.org/es/docs/Web/JavaScript/Referencia/Objetos_globales/String/padStart
If you want your own function check this example:
const myString = 'Welcome to my house';
String.prototype.padLeft = function(times = 0, str = ' ') {
return (Array(times).join(str) + this);
}
console.log(myString.padLeft(12, ':'));
//:::::::::::Welcome to my house

Here is a build in method you can use -
str1.padStart(2, '0')

Here's a simple function that I use.
var pad=function(num,field){
var n = '' + num;
var w = n.length;
var l = field.length;
var pad = w < l ? l-w : 0;
return field.substr(0,pad) + n;
};
For example:
pad (20,' '); // 20
pad (321,' '); // 321
pad (12345,' '); //12345
pad ( 15,'00000'); //00015
pad ( 999,'*****'); //**999
pad ('cat','_____'); //__cat

A short way:
(x=>(new Array(int-x.length+1)).join(char)+x)(String)
Example:
(x=>(new Array(6-x.length+1)).join("0")+x)("1234")
return: "001234"

Here is a simple answer in basically one line of code.
var value = 35 // the numerical value
var x = 5 // the minimum length of the string
var padded = ("00000" + value).substr(-x);
Make sure the number of characters in you padding, zeros here, is at least as many as your intended minimum length. So really, to put it into one line, to get a result of "00035" in this case is:
var padded = ("00000" + 35).substr(-5);

ES7 is just drafts and proposals right now, but if you wanted to track compatibility with the specification, your pad functions need:
Multi-character pad support.
Don't truncate the input string
Pad defaults to space
From my polyfill library, but apply your own due diligence for prototype extensions.
// Tests
'hello'.lpad(4) === 'hello'
'hello'.rpad(4) === 'hello'
'hello'.lpad(10) === ' hello'
'hello'.rpad(10) === 'hello '
'hello'.lpad(10, '1234') === '41234hello'
'hello'.rpad(10, '1234') === 'hello12341'
String.prototype.lpad || (String.prototype.lpad = function(length, pad)
{
if(length < this.length)
return this;
pad = pad || ' ';
let str = this;
while(str.length < length)
{
str = pad + str;
}
return str.substr( -length );
});
String.prototype.rpad || (String.prototype.rpad = function(length, pad)
{
if(length < this.length)
return this;
pad = pad || ' ';
let str = this;
while(str.length < length)
{
str += pad;
}
return str.substr(0, length);
});

Array manipulations are really slow compared to simple string concat. Of course, benchmark for your use case.
function(string, length, pad_char, append) {
string = string.toString();
length = parseInt(length) || 1;
pad_char = pad_char || ' ';
while (string.length < length) {
string = append ? string+pad_char : pad_char+string;
}
return string;
};

A variant of #Daniel LaFavers' answer.
var mask = function (background, foreground) {
bg = (new String(background));
fg = (new String(foreground));
bgl = bg.length;
fgl = fg.length;
bgs = bg.substring(0, Math.max(0, bgl - fgl));
fgs = fg.substring(Math.max(0, fgl - bgl));
return bgs + fgs;
};
For example:
mask('00000', 11 ); // '00011'
mask('00011','00' ); // '00000'
mask( 2 , 3 ); // '3'
mask('0' ,'111'); // '1'
mask('fork' ,'***'); // 'f***'
mask('_____','dog'); // '__dog'

If you don't mind including a utility library, lodash library has _.pad, _.padLeft and _.padRight functions.

I think its better to avoid recursion because its costly.
function padLeft(str,size,padwith) {
if(size <= str.length) {
// not padding is required.
return str;
} else {
// 1- take array of size equal to number of padding char + 1. suppose if string is 55 and we want 00055 it means we have 3 padding char so array size should be 3 + 1 (+1 will explain below)
// 2- now join this array with provided padding char (padwith) or default one ('0'). so it will produce '000'
// 3- now append '000' with orginal string (str = 55), will produce 00055
// why +1 in size of array?
// it is a trick, that we are joining an array of empty element with '0' (in our case)
// if we want to join items with '0' then we should have at least 2 items in the array to get joined (array with single item doesn't need to get joined).
// <item>0<item>0<item>0<item> to get 3 zero we need 4 (3+1) items in array
return Array(size-str.length+1).join(padwith||'0')+str
}
}
alert(padLeft("59",5) + "\n" +
padLeft("659",5) + "\n" +
padLeft("5919",5) + "\n" +
padLeft("59879",5) + "\n" +
padLeft("5437899",5));

It's 2014, and I suggest a JavaScript string-padding function. Ha!
Bare-bones: right-pad with spaces
function pad (str, length) {
var padding = (new Array(Math.max(length - str.length + 1, 0))).join(" ");
return str + padding;
}
Fancy: pad with options
/**
* #param {*} str Input string, or any other type (will be converted to string)
* #param {number} length Desired length to pad the string to
* #param {Object} [opts]
* #param {string} [opts.padWith=" "] Character to use for padding
* #param {boolean} [opts.padLeft=false] Whether to pad on the left
* #param {boolean} [opts.collapseEmpty=false] Whether to return an empty string if the input was empty
* #returns {string}
*/
function pad(str, length, opts) {
var padding = (new Array(Math.max(length - (str + "").length + 1, 0))).join(opts && opts.padWith || " "),
collapse = opts && opts.collapseEmpty && !(str + "").length;
return collapse ? "" : opts && opts.padLeft ? padding + str : str + padding;
}
Usage (fancy):
pad("123", 5);
// Returns "123 "
pad(123, 5);
// Returns "123 " - non-string input
pad("123", 5, { padWith: "0", padLeft: true });
// Returns "00123"
pad("", 5);
// Returns " "
pad("", 5, { collapseEmpty: true });
// Returns ""
pad("1234567", 5);
// Returns "1234567"

/**************************************************************************************************
Pad a string to pad_length fillig it with pad_char.
By default the function performs a left pad, unless pad_right is set to true.
If the value of pad_length is negative, less than, or equal to the length of the input string, no padding takes place.
**************************************************************************************************/
if(!String.prototype.pad)
String.prototype.pad = function(pad_char, pad_length, pad_right)
{
var result = this;
if( (typeof pad_char === 'string') && (pad_char.length === 1) && (pad_length > this.length) )
{
var padding = new Array(pad_length - this.length + 1).join(pad_char); //thanks to http://stackoverflow.com/questions/202605/repeat-string-javascript/2433358#2433358
result = (pad_right ? result + padding : padding + result);
}
return result;
}
And then you can do:
alert( "3".pad("0", 3) ); //shows "003"
alert( "hi".pad(" ", 3) ); //shows " hi"
alert( "hi".pad(" ", 3, true) ); //shows "hi "

If you just want a very simple hacky one-liner to pad, just make a string of the desired padding character of the desired max padding length and then substring it to the length of what you want to pad.
Example: padding the string store in e with spaces to 25 characters long.
var e = "hello"; e = e + " ".substring(e.length)
Result: "hello "
If you want to do the same with a number as input just call .toString() on it before.

A friend asked about using a JavaScript function to pad left. It turned into a little bit of an endeavor between some of us in chat to code golf it. This was the result:
function l(p,t,v){
v+="";return v.length>=t?v:l(p,t,p+v);
}
It ensures that the value to be padded is a string, and then if it isn't the length of the total desired length it will pad it once and then recurse. Here is what it looks like with more logical naming and structure
function padLeft(pad, totalLength, value){
value = value.toString();
if( value.length >= totalLength ){
return value;
}else{
return padLeft(pad, totalLength, pad + value);
}
}
The example we were using was to ensure that numbers were padded with 0 to the left to make a max length of 6. Here is an example set:
function l(p,t,v){v+="";return v.length>=t?v:l(p,t,p+v);}
var vals = [6451,123,466750];
var pad = l(0,6,vals[0]);// pad with 0's, max length 6
var pads = vals.map(function(i){ return l(0,6,i) });
document.write(pads.join("<br />"));

A little late, but thought I might share anyway. I found it useful to add a prototype extension to Object. That way I can pad numbers and strings, left or right. I have a module with similar utilities I include in my scripts.
// include the module in your script, there is no need to export
var jsAddOns = require('<path to module>/jsAddOns');
~~~~~~~~~~~~ jsAddOns.js ~~~~~~~~~~~~
/*
* method prototype for any Object to pad it's toString()
* representation with additional characters to the specified length
*
* #param padToLength required int
* entire length of padded string (original + padding)
* #param padChar optional char
* character to use for padding, default is white space
* #param padLeft optional boolean
* if true padding added to left
* if omitted or false, padding added to right
*
* #return padded string or
* original string if length is >= padToLength
*/
Object.prototype.pad = function(padToLength, padChar, padLeft) {
// get the string value
s = this.toString()
// default padToLength to 0
// if omitted, original string is returned
padToLength = padToLength || 0;
// default padChar to empty space
padChar = padChar || ' ';
// ignore padding if string too long
if (s.length >= padToLength) {
return s;
}
// create the pad of appropriate length
var pad = Array(padToLength - s.length).join(padChar);
// add pad to right or left side
if (padLeft) {
return pad + s;
} else {
return s + pad;
}
};

Never insert data somewhere (especially not at beginning, like str = pad + str;), since the data will be reallocated everytime. Append always at end!
Don't pad your string in the loop. Leave it alone and build your pad string first. In the end concatenate it with your main string.
Don't assign padding string each time (like str += pad;). It is much faster to append the padding string to itself and extract first x-chars (the parser can do this efficiently if you extract from first char). This is exponential growth, which means that it wastes some memory temporarily (you should not do this with extremely huge texts).
if (!String.prototype.lpad) {
String.prototype.lpad = function(pad, len) {
while (pad.length < len) {
pad += pad;
}
return pad.substr(0, len-this.length) + this;
}
}
if (!String.prototype.rpad) {
String.prototype.rpad = function(pad, len) {
while (pad.length < len) {
pad += pad;
}
return this + pad.substr(0, len-this.length);
}
}

Here is a JavaScript function that adds a specified number of paddings with a custom symbol. The function takes three parameters.
padMe --> string or number to left pad
pads --> number of pads
padSymble --> custom symbol, default is "0"
function leftPad(padMe, pads, padSymble) {
if(typeof padMe === "undefined") {
padMe = "";
}
if (typeof pads === "undefined") {
pads = 0;
}
if (typeof padSymble === "undefined") {
padSymble = "0";
}
var symble = "";
var result = [];
for(var i=0; i < pads; i++) {
symble += padSymble;
}
var length = symble.length - padMe.toString().length;
result = symble.substring(0, length);
return result.concat(padMe.toString());
}
Here are some results:
> leftPad(1)
"1"
> leftPad(1, 4)
"0001"
> leftPad(1, 4, "0")
"0001"
> leftPad(1, 4, "#")
"###1"

Yet another take at with combination of a couple of solutions:
/**
* pad string on left
* #param {number} number of digits to pad, default is 2
* #param {string} string to use for padding, default is '0' *
* #returns {string} padded string
*/
String.prototype.paddingLeft = function (b, c) {
if (this.length > (b||2))
return this + '';
return (this || c || 0) + '', b = new Array((++b || 3) - this.length).join(c || 0), b + this
};
/**
* pad string on right
* #param {number} number of digits to pad, default is 2
* #param {string} string to use for padding, default is '0' *
* #returns {string} padded string
*/
String.prototype.paddingRight = function (b, c) {
if (this.length > (b||2))
return this + '';
return (this||c||0) + '', b = new Array((++b || 3) - this.length).join(c || 0), this + b
};

Convert Number To String (Charset Supplied) JavaScript

Update:
The following code works perfectly until $char.to_text encounters an integer greater than 55,834,574,847.
alpha="abcdefghijklmnopqrstuvwxyz";
$char={
to_num:function(s,c){
var l=c.length,o={};
c.split('').forEach(function(a,i){
o[a]=i
});
return s.split('').reduce(function(r,a){
return r*l+o[a]
},0)
},
to_text:function(i,c){
var l=c.length,s='';
do{
s=c[i%l]+s; // i%l
i/=l;
i|=0
}while(i!==0);
return s
}
};
Here is a quick snip:
$char.to_num("military",alpha) => 98987733674
$char.to_text(98987733674,alpha) => "undefinedundefinedundefinedundefinedundefinedundefinedundefinedy"
Manually iterating the above code should generate a normal response, why does it yield this "undefined..." string, is it simply because it's a large number operation for JS?

This is a proposal with a rewritten hash function, which uses an object o as simplified indexOf and a simple loop for the return value.
The wanted function ihash uses a single do ... until loop. It uses the remainder of the value and the length as index of the given charcter set. The value is then divided by the lenght of the caracter set and the integer part is taken for the next iteration, if not equal zero.
function hash(s) {
var c = '0abcdefghijklmnopqrstuvwxyz',
l = c.length,
o = {};
c.split('').forEach(function (a, i) {
o[a] = i;
});
return s.split('').reduce(function (r, a) {
return r * l + o[a];
}, 0);
}
function ihash(i) {
var c = '0abcdefghijklmnopqrstuvwxyz',
l = c.length,
s = '';
do {
s = c[i % l] + s;
i = Math.floor(i / l);
} while (i !== 0);
return s;
}
document.write(hash('0') + '<br>'); // => 0
document.write(hash('a') + '<br>'); // => 1
document.write(hash('hi') + '<br>'); // => 225
document.write(hash('world') + '<br>'); // => 12531838
document.write(hash('freecode') + '<br>'); // => 69810159857
document.write(ihash(0) + '<br>'); // => '0'
document.write(ihash(1) + '<br>'); // => 'a'
document.write(ihash(225) + '<br>'); // => 'hi'
document.write(ihash(12531838) + '<br>'); // => 'world'
document.write(ihash(69810159857) + '<br>'); // => 'freecode'

Here is a pseudo code for getting the string back. It's similar to convert numbers of different base.
var txt = function(n, charset) {
var s =""
while (n > 0) {
var r = n % charset.length;
n = n / charset.length;
s += charset[r];
}
return s;
}

How to reduce consecutive integers in an array to hyphenated range expressions?

In JavaScript, how can I convert a sequence of numbers in an array to a range of numbers? In other words, I want to express consecutive occurring integers (no gaps) as hyphenated ranges.
[2,3,4,5,10,18,19,20] would become [2-5,10,18-20]
[1,6,7,9,10,12] would become [1,6-7,9-10,12]
[3,5,99] would remain [3,5,99]
[5,6,7,8,9,10,11] would become [5-11]

Here is an algorithm that I made some time ago, originally written for C#, now I ported it to JavaScript:
function getRanges(array) {
var ranges = [], rstart, rend;
for (var i = 0; i < array.length; i++) {
rstart = array[i];
rend = rstart;
while (array[i + 1] - array[i] == 1) {
rend = array[i + 1]; // increment the index if the numbers sequential
i++;
}
ranges.push(rstart == rend ? rstart+'' : rstart + '-' + rend);
}
return ranges;
}
getRanges([2,3,4,5,10,18,19,20]);
// returns ["2-5", "10", "18-20"]
getRanges([1,2,3,5,7,9,10,11,12,14 ]);
// returns ["1-3", "5", "7", "9-12", "14"]
getRanges([1,2,3,4,5,6,7,8,9,10])
// returns ["1-10"]

Just having fun with solution from CMS :
function getRanges (array) {
for (var ranges = [], rend, i = 0; i < array.length;) {
ranges.push ((rend = array[i]) + ((function (rstart) {
while (++rend === array[++i]);
return --rend === rstart;
})(rend) ? '' : '-' + rend));
}
return ranges;
}

Very nice question: here's my attempt:
function ranges(numbers){
var sorted = numbers.sort(function(a,b){return a-b;});
var first = sorted.shift();
return sorted.reduce(function(ranges, num){
if(num - ranges[0][1] <= 1){
ranges[0][1] = num;
} else {
ranges.unshift([num,num]);
}
return ranges;
},[[first,first]]).map(function(ranges){
return ranges[0] === ranges[1] ?
ranges[0].toString() : ranges.join('-');
}).reverse();
}
Demo on JSFiddler

I needed TypeScript code today to solve this very problem -- many years after the OP -- and decided to try a version written in a style more functional than the other answers here. Of course, only the parameter and return type annotations distinguish this code from standard ES6 JavaScript.
function toRanges(values: number[],
separator = '\u2013'): string[] {
return values
.slice()
.sort((p, q) => p - q)
.reduce((acc, cur, idx, src) => {
if ((idx > 0) && ((cur - src[idx - 1]) === 1))
acc[acc.length - 1][1] = cur;
else acc.push([cur]);
return acc;
}, [])
.map(range => range.join(separator));
}
Note that slice is necessary because sort sorts in place and we can't change the original array.

Here's my take on this...
function getRanges(input) {
//setup the return value
var ret = [], ary, first, last;
//copy and sort
var ary = input.concat([]);
ary.sort(function(a,b){
return Number(a) - Number(b);
});
//iterate through the array
for (var i=0; i<ary.length; i++) {
//set the first and last value, to the current iteration
first = last = ary[i];
//while within the range, increment
while (ary[i+1] == last+1) {
last++;
i++;
}
//push the current set into the return value
ret.push(first == last ? first : first + "-" + last);
}
//return the response array.
return ret;
}

Using ES6, a solution is:
function display ( vector ) { // assume vector sorted in increasing order
// display e.g.vector [ 2,4,5,6,9,11,12,13,15 ] as "2;4-6;9;11-13;15"
const l = vector.length - 1; // last valid index of vector array
// map [ 2,4,5,6,9,11,12,13,15 ] into array of strings (quote ommitted)
// --> [ "2;", "4-", "-", "6;", "9;", "11-", "-", "13;", "15;" ]
vector = vector.map ( ( n, i, v ) => // n is current number at index i of vector v
i < l && v [ i + 1 ] - n === 1 ? // next number is adjacent ?
`${ i > 0 && n - v [ i - 1 ] === 1 ? "" : n }-` :
`${ n };`
);
return vector.join ( "" ). // concatenate all strings in vector array
replace ( /-+/g, "-" ). // replace multiple dashes by single dash
slice ( 0, -1 ); // remove trailing ;
}
If you want to add extra spaces for readability, just add extra calls to string.prototype.replace().
If the input vector is not sorted, you can add the following line right after the opening brace of the display() function:
vector.sort ( ( a, b ) => a - b ); // sort vector in place, in increasing order.
Note that this could be improved to avoid testing twice for integer adjacentness (adjacenthood? I'm not a native English speaker;-).
And of course, if you don't want a single string as output, split it with ";".

Rough outline of the process is as follows:
Create an empty array called ranges
For each value in sorted input array
If ranges is empty then insert the item {min: value, max: value}
Else if max of last item in ranges and the current value are consecutive then set max of last item in ranges = value
Else insert the item {min: value, max: value}
Format the ranges array as desired e.g. by combining min and max if same
The following code uses Array.reduce and simplifies the logic by combining step 2.1 and 2.3.
function arrayToRange(array) {
return array
.slice()
.sort(function(a, b) {
return a - b;
})
.reduce(function(ranges, value) {
var lastIndex = ranges.length - 1;
if (lastIndex === -1 || ranges[lastIndex].max !== value - 1) {
ranges.push({ min: value, max: value });
} else {
ranges[lastIndex].max = value;
}
return ranges;
}, [])
.map(function(range) {
return range.min !== range.max ? range.min + "-" + range.max : range.min.toString();
});
}
console.log(arrayToRange([2, 3, 4, 5, 10, 18, 19, 20]));

If you simply want a string that represents a range, then you'd find the mid-point of your sequence, and that becomes your middle value (10 in your example). You'd then grab the first item in the sequence, and the item that immediately preceded your mid-point, and build your first-sequence representation. You'd follow the same procedure to get your last item, and the item that immediately follows your mid-point, and build your last-sequence representation.
// Provide initial sequence
var sequence = [1,2,3,4,5,6,7,8,9,10];
// Find midpoint
var midpoint = Math.ceil(sequence.length/2);
// Build first sequence from midpoint
var firstSequence = sequence[0] + "-" + sequence[midpoint-2];
// Build second sequence from midpoint
var lastSequence = sequence[midpoint] + "-" + sequence[sequence.length-1];
// Place all new in array
var newArray = [firstSequence,midpoint,lastSequence];
alert(newArray.join(",")); // 1-4,5,6-10
Demo Online: http://jsbin.com/uvahi/edit

; For all cells of the array
;if current cell = prev cell + 1 -> range continues
;if current cell != prev cell + 1 -> range ended
int[] x = [2,3,4,5,10,18,19,20]
string output = '['+x[0]
bool range = false; --current range
for (int i = 1; i > x[].length; i++) {
if (x[i+1] = [x]+1) {
range = true;
} else { //not sequential
if range = true
output = output || '-'
else
output = output || ','
output.append(x[i]','||x[i+1])
range = false;
}
}
Something like that.

An adaptation of CMS's javascript solution for Cold Fusion
It does sort the list first so that 1,3,2,4,5,8,9,10 (or similar) properly converts to 1-5,8-10.
<cfscript>
function getRanges(nArr) {
arguments.nArr = listToArray(listSort(arguments.nArr,"numeric"));
var ranges = [];
var rstart = "";
var rend = "";
for (local.i = 1; i <= ArrayLen(arguments.nArr); i++) {
rstart = arguments.nArr[i];
rend = rstart;
while (i < ArrayLen(arguments.nArr) and (val(arguments.nArr[i + 1]) - val(arguments.nArr[i])) == 1) {
rend = val(arguments.nArr[i + 1]); // increment the index if the numbers sequential
i++;
}
ArrayAppend(ranges,rstart == rend ? rstart : rstart & '-' & rend);
}
return arraytolist(ranges);
}
</cfscript>

Tiny ES6 module for you guys. It accepts a function to determine when we must break the sequence (breakDetectorFunc param - default is the simple thing for integer sequence input).
NOTICE: since input is abstract - there's no auto-sorting before processing, so if your sequence isn't sorted - do it prior to calling this module
function defaultIntDetector(a, b){
return Math.abs(b - a) > 1;
}
/**
* #param {Array} valuesArray
* #param {Boolean} [allArraysResult=false] if true - [1,2,3,7] will return [[1,3], [7,7]]. Otherwise [[1.3], 7]
* #param {SequenceToIntervalsBreakDetector} [breakDetectorFunc] must return true if value1 and value2 can't be in one sequence (if we need a gap here)
* #return {Array}
*/
const sequenceToIntervals = function (valuesArray, allArraysResult, breakDetectorFunc) {
if (!breakDetectorFunc){
breakDetectorFunc = defaultIntDetector;
}
if (typeof(allArraysResult) === 'undefined'){
allArraysResult = false;
}
const intervals = [];
let from = 0, to;
if (valuesArray instanceof Array) {
const cnt = valuesArray.length;
for (let i = 0; i < cnt; i++) {
to = i;
if (i < cnt - 1) { // i is not last (to compare to next)
if (breakDetectorFunc(valuesArray[i], valuesArray[i + 1])) {
// break
appendLastResult();
}
}
}
appendLastResult();
} else {
throw new Error("input is not an Array");
}
function appendLastResult(){
if (isFinite(from) && isFinite(to)) {
const vFrom = valuesArray[from];
const vTo = valuesArray[to];
if (from === to) {
intervals.push(
allArraysResult
? [vFrom, vTo] // same values array item
: vFrom // just a value, no array
);
} else if (Math.abs(from - to) === 1) { // sibling items
if (allArraysResult) {
intervals.push([vFrom, vFrom]);
intervals.push([vTo, vTo]);
} else {
intervals.push(vFrom, vTo);
}
} else {
intervals.push([vFrom, vTo]); // true interval
}
from = to + 1;
}
}
return intervals;
};
module.exports = sequenceToIntervals;
/** #callback SequenceToIntervalsBreakDetector
#param value1
#param value2
#return bool
*/
first argument is the input sequence sorted array, second is a boolean flag controlling the output mode: if true - single item (outside the intervals) will be returned as arrays anyway: [1,7],[9,9],[10,10],[12,20], otherwise single items returned as they appear in the input array
for your sample input
[2,3,4,5,10,18,19,20]
it will return:
sequenceToIntervals([2,3,4,5,10,18,19,20], true) // [[2,5], [10,10], [18,20]]
sequenceToIntervals([2,3,4,5,10,18,19,20], false) // [[2,5], 10, [18,20]]
sequenceToIntervals([2,3,4,5,10,18,19,20]) // [[2,5], 10, [18,20]]

Here's a version in Coffeescript
getRanges = (array) ->
ranges = []
rstart
rend
i = 0
while i < array.length
rstart = array[i]
rend = rstart
while array[i + 1] - array[i] is 1
rend = array[i + 1] # increment the index if the numbers sequential
i = i + 1
if rstart == rend
ranges.push rstart + ''
else
ranges.push rstart + '-' + rend
i = i + 1
return ranges

I've written my own method that's dependent on Lo-Dash, but doesn't just give you back an array of ranges, rather, it just returns an array of range groups.
[1,2,3,4,6,8,10] becomes:
[[1,2,3,4],[6,8,10]]
http://jsfiddle.net/mberkom/ufVey/