Lilah has a string, s, of lowercase English letters that she repeated infinitely many times.
Given an integer, n, find and print the number of letter 'a' in the first n letters of Lilah's infinite string.
This is my solution, but it is not correct, and I'm struggling to figure out why:
function repeatedString(s, n) {
let counter = 0;
const remainder = n % s.length;
const substring = s.substring(0, remainder);
const concatString = s + substring;
for (let letter of concatString) {
if (letter === 'a') {
counter++;
}
}
return (counter * n);
}
const str = "dhfgjhdfoiahwiuerhiguhzlkjvxzlkjghatriaeriaauih";
console.log(
repeatedString(str, 20)
);
I think it may be the
const concatString = s + substring;
would you please just reference the substring instead...
for (let letter of substring) {
if (letter === 'a') {
counter++;
}
}
return counter
You just need to loop through the s.substring(0, n) and return the counter value, (counter * n) doesn't make any sense.
function repeatedString(s, n) {
let counter = 0;
//const remainder = n % s.length;
const substring = s.substring(0, n);
console.log(substring);
//const concatString = s + substring;
for (let letter of substring) {
if (letter === 'a') {
counter++;
}
}
return counter;
}
const str = "dhfgjhdfoiahwiuerhiguhzlkjvxzlkjghatriaeriaauih";
console.log(
repeatedString(str, 20)
);
You can do that in following steps:
Get the count of the letter in given string. i.e c1.
Then get the part of the substring using slice(). The part will start of 0 up the the remainder of n and length of string.(c2)
Then multiply c1 with the number of times the given string will be in string of length n. i.e c1 * Math.floor(n/str.length)
Add the other count of remaining part c2 to the result and return
You can do that using filter() and check the count of given letter in given string. And then multiply it with no of times the string will repeat in for length n.
function func(str,l,n){
let c1 = [...str].filter(x => x === l).length;
let c2 = [...str.slice(0,n%str.length)].filter(x => x === l).length;
return (c1 * Math.floor(n/str.length)) + c2;
}
console.log(func('abcac','a',10))
This should give you the number of times a appears in the numbers of n length
const input = s;
var subStr = input.substr(0,n).split('');
console.log(subStr);
var output = 0;
subStr.forEach((e) => {
if (e === 'a') {
console.log(e);
output++;
}
})
console.log(output);
Using the next link as the source of the question:
https://medium.com/#yashka.troy/want-to-know-the-major-player-in-programming-18f2d35d91f7
Where it is explained better:
Lilah has a string, s, of lowercase English letters that she repeated infinitely many times.
Given an integer, n, find and print the number of letter a's in the first letters of Lilah's infinite string.
For example, if the string s=’abcac’ and n=10, the sub string we consider is ‘abcacabcac’, the first 10 characters of her infinite string. There are 4 occurrences of a in the substring.
A solution could be:
function repeatedString(s, n)
{
let res = 0;
const remainder = s.slice(0, n % s.length);
// Get number of times 'a' is present on "n / s.length" repetitions.
for (let letter of s)
{
if (letter === 'a') res++;
}
res *= Math.floor(n / s.length);
// Add the number of times 'a' is present on the remainder.
for (let letter of remainder)
{
if (letter === 'a') res++;
}
return res;
}
const str = "abca";
console.log(`${repeatedString(str, 10)} a's on first 10 letters:`);
console.log(`${repeatedString(str, 4)} a's on first 4 letters:`);
console.log(`${repeatedString(str, 0)} a's on first 0 letters:`);
console.log(`${repeatedString(str, 22)} a's on first 22 letters:`);
.as-console {background-color:black !important; color:lime;}
.as-console-wrapper {max-height:100% !important; top:0;}
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
};
the purpose of this code is to to write a function which increments a string, to create a new string. If the string already ends with a number, the number should be incremented by 1. If the string does not end with a number the number 1 should be appended to the new string. e.g. "foo123" --> "foo124" or "foo" --> "foo1".
With my code below, pretty much all my test cases are passed except a corner case for "foo999" did not print out "foo1000". I know that there should be a way to do with regex to fix my problem, but I am not too familiar with it. Can anyone please help?
function incrementString (input) {
var reg = /[0-9]/;
var result = "";
if(reg.test(input[input.length - 1]) === true){
input = input.split("");
for(var i = 0; i < input.length; i++){
if(parseInt(input[i]) === NaN){
result += input[i];
}
else if(i === input.length - 1){
result += (parseInt(input[i]) + 1).toString();
}
else{
result += input[i];
}
}
return result;
}
else if (reg.test(input[input.length - 1]) === false){
return input += 1;
}
}
You can use replace with a callback:
'foo'.replace(/(\d*)$/, function($0, $1) { return $1*1+1; });
//=> "foo1"
'foo999'.replace(/(\d*)$/, function($0, $1) { return $1*1+1; });
//=> "foo1000"
'foo123'.replace(/(\d*)$/, function($0, $1) { return $1*1+1; });
//=> "foo124"
Explanation:
/(\d*)$/ # match 0 or more digits at the end of string
function($0, $1) {...} # callback function with 2nd parameter as matched group #1
return $1*1+1; # return captured number+1. $1*1 is a trick to convert
# string to number
The most concise way that also accounts for leading zeros, non-numeric endings, and empty strings I've seen is:
''.replace(/[0-8]?9*$/, w => ++w)
//=> 1
'foo'.replace(/[0-8]?9*$/, w => ++w)
//=> foo1
'foo099'.replace(/[0-8]?9*$/, w => ++w)
//=> foo100
'foo999'.replace(/[0-8]?9*$/, w => ++w)
//=> foo1000
function pad(number, length, filler) {
number = number + "";
if (number.length < length) {
for (var i = number.length; i < length; i += 1) {
number = filler + number;
}
}
return number;
}
function incrementString (input) {
var orig = input.match(/\d+$/);
if (orig.length === 1) {
orig = pad(parseInt(orig[0]) + 1, orig[0].length, '0');
input = input.replace(/\d+$/, orig);
return input;
}
return input + "1";
}
What does it do?
It first checks if there is a trailing number. If yes, increment it and pad left it with zeros (with the function "pad" which you'd be able to sort it out yourself).
string.replace is a function which works with argument 1 the substring to search (string, regex), argument 2 the element to replace with (string, function).
In this case I've used a regex as first argument and the incremented, padded number.
The regex is pretty simple: \d means "integer" and + means "one or more of the preceeding", which means one or more digits. $ means the end of the string.
More info about regular expressions (in JavaScript): https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/RegExp (thanks #Casimir et Hippolyte for the link)
I think you can simplify your code greatly:
function incrementString(input) {
var splits = input.split(/(\d+)$/),
num = 1;
if (splits[1] !== undefined) num = parseInt(splits[1]) + 1;
return splits[0] + num;
}
This checks for any number of digits at the end of your string.
function incrementString (strng) {
//separateing number from string
let x = (strng).replace( /^\D+/g, '');
//getting the length of original number from the string
let len = x.length;
//getting the string part from strng
str = strng.split(x);
//incrementing number by 1
let number = Number(x) + 1 + '';
//padding the number with 0 to make it's length exactly to orignal number
while(number.length < len){
number = '0' + number;
}
//new string by joining string and the incremented number
str = (str + number).split(',').join('');
//return new string
return str;
}
My quick answer will be :
let newStr = string.split('');
let word = [];
let num = [];
for (let i = 0 ; i<string.length ;i++){
isNaN(string[i])? word.push(string[i]) : num.push(string[i])
}
let l=num.length-1;
let pureNum=0;
for (let i = 0 ; i<num.length ;i++){
pureNum += num[i] * Math.pow(10,l);
l--;
}
let wordNum = (pureNum+1).toString().split('');
for (let i = wordNum.length ; i<num.length ;i++){
wordNum.unshift("0");
}
return word.join("")+wordNum.join("");
}
I'm creating some client side functions for a mappable spreadsheet export feature.
I'm using jQuery to manage the sort order of the columns, but each column is ordered like an Excel spreadsheet i.e. a b c d e......x y z aa ab ac ad etc etc
How can I generate a number as a letter? Should I define a fixed array of values? Or is there a dynamic way to generate this?
I think you're looking for something like this
function colName(n) {
var ordA = 'a'.charCodeAt(0);
var ordZ = 'z'.charCodeAt(0);
var len = ordZ - ordA + 1;
var s = "";
while(n >= 0) {
s = String.fromCharCode(n % len + ordA) + s;
n = Math.floor(n / len) - 1;
}
return s;
}
// Example:
for(n = 0; n < 125; n++)
document.write(n + ":" + colName(n) + "<br>");
This is a very easy way:
function numberToLetters(num) {
let letters = ''
while (num >= 0) {
letters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'[num % 26] + letters
num = Math.floor(num / 26) - 1
}
return letters
}
function getColumnDescription(i) {
const m = i % 26;
const c = String.fromCharCode(65 + m);
const r = i - m;
return r > 0
? `${getColumnDescription((r - 1) / 26)}${c}`
: `Column ${c}`
}
Usage:
getColumnDescription(15)
"Column P"
getColumnDescription(26)
"Column AA"
getColumnDescription(4460)
"Column FOO"
If you have your data in a two-dimensional array, e.g.
var data = [
['Day', 'score],
['Monday', 99],
];
you can map the rows/columns to spreadsheet cell numbers as follows (building on the code examples above):
function getSpreadSheetCellNumber(row, column) {
let result = '';
// Get spreadsheet column letter
let n = column;
while (n >= 0) {
result = String.fromCharCode(n % 26 + 65) + result;
n = Math.floor(n / 26) - 1;
}
// Get spreadsheet row number
result += `${row + 1}`;
return result;
};
E.g. the 'Day' value from data[0][0] would go in spreadsheet cell A1.
> getSpreadSheetCellNumber(0, 0)
> "A1"
This also works when you have 26+ columns:
> getSpreadSheetCellNumber(0, 26)
> "AA1"
You can use code like this, assuming that numbers contains the numbers of your columns. So after this code you'll get the string names for your columns:
var letters = ['a', 'b', 'c', ..., 'z'];
var numbers = [1, 2, 3, ...];
var columnNames = [];
for(var i=0;i<numbers.length;i++) {
var firstLetter = parseInt(i/letters.length) == 0 ? '' : letters[parseInt(i/letters.length)];
var secondLetter = letters[i%letters.length-1];
columnNames.push(firstLetter + secondLetter);
}
Simple recursive solution:
function numberToColumn(n) {
const res = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'[n % 26];
return n >= 26 ? numberToColumn(Math.floor(n / 26) - 1) + res : res;
}
Here is an alternative approach that relies on .toString(26). It uses conversion to base-26 and then translates the characters so they are in the a..z range:
const conv = ((base, alpha) => { // Closure for preparing the function
const map = Object.fromEntries(Array.from(alpha, (c, i) => [c, alpha[i + 10]]));
return n => (n + base).toString(26).replace(/o*p/, "").replace(/./g, m => map[m]);
})(parseInt("ooooooooop0", 26), "0123456789abcdefghijklmnopqrstuvwxyz");
// Example:
for (let n = 0; n < 29; n++) console.log(n, conv(n));
console.log("...");
for (let n = 690; n < 705; n++) console.log(n, conv(n));
About the magical number
The magical value "ooooooooop0" is derived as follows:
It is a number expressed in radix 26, in the standard way, i.e. where the ten digits also play a role, and then the first letters of the alphabet.
The greatest "digit" in this radix 26 is "p" (the 16th letter of the Latin alphabet), and "o" is the second greatest.
The magical value is formed by a long enough series of the one-but-greatest digit, followed by the greatest digit and ended by a 0.
As JavaScript integer numbers max out around Number.MAX_SAFE_INTEGER (greater integers numbers would suffer from rounding errors), there is no need to have a longer series of "o" than was selected. We can see that Number.MAX_SAFE_INTEGER.toString(26) has 12 digits, so precision is ensured up to 11 digits in radix 26, meaning we need 9 "o".
This magical number ensures that if we add units to it (in radix 26), we will always have a representation which starts with a series of "o" and then a "p". That is because at some point the last digit will wrap around to 0 again, and the "p" will also wrap around to 0, bringing the preceding "o" to "p". And so we have this invariant that the number always starts with zero or more "o" and then a "p".
More generic
The above magic number could be derived via code, and we could make it more generic by providing the target alphabet. The length of that target alphabet then also directly determines the radix (i.e. the number of characters in that string).
Here is the same output generated as above, but with a more generic function:
function createConverter(targetDigits) {
const radix = targetDigits.length,
alpha = "0123456789abcdefghijklmnopqrstuvwxyz",
map = Object.fromEntries(Array.from(alpha,
(src, i) => [src, targetDigits[i]]
)),
base = parseInt((alpha[radix-1]+'0').padStart(
Number.MAX_SAFE_INTEGER.toString(radix).length - 1, alpha[radix-2]
), radix),
trimmer = RegExp("^" + alpha[radix-2] + "*" + alpha[radix-1]);
return n => (n + base).toString(radix)
.replace(trimmer, "")
.replace(/./g, m => map[m]);
}
// Example:
const conv = createConverter("abcdefghijklmnopqrstuvwxyz");
for (let n = 0; n < 29; n++) console.log(n, conv(n));
console.log("...");
for (let n = 690; n < 705; n++) console.log(n, conv(n));
This can now easily be adapted to use a more reduced target alphabet (like without the letters "l" and "o"), giving a radix of 24 instead of 26:
function createConverter(targetDigits) {
const radix = targetDigits.length,
alpha = "0123456789abcdefghijklmnopqrstuvwxyz",
map = Object.fromEntries(Array.from(alpha,
(src, i) => [src, targetDigits[i]]
)),
base = parseInt((alpha[radix-1]+'0').padStart(
Number.MAX_SAFE_INTEGER.toString(radix).length - 1, alpha[radix-2]
), radix),
trimmer = RegExp("^" + alpha[radix-2] + "*" + alpha[radix-1]);
return n => (n + base).toString(radix)
.replace(trimmer, "")
.replace(/./g, m => map[m]);
}
// Example without "l" and "o" in target alphabet:
const conv = createConverter("abcdefghijkmnpqrstuvwxyz");
for (let n = 0; n < 29; n++) console.log(n, conv(n));
console.log("...");
for (let n = 690; n < 705; n++) console.log(n, conv(n));
This covers the range from 1 to 1000. Beyond that I haven't checked.
function colToletters(num) {
let a = " ABCDEFGHIJKLMNOPQRSTUVWXYZ";
if (num < 27) return a[num % a.length];
if (num > 26) {
num--;
let letters = ''
while (num >= 0) {
letters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'[num % 26] + letters
num = Math.floor(num / 26) - 1
}
return letters;
}
}
I could be wrong but I've checked the other functions in this answer and they seem to fail at 26 which should be Z. Remember there are 26 letters in the alphabet not 25.