Develop Reference

JS: Return a number to a specified power of 10

I'm working on a problem that requires me to return a number rounded to a specified power of 10. For example:
1234, specified power = 2 => 1200
1234, specified power = 3 => 1000
I did find a solution to the problem with this function:
const roundToPower = (num, pow) => {
return Math.round(num / Math.pow(10, pow)) * Math.pow(10,pow)
};
However, I am not exactly sure how and why it works.
Is someone able to break it down for me? Thanks!

Lets divide the upper function into three parts
num / Math.pow(10, pow) will divide the given number by the power of 10 which is given. e.g for pow = 3 num will be divided by 1000
It then uses use Math.round() on that decimal.
Then again to equalize the division we did in first step it multiples again with power of ten Math.pow(10,pow)
Example
For pow = 3 and num = 1230
=> Math.round(1230 / Math.pow(10, 3)) * Math.pow(10, 3)
=> Math.round(1230 / 1000 )) * 1000
=> Math.round( 1.230 )) * 1000
=> 1 * 1000
=> 1000

I know that question is not about how to convert numbers to a specified power of 10. I know a better and simple way to do this. Maybe it can help.
let a = 1234;
a = convert(a,2);
console.log(a); // 1200
let b = 123456;
b = convert(b,5);
console.log(b); //123460
function convert(num, power){
if(power < String(num).length){
num = num.toPrecision(power);
num = Number(num).toFixed();
return num;
}
else{
console.log("Invalid power for number: " + num);
}
}

Rounding up to the nearest 0.05 in JavaScript

Question
Does anyone know of a way to round a float to the nearest 0.05 in JavaScript?
Example
BEFORE | AFTER
2.51 | 2.55
2.50 | 2.50
2.56 | 2.60
Current Code
var _ceil = Math.ceil;
Math.ceil = function(number, decimals){
if (arguments.length == 1)
return _ceil(number);
multiplier = Math.pow(10, decimals);
return _ceil(number * multiplier) / multiplier;
}
Then elsewhere...
return (Math.ceil((amount - 0.05), 1) + 0.05).toFixed(2);
Which is resulting in...
BEFORE | AFTER
2.51 | 2.55
2.50 | 2.55
2.56 | 2.65

Multiply by 20, then divide by 20:
(Math.ceil(number*20)/20).toFixed(2)

Rob's answer with my addition:
(Math.ceil(number*20 - 0.5)/20).toFixed(2)
Otherwise it always rounds up to the nearest 0.05.
** UPDATE **
Sorry has been pointed out this is not what the orig poster wanted.

I would go for the standard of actually dividing by the number you're factoring it to, and rounding that and multiplying it back again after. That seems to be a proper working method which you can use with any number and maintain the mental image of what you are trying to achieve.
var val = 26.14,
factor = 0.05;
val = Math.round(val / factor) * factor;
This will work for tens, hundreds or any number. If you are specifically rounding to the higher number then use Math.ceil instead of Math.round.
Another method specifically for rounding just to 1 or more decimal places (rather than half a place) is the following:
Number(Number(1.5454545).toFixed(1));
It creates a fixed number string and then turns it into a real Number.

I would write a function that does it for you by
move the decimal over two places (multiply by 100)
then mod (%) that inflatedNumber by 5 and get the remainder
subtract the remainder from 5 so that you know what the 'gap'(ceilGap) is between your number and the next closest .05
finally, divide your inflatedNumber by 100 so that it goes back to your original float, and voila, your num will be rounded up to the nearest .05.
function calcNearestPointZeroFive(num){
var inflatedNumber = num*100,
remainder = inflatedNumber % 5;
ceilGap = 5 - remainder
return (inflatedNumber + ceilGap)/100
}
If you want to leave numbers like 5.50 untouched you can always add this checker:
if (remainder===0){
return num
} else {
var ceilGap = 5 - remainder
return (inflatedNumber + ceilGap)/100
}

You need to put -1 to round half down and after that multiply by -1 like the example down bellow.
<script type="text/javascript">
function roundNumber(number, precision, isDown) {
var factor = Math.pow(10, precision);
var tempNumber = number * factor;
var roundedTempNumber = 0;
if (isDown) {
tempNumber = -tempNumber;
roundedTempNumber = Math.round(tempNumber) * -1;
} else {
roundedTempNumber = Math.round(tempNumber);
}
return roundedTempNumber / factor;
}
</script>
<div class="col-sm-12">
<p>Round number 1.25 down: <script>document.write(roundNumber(1.25, 1, true));</script>
</p>
<p>Round number 1.25 up: <script>document.write(roundNumber(1.25, 1, false));</script></p>
</div>

I ended up using this function in my project, successfully:
roundToNearestFiveCents( number: any ) {
return parseFloat((Math.round(number / 0.05) * 0.05).toFixed(2));
}
Might be of use to someone wanting to simply round to the nearest 5 cents on their monetary results, keeps the result a number, so if you perform addition on it further it won't result in string concatenation; also doesn't unnecessarily round up as a few of the other answers pointed out. Also limits it to two decimals, which is customary with finance.

My solution and test:
let round = function(number, precision = 2, rounding = 0.05) {
let multiply = 1 / rounding;
return parseFloat((Math.round(number * multiply) / multiply)).toFixed(precision);
};
https://jsfiddle.net/maciejSzewczyk/7r1tvhdk/40/

Even though the OP is not explicit about banker rounding, rounding up to the nearest $0.05 (5 cents) should be compatible with banker rounding. What suggested by Arth is more accurate than the accepted answer by Rob W.
(Math.ceil(number*20 - 0.5)/20).toFixed(2)
With banker rounding, you need a basic banker rounding function as suggested at Gaussian/banker's rounding in JavaScript, and I rewrite in TypeScript:
static bankerRound(num: number, decimalPlaces?: number) {
const d = decimalPlaces || 0;
const m = Math.pow(10, d);
const n = +(d ? num * m : num).toFixed(8);
const i = Math.floor(n), f = n - i;
const e = 1e-8;
const r = (f > 0.5 - e && f < 0.5 + e) ?
((i % 2 === 0) ? i : i + 1) : Math.round(n);
return d ? r / m : r;
}
static roundTo5cents(num: number) {
const r = bankerRound(Math.ceil(num * 20 - 0.5) / 20, 2);
return r;
}
The correctness of this algorithm could be verified through MBS Online, e.g. http://www9.health.gov.au/mbs/ready_reckoner.cfm?item_num=60

Round money to nearest 10 dollars in Javascript

How can I round a decimal number in Javascript to the nearest 10? My math is pretty rubbish today, it could be the 2 hour sleep :/
Some sample cases
$2823.66 = $2820
$142.11 = $140
$9.49 = $10
I understand I probably need a combination of Math.round/floor but I can't seem to get expected result.
Any help/pointers appreciated!
M

Try
Math.round(val / 10) * 10;

Use this function:
function roundTen(number)
{
return Math.round(number/10)*10;
}
alert(roundTen(2823.66));

To round a number to the nearest 10, first divide it by 10, then round it to the nearest 1, then multiply it by 10 again:
val = Math.round(val/10)*10;
This page has some details. They go the other way (e.g., rounding to the nearest 0.01) but the theory and practice are identical - multiply (or divide), round, then divide (or multiply).

10 * Math.round(val / 10)

function round(number, multiplier) {
multiplier = multiplier || 1;
return Math.round(number / multiplier) * multiplier;
}
var num1 = 2823.66;
var num2 = 142.11;
var num3 = 9.49;
console.log(
"%s\n%s\n%s", // just a formating thing
round(num1, 10), // 2820
round(num2, 10), // 140
round(num3, 10) // 10
);

JavaScript, Generate a Random Number that is 9 numbers in length

I'm looking for an efficient, elegant way to generate a JavaScript variable that is 9 digits in length:
Example: 323760488

You could generate 9 random digits and concatenate them all together.
Or, you could call random() and multiply the result by 1000000000:
Math.floor(Math.random() * 1000000000);
Since Math.random() generates a random double precision number between 0 and 1, you will have enough digits of precision to still have randomness in your least significant place.
If you want to ensure that your number starts with a nonzero digit, try:
Math.floor(100000000 + Math.random() * 900000000);
Or pad with zeros:
function LeftPadWithZeros(number, length)
{
var str = '' + number;
while (str.length < length) {
str = '0' + str;
}
return str;
}
Or pad using this inline 'trick'.

why don't just extract digits from the Math.random() string representation?
Math.random().toString().slice(2,11);
/*
Math.random() -> 0.12345678901234
.toString() -> "0.12345678901234"
.slice(2,11) -> "123456789"
*/
(requirement is that every javascript implementation Math.random()'s precision is at least 9 decimal places)

Also...
function getRandom(length) {
return Math.floor(Math.pow(10, length-1) + Math.random() * 9 * Math.pow(10, length-1));
}
getRandom(9) => 234664534

Three methods I've found in order of efficiency:
(Test machine running Firefox 7.0 Win XP)
parseInt(Math.random()*1000000000, 10)
1 million iterations: ~626ms. By far the fastest - parseInt is a native function vs calling the Math library again. NOTE: See below.
Math.floor(Math.random()*1000000000)
1 million iterations: ~1005ms. Two function calls.
String(Math.random()).substring(2,11)
1 million iterations: ~2997ms. Three function calls.
And also...
parseInt(Math.random()*1000000000)
1 million iterations: ~362ms.
NOTE: parseInt is usually noted as unsafe to use without radix parameter. See https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/parseInt or google "JavaScript: The Good Parts". However, it seems the parameter passed to parseInt will never begin with '0' or '0x' since the input is first multiplied by 1000000000. YMMV.

Math.random().toFixed(length).split('.')[1]
Using toFixed alows you to set the length longer than the default (seems to generate 15-16 digits after the decimal. ToFixed will let you get more digits if you need them.

In one line(ish):
var len = 10;
parseInt((Math.random() * 9 + 1) * Math.pow(10,len-1), 10);
Steps:
We generate a random number that fulfil 1 ≤ x < 10.
Then, we multiply by Math.pow(10,len-1) (number with a length len).
Finally, parseInt() to remove decimals.

Thought I would take a stab at your question. When I ran the following code it worked for me.
<script type="text/javascript">
function getRandomInt(min, max) {
return Math.floor(Math.random() * (max - min)) + min;
} //The maximum is exclusive and the minimum is inclusive
$(document).ready(function() {
$("#random-button").on("click", function() {
var randomNumber = getRandomInt(100000000, 999999999);
$("#random-number").html(randomNumber);
});
</script>

Does this already have enough answers?
I guess not. So, this should reliably provide a number with 9 digits, even if Math.random() decides to return something like 0.000235436:
Math.floor((Math.random() + Math.floor(Math.random()*9)+1) * Math.pow(10, 8))

Screen scrape this page:
9 random numbers

function rand(len){var x='';
for(var i=0;i<len;i++){x+=Math.floor(Math.random() * 10);}
return x;
}
rand(9);

If you mean to generate random telephone number, then they usually are forbidden to start with zero.
That is why you should combine few methods:
Math.floor(Math.random()*8+1)+Math.random().toString().slice(2,10);
this will generate random in between 100 000 000 to 999 999 999
With other methods I had a little trouble to get reliable results as leading zeroes was somehow a problem.

I know the answer is old, but I want to share this way to generate integers or float numbers from 0 to n. Note that the position of the point (float case) is random between the boundaries. The number is an string because the limitation of the MAX_SAFE_INTEGER that is now 9007199254740991
Math.hRandom = function(positions, float = false) {
var number = "";
var point = -1;
if (float) point = Math.floor(Math.random() * positions) + 1;
for (let i = 0; i < positions; i++) {
if (i == point) number += ".";
number += Math.floor(Math.random() * 10);
}
return number;
}
//integer random number 9 numbers
console.log(Math.hRandom(9));
//float random number from 0 to 9e1000 with 1000 numbers.
console.log(Math.hRandom(1000, true));

function randomCod(){
let code = "";
let chars = 'abcdefghijlmnopqrstuvxwz';
let numbers = '0123456789';
let specialCaracter = '/{}$%&#*/()!-=?<>';
for(let i = 4; i > 1; i--){
let random = Math.floor(Math.random() * 99999).toString();
code += specialCaracter[random.substring(i, i-1)] + ((parseInt(random.substring(i, i-1)) % 2 == 0) ? (chars[random.substring(i, i-1)].toUpperCase()) : (chars[random.substring(i, i+1)])) + (numbers[random.substring(i, i-1)]);
}
code = (code.indexOf("undefined") > -1 || code.indexOf("NaN") > -1) ? randomCod() : code;
return code;
}

With max exclusive: Math.floor(Math.random() * max);
With max inclusive: Math.round(Math.random() * max);

To generate a number string with length n, thanks to #nvitaterna, I came up with this:
1 + Math.floor(Math.random() * 9) + Math.random().toFixed(n - 1).split('.')[1]
It prevents first digit to be zero.
It can generate string with length ~ 50 each time you call it.

var number = Math.floor(Math.random() * 900000000) + 100000000

var number = Math.floor(Math.random()*899999999 + 100000000)

For a number of 10 characters
Math.floor(Math.random() * 9000000000) + 1000000000
From https://gist.github.com/lpf23/9762508
This answer is intended for people who are looking to generate a 10 digit number (without a country code)

Round a number to nearest .25 in JavaScript

I want to convert all numbers to the nearest .25
So...
5 becomes 5.00
2.25 becomes 2.25
4 becomes 4.00
3.5 becomes 3.50

Here’s an implementation of what rslite said:
var number = 5.12345;
number = (Math.round(number * 4) / 4).toFixed(2);

Multiply by 4, round to integer, divide by 4 and format with two decimals.

If speed is your concern, note that you can get about a 30% speed improvement by using:
var nearest = 4;
var rounded = number + nearest/2 - (number+nearest/2) % nearest;
From my website: http://phrogz.net/round-to-nearest-via-modulus-division
Performance tests here: http://jsperf.com/round-to-nearest

Here is a generic function to do rounding. In the examples above, 4 was used because that is in the inverse of .25. This function allows the user to ignore that detail. It doesn't currently support preset precision, but that can easily be added.
function roundToNearest(numToRound, numToRoundTo) {
numToRoundTo = 1 / (numToRoundTo);
return Math.round(numToRound * numToRoundTo) / numToRoundTo;
}

Here is #Gumbo's answer in a form of a function:
var roundNearQtr = function(number) {
return (Math.round(number * 4) / 4).toFixed(2);
};
You can now make calls:
roundNearQtr(5.12345); // 5.00
roundNearQtr(3.23); // 3.25
roundNearQtr(3.13); // 3.25
roundNearQtr(3.1247); // 3.00

function roundToInc(num, inc) {
const diff = num % inc;
return diff>inc/2?(num-diff+inc):num-diff;
}
> roundToInc(233223.2342343, 0.01)
233223.23
> roundToInc(505, 5)
505
> roundToInc(507, 5)
505
> roundToInc(508, 5)
510

Use below function, hope it helps
function roundByQuarter(value) {
var inv = 1.0 / 0.25;
return Math.round(value * inv) / inv;
}
Call the function as below, will result the nearest Quarter value, that is it will not return .32, .89, .56 but will return .25, .75, .50 decimals only.
roundByQuarter(2.74) = 2.75
roundByQuarter(2.34) = 2.25
roundByQuarter(2.94) = 3.00
roundByQuarter(2.24) = 2.25

A very good approximation for rounding:
function Rounding (number, precision){
var newNumber;
var sNumber = number.toString();
var increase = precision + sNumber.length - sNumber.indexOf('.') + 1;
if (number < 0)
newNumber = (number - 5 * Math.pow(10,-increase));
else
newNumber = (number + 5 * Math.pow(10,-increase));
var multiple = Math.pow(10,precision);
return Math.round(newNumber * multiple)/multiple;
}