I have a set of string numbers having decimals, for example: 23.456, 9.450, 123.01... I need to retrieve the number of decimals for each number, knowing that they have at least 1 decimal.
In other words, the retr_dec() method should return the following:
retr_dec("23.456") -> 3
retr_dec("9.450") -> 3
retr_dec("123.01") -> 2
Trailing zeros do count as a decimal in this case, unlike in this related question.
Is there an easy/delivered method to achieve this in Javascript or should I compute the decimal point position and compute the difference with the string length? Thanks
function decimalPlaces(num) {
var match = (''+num).match(/(?:\.(\d+))?(?:[eE]([+-]?\d+))?$/);
if (!match) { return 0; }
return Math.max(
0,
// Number of digits right of decimal point.
(match[1] ? match[1].length : 0)
// Adjust for scientific notation.
- (match[2] ? +match[2] : 0));
}
The extra complexity is to handle scientific notation so
decimalPlaces('.05')
2
decimalPlaces('.5')
1
decimalPlaces('1')
0
decimalPlaces('25e-100')
100
decimalPlaces('2.5e-99')
100
decimalPlaces('.5e1')
0
decimalPlaces('.25e1')
1
function retr_dec(num) {
return (num.split('.')[1] || []).length;
}
function retr_dec(numStr) {
var pieces = numStr.split(".");
return pieces[1].length;
}
Since there is not already a regex-based answer:
/\d*$/.exec(strNum)[0].length
Note that this "fails" for integers, but per the problem specification they will never occur.
You could get the length of the decimal part of your number this way:
var value = 192.123123;
stringValue = value.toString();
length = stringValue.split('.')[1].length;
It makes the number a string, splits the string in two (at the decimal point) and returns the length of the second element of the array returned by the split operation and stores it in the 'length' variable.
Try using String.prototype.match() with RegExp /\..*/ , return .length of matched string -1
function retr_decs(args) {
return /\./.test(args) && args.match(/\..*/)[0].length - 1 || "no decimal found"
}
console.log(
retr_decs("23.456") // 3
, retr_decs("9.450") // 3
, retr_decs("123.01") // 2
, retr_decs("123") // "no decimal found"
)
I had to deal with very small numbers so I created a version that can handle numbers like 1e-7.
Number.prototype.getPrecision = function() {
var v = this.valueOf();
if (Math.floor(v) === v) return 0;
var str = this.toString();
var ep = str.split("e-");
if (ep.length > 1) {
var np = Number(ep[0]);
return np.getPrecision() + Number(ep[1]);
}
var dp = str.split(".");
if (dp.length > 1) {
return dp[1].length;
}
return 0;
}
document.write("NaN => " + Number("NaN").getPrecision() + "<br>");
document.write("void => " + Number("").getPrecision() + "<br>");
document.write("12.1234 => " + Number("12.1234").getPrecision() + "<br>");
document.write("1212 => " + Number("1212").getPrecision() + "<br>");
document.write("0.0000001 => " + Number("0.0000001").getPrecision() + "<br>");
document.write("1.12e-23 => " + Number("1.12e-23").getPrecision() + "<br>");
document.write("1.12e8 => " + Number("1.12e8").getPrecision() + "<br>");
A slight modification of the currently accepted answer, this adds to the Number prototype, thereby allowing all number variables to execute this method:
if (!Number.prototype.getDecimals) {
Number.prototype.getDecimals = function() {
var num = this,
match = ('' + num).match(/(?:\.(\d+))?(?:[eE]([+-]?\d+))?$/);
if (!match)
return 0;
return Math.max(0, (match[1] ? match[1].length : 0) - (match[2] ? +match[2] : 0));
}
}
It can be used like so:
// Get a number's decimals.
var number = 1.235256;
console.debug(number + " has " + number.getDecimals() + " decimal places.");
// Get a number string's decimals.
var number = "634.2384023";
console.debug(number + " has " + parseFloat(number).getDecimals() + " decimal places.");
Utilizing our existing code, the second case could also be easily added to the String prototype like so:
if (!String.prototype.getDecimals) {
String.prototype.getDecimals = function() {
return parseFloat(this).getDecimals();
}
}
Use this like:
console.debug("45.2342".getDecimals());
A bit of a hybrid of two others on here but this worked for me. Outside cases in my code weren't handled by others here. However, I had removed the scientific decimal place counter. Which I would have loved at uni!
numberOfDecimalPlaces: function (number) {
var match = ('' + number).match(/(?:\.(\d+))?(?:[eE]([+-]?\d+))?$/);
if (!match || match[0] == 0) {
return 0;
}
return match[0].length;
}
Based on Liam Middleton's answer, here's what I did (without scientific notation):
numberOfDecimalPlaces = (number) => {
let match = (number + "").match(/(?:\.(\d+))?$/);
if (!match || !match[1]) {
return 0;
}
return match[1].length;
};
alert(numberOfDecimalPlaces(42.21));
function decimalPlaces(n) {
if (n === NaN || n === Infinity)
return 0;
n = ('' + n).split('.');
if (n.length == 1) {
if (Boolean(n[0].match(/e/g)))
return ~~(n[0].split('e-'))[1];
return 0;
}
n = n[1].split('e-');
return n[0].length + ~~n[1];
}
This question already has answers here:
Javascript toFixed Not Rounding
(23 answers)
Closed 2 years ago.
The default implementation of javascript's "Number.toFixed" appears to be a bit broken.
console.log((8.555).toFixed(2)); // returns 8.56
console.log((8.565).toFixed(2)); // returns 8.57
console.log((8.575).toFixed(2)); // returns 8.57
console.log((8.585).toFixed(2)); // returns 8.59
I need a rounding method that is more consistent than that.
In the range between 8.500 and 8.660 the following numbers don't round up correctly.
8.575
8.635
8.645
8.655
I've tried to fix the prototype implementation as follows, but it's only half way there. Can anyone suggest any change that would make it work more consistently?
Number.prototype.toFixed = function(decimalPlaces) {
var factor = Math.pow(10, decimalPlaces || 0);
var v = (Math.round(this * factor) / factor).toString();
if (v.indexOf('.') >= 0) {
return v + factor.toString().substr(v.length - v.indexOf('.'));
}
return v + '.' + factor.toString().substr(1);
};
This is because of floating-point errors.
Compare (8.575).toFixed(20) with (8.575).toFixed(3) and imagine this proposition: 8.575 < real("8.575"), where real is an imaginary function that creates a real number with infinite precision.
That is, the original number is not as expected and the inaccuracy has already been introduced.
One quick "workabout" I can think of is: Multiply by 1000 (or as appropriate), get the toFixed(0) of that (still has a limit, but it's absurd), then shove back in the decimal form.
Happy coding.
Thanks for the answer pst. My implementation almost worked, but didn't in some cases because of floating point errors.
this line in my function is the culprit:
Math.round(this * factor)
(it's on the Number.prototype, so "this" is the number);
8.575 * 100 comes out to 857.4999999999999, which in turn rounds down.
this is corrected by changing the line to read as follows:
Math.round(Math.round(this * factor * 100) / 100)
My entire workaround is now changed to:
Number.prototype.toFixed = function(decimalPlaces) {
var factor = Math.pow(10, decimalPlaces || 0);
var v = (Math.round(Math.round(this * factor * 100) / 100) / factor).toString();
if (v.indexOf('.') >= 0) {
return v + factor.toString().substr(v.length - v.indexOf('.'));
}
return v + '.' + factor.toString().substr(1);
};
A consistent solution would be to add a fixed tolerance (epsilon) to each number before rounding. It should be small, but not too small.
For example, with an eps = 1e-9, this:
console.log((8.555).toFixed(2)); // returns 8.56
console.log((8.565).toFixed(2)); // returns 8.57
console.log((8.575).toFixed(2)); // returns 8.57
console.log((8.585).toFixed(2)); // returns 8.59
Becomes this:
console.log((8.555 + eps).toFixed(2)); // returns 8.56
console.log((8.565 + eps).toFixed(2)); // returns 8.57
console.log((8.575 + eps).toFixed(2)); // returns 8.58
console.log((8.585 + eps).toFixed(2)); // returns 8.59
Maybe it will help someone, this is fixed popular formatMoney() function, but with correct roundings.
Number.prototype.formatMoney = function() {
var n = this,
decPlaces = 2,
decSeparator = ",",
thouSeparator = " ",
sign = n < 0 ? "-" : "",
i = parseInt(n = Math.abs(+n || 0)) + "",
j = (j = i.length) > 3 ? j % 3 : 0,
decimals = Number(Math.round(n +'e'+ decPlaces) +'e-'+ decPlaces).toFixed(decPlaces),
result = sign + (j ? i.substr(0, j) + thouSeparator : "") + i.substr(j).replace(/(\d{3})(?=\d)/g, "$1" + thouSeparator) + (decPlaces ? decSeparator + Math.abs(decimals-i).toFixed(decPlaces).slice(2) : "");
return result;
};
(9.245).formatMoney(); // returns 9,25
(7.5).formatMoney(); // returns 7,50
(8.575).formatMoney(); // returns 8,58
Check my answer
function toFixed( num, precision ) {
return (+(Math.round(+(num + 'e' + precision)) + 'e' + -precision)).toFixed(precision);
}
What is the recommended way to zerofill a value in JavaScript? I imagine I could build a custom function to pad zeros on to a typecasted value, but I'm wondering if there is a more direct way to do this?
Note: By "zerofilled" I mean it in the database sense of the word (where a 6-digit zerofilled representation of the number 5 would be "000005").
I can't believe all the complex answers on here... Just use this:
var zerofilled = ('0000'+n).slice(-4);
let n = 1
var zerofilled = ('0000'+n).slice(-4);
console.log(zerofilled)
Simple way. You could add string multiplication for the pad and turn it into a function.
var pad = "000000";
var n = '5';
var result = (pad+n).slice(-pad.length);
As a function,
function paddy(num, padlen, padchar) {
var pad_char = typeof padchar !== 'undefined' ? padchar : '0';
var pad = new Array(1 + padlen).join(pad_char);
return (pad + num).slice(-pad.length);
}
var fu = paddy(14, 5); // 00014
var bar = paddy(2, 4, '#'); // ###2
Since ECMAScript 2017 we have padStart:
const padded = (.1 + "").padStart(6, "0");
console.log(`-${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