I'm totally stuck at one excersise question. Can someone help me out with this question?
Create a function sumRangeNumbers() that returns the sum of all numbers
between two chosen numbers. The function should take two arguments, one
representing the lowest boundary and one that represents the highest
boundary. For example, the arguments 10 and 20 should return the sum of
10+11+12+13...+20.
for (var i = 0; i < 82; i++) {
document.write(i + i + '+');
}
How do I write the loop that sums all the numbers with an function?
The answer of DCR already provides a nice implementation and is probably what you were looking for. However, with a little mathematical knowledge you can make the function a little easier.
We know that the sum of 1 to n is n(n+1)/2 by looking at this wikipedia page.
The sum of a to b, is simply the sum of 1 to b minus the sum of 1 to a - 1 (we also want to include a itself).
The sum between a and b is then b(b + 1)/2 - (a - 1)(a)/2 and therefore your function becomes:
const a = 10
const b = 20
function sumRangeNumbers(a, b) {
const high = Math.max(a, b);
const low = Math.min(a, b);
return high * (high + 1) / 2 - (low - 1) * (low) / 2;
}
console.log(sumRangeNumbers(a, b)); // 165
console.log(sumRangeNumbers(b, a)); // 165
console.log(sumRangeNumbers(5, 7)); // 18
function sumRangeNumber (num1, num2) {
let total = 0
for (let i = num1; i <= num2; i++) {
total += i
}
return total
}
You are on the right track with a for loop. What we did here was in place of declaring i as zero we passed the low value and in the comparison we pass the high value. This creates the range ie 10-20. From there each loop we add I too total which is declared outside fo the loop so as to not have it reset and we add to it.
As a previous comment mentioned, this is kinda doing your HW for you, so give the above function a shot and play around with it and change things to make sure you understand whats happening.
you need to first create a function and then you need to call it.
function sum(x,y){
var sum = 0;
for(let i = x;i<=y;i++){
sum = sum + i;
}
console.log(sum)
}
sum(1,10);
const sumRange = (num1, num2) => (
min = Math.min(num1, num2),
Array(Math.abs(num1 - num2) + 1)
.fill().map((_, i) => i + min)
.reduce((sum, el) => sum + el, 0)
);
console.log(sumRange(20, 10));
console.log(sumRange(10, 20));
function sumRangeNumbers(lower, upper) {
let total = 0;
for (let index=lower; index<=upper; index++) {
total = total + index;
}
return total;
}
document.write(sumRangeNumbers(10,20));
Simple:
function sumRangeNumbers(from, to) {
let result = 0;
for (let i = from; i <= to; i++) {
result += i;
}
return result;
}
If the numbers belong to range of natural numbers, then why do you need a loop. Just use the fact that sum from low to high=
((high-low+1) * (low + high)) / 2
Give this a shot:
function getSum(x,y) {
sum += x
console.log(sum, x)
if (x == y) { return sum }
else { return getSum(x+1,y) }
}
sum = 0
Here's a simple example using your current attempt. Keep in mind, you'll want to some error handling for cases where they give you an invalid high/low number.
IE:
if (lowNum >= highNum) { console.error('invalid range'); }
and maybe this too
if (typeof lowNum !== 'number' && typeof highNum !== 'number') { console.error('Invalid inputs'); }
function sumUp(lowNum, highNum) {
if (lowNum >= highNum) { throw new Error('Invalid Range'); }
// Initialize total at 0
let total = 0;
// Loop through numbers between lowNum and highNum.
// Do not include lowNum and highNum in the addition
for (let i = lowNum + 1; i < highNum; i++) {
// Increment the total with the 'in-between number (i)'
total += i;
}
// Return the result
return total;
}
// Test 1 (should be 44)
console.log(2 + 3 + 4 + 5 + 6 + 7 + 8 + 9, sumUp(1, 10));
// Test 2 (should be 315)
console.log(50 + 51 + 52 + 53 + 54 + 55, sumUp(49, 56));
// If you really want to do document.write
document.write(sumUp(49, 56));
// Test 3 (should fail)
console.log(sumUp(15, 3));
The question as per the practice course is :
Write a JavaScript program to find the maximum integer n such that (1 + 2 + ... + n <= given integer ) is true. For eg. If a given integer is 10, value of maximum integer n is 4 so that 1+2+3+4 <= 10 is true. Your output code should be in the format console.log("Value of n is ", variableName)
My code is :
var num = prompt("Enter a number");
function test(x) {
var sum = 1,
n = 1,
a = 0;
while (sum <= x) {
sum += n;
n = n + 1;
a += 1;
}
return a;
}
var output = test(num);
console.log("Result is :", output);
I'm getting the correct outputs as per the test cases I've entered(10-4,15-5,16-6,17-6) but the website says there is something wrong with the program.
What am i doing wrong?
Better answer than looping: exploit maths. Starting with Triangular number formula:
1 + 2 + ... + n = n * (n + 1) / 2
Thus, for input x, you need to find n such that
n * (n + 1) / 2 <= x
To solve this, we need to clean up the inequality, then use the quadratic equation formula:
n^2 + n <= 2x
n^2 + n - 2x <= 0
n <= (-1 + sqrt(1 + 8x)) / 2
as the final solution. e.g. for
x = 10: n <= (-1 + sqrt(81)) / 2; n <= 4
x = 16: n <= (-1 + sqrt(128)) / 2; n <= 5.156854249492381
Round the upper limit down, and you have the largest allowed integer. Translated into JavaScript:
function test(x) {
return Math.floor((Math.sqrt(8 * x + 1) - 1) / 2);
}
var num = prompt("Enter a number");
console.log("Result is :", test(num));
Consider if the passed value is 11. Then, the maximum integer n should be 4, because 1+2+3+4 < 11 is true, while 1+2+3+4+5 < 11 is false. Your current code outputs 5 for an input of 11, though, which is incorrect; your while loop is sometimes overshooting sum.
You also need to initialize sum to start at 0, not at 1.
Subtract one from a before returning it:
function test(x) {
var sum = 0,
n = 1,
a = 0;
while (sum <= x) {
sum += n;
n = n + 1;
a += 1;
console.log(a, sum);
}
return a - 1;
}
console.log(test(10));
console.log(test(11));
var num = prompt("Enter a number");
var output = test(num);
console.log("Result is :", output);
The code below should work for you. Basically, what I did was that if the input is 10, and your sum is 9, it will still go into the while loop. Then it will add n again and now your number is greater than your input (which is 10), but you still return it. Here what I did is that at the end of the while loop, if your sum is greater than your input, subtract one from a. That way it will still execute, but it will fix the problem.
Also another error I noticed was that sum started at 1, and n started at 1. You wanted 1+2+3+...+n, however using your previous method, you got 1+1+2+3+...+n.
var num = prompt("Enter a number");
function test(x) {
var sum = 0,
n = 1,
tempSum = 1,
a = 0;
while (sum <= x) {
sum += n;
n++;
a++;
if (sum > x) {
a--;
}
}
return a;
}
var output = test(num);
console.log("Result is :", output);
Your order of operation is a little funky; all you have to do is add the incrementor. The while false case will make sure the sum only passes over the number once. So when you return, reduce the number by one:
var num = prompt("Enter a number");
var output = test(num);
console.log("Result is :", output);
function test(num){
let sum = 0
let inc = 0
while(sum<=num)
sum+=++inc
return --inc;
}
This is a reduced version of your code, basically we increment first the number to add (n) in each iteration, and then we add it to the variable holding the sum. When the loop conditions evaluates to false you need to decrement one to n to get your value:
var num = prompt("Enter a number");
function test(x)
{
var sum = 0, n = 0;
while (sum <= x)
{
sum += (++n);
}
return --n;
}
var output = test(num);
console.log("Result is :", output);
I think this will work for you:
var num = prompt("Enter a number");
function test(x) {
var sum = 1,
n = 0;
while ((sum+n) <= x) {
n = n + 1;
sum += n;
}
return n;
}
var output = test(num);
console.log("Result is :", output);
Try below function to find max Number
function maxNumber(a){
var i=1,sum=0,maxNumber=0;
while(sum<=a) {
sum=sum+i;
if(sum<=a)
{
maxNumber=i;
}
i+=1;
}
return maxNumber;
}
doubled checked condition sum<=a to preserve the previous loop value and if condition not satisfied that means current loop value is not useful so returned preserved value of previous loop
Output tested :
Below will help you get the job done.
var num = prompt("Enter a number");
function findMaxNumber(num){
var sum = 0;
var counter = 0;
while(sum < num){
if(sum + counter > num){
break; // Exit loop
}
sum = sum + counter;
counter++;
}
return --counter; // Loop will cause this to be 1 higher than the max int.
}
console.log('Result is: ' + findMaxNumber(num));
i'm doing some coding exercises and i'm not being able to solve this one.
Find the sum of all divisors of a given integer.
For n = 12, the input should be
sumOfDivisors(n) = 28.
example: 1 + 2 + 3 + 4 + 6 + 12 = 28.
Constraints:
1 ≤ n ≤ 15.
how can i solve this exercise? i'm not being able to.
function(n){
var arr = [],
finalSum;
if(n <= 1 || n => 16){
return false ;
}
for(var i = 0; i < n; i++){
var tmp= n/2;
arr.push(tmp)
// i need to keep on dividing n but i can't get the way of how to
}
return finalSum;
}
This is another way to do it:
var divisors = n=>[...Array(n+1).keys()].slice(1)
.reduce((s, a)=>s+(!(n % a) && a), 0);
console.log(divisors(12));
JSFiddle: https://jsfiddle.net/32n5jdnb/141/
Explaining:
n=> this is an arrow function, the equivalent to function(n) {. You don't need the () if there's only one parameter.
Array(n+1) creates an empty array of n+1 elements
.keys() gets the keys of the empty array (the indexes i.e. 0, 1, 2) so this is a way to create a numeric sequence
[...Array(n+1)].keys()] uses the spread (...) operator to transform the iterator in another array so creating an array with the numeric sequence
.slice(1) removes the first element thus creating a sequence starting with 1. Remember the n+1 ?
.reduce() is a method that iterates though each element and calculates a value in order to reduce the array to one value. It receives as parameter a callback function to calculate the value and the initial value of the calculation
(s, a)=> is the callback function for reduce. It's an arrow function equivalent to function(s, a) {
s+(!(n % a) && a) is the calculation of the value.
s+ s (for sum) or the last value calculated +
!(n % a) this returns true only for the elements that have a 0 as modular value.
(!(n % a) && a) is a js 'trick'. The case is that boolean expressions in javascript don't return true or false. They return a 'truthy' or 'falsy' value which is then converted to boolean. So the actual returned value is the right value for && (considering both have to be truthy) and the first thuthy value found for || (considering only one need to be truthy). So this basically means: if a is a modular value (i.e. != 0) return a to add to the sum, else return 0.
, 0 is the initial value for the reduce calculation.
Reduce documentation: https://developer.mozilla.org/pt-BR/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce
Edit
Answering to Tristan Forward:
var divisorsList = [];
var divisors = (n)=>[...Array(n+1).keys()].slice(1)
.reduce((s, a)=>{
var divisor = !(n % a) && a;
if (divisor) divisorsList.push(divisor);
return s+divisor;
}, 0);
console.log('Result:', divisors(12));
console.log('Divisors:', divisorsList);
You have to check if specified number is or not a divisor of given integer. You can use modulo % - if there's no rest, specified number is the divisor of the given integer - add it to the sum.
function sumDivisors(num){
var sum = 0;
for (var i = 1; i <= num; i++){
if (!(num % i)) {
sum += i;
}
}
console.log(sum);
}
sumDivisors(6);
sumDivisors(10);
Here is a solution with better algorithm performance (O(sqrt(largest prime factor of n)))
divisors = n => {
sum = 1
for (i = 2; n > 1; i++) {
i * i > n ? i = n : 0
b = 0
while (n % i < 1) {
c = sum * i
sum += c - b
b = c
n /= i
}
}
return sum
}
since n / i is also a devisor this can be done more efficiently.
function sumDivisors(num) {
let sum = 1;
for (let i = 2; i < num / i; i++) {
if (num % i === 0) {
sum += i + num / i;
}
}
const sqrt = Math.sqrt(num);
return num + (num % sqrt === 0 ? sum + sqrt : sum);
}
function countDivisors(n){
let counter = 1;
for(let i=1; i <= n/2; i++){
n % i === 0 ? counter++ : null;
}
return counter
}
in this case, we consider our counter as starting with 1 since by default all numbers are divisible by 1. Then we half the number since numbers that can be able to divide n are less or equal to half its value
Suppose I have a value of 15.7784514, I want to display it 15.77 with no rounding.
var num = parseFloat(15.7784514);
document.write(num.toFixed(1)+"<br />");
document.write(num.toFixed(2)+"<br />");
document.write(num.toFixed(3)+"<br />");
document.write(num.toFixed(10));
Results in -
15.8
15.78
15.778
15.7784514000
How do I display 15.77?
Convert the number into a string, match the number up to the second decimal place:
function calc(theform) {
var num = theform.original.value, rounded = theform.rounded
var with2Decimals = num.toString().match(/^-?\d+(?:\.\d{0,2})?/)[0]
rounded.value = with2Decimals
}
<form onsubmit="return calc(this)">
Original number: <input name="original" type="text" onkeyup="calc(form)" onchange="calc(form)" />
<br />"Rounded" number: <input name="rounded" type="text" placeholder="readonly" readonly>
</form>
The toFixed method fails in some cases unlike toString, so be very careful with it.
Update 5 Nov 2016
New answer, always accurate
function toFixed(num, fixed) {
var re = new RegExp('^-?\\d+(?:\.\\d{0,' + (fixed || -1) + '})?');
return num.toString().match(re)[0];
}
As floating point math in javascript will always have edge cases, the previous solution will be accurate most of the time which is not good enough.
There are some solutions to this like num.toPrecision, BigDecimal.js, and accounting.js.
Yet, I believe that merely parsing the string will be the simplest and always accurate.
Basing the update on the well written regex from the accepted answer by #Gumbo, this new toFixed function will always work as expected.
Old answer, not always accurate.
Roll your own toFixed function:
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}
Another single-line solution :
number = Math.trunc(number*100)/100
I used 100 because you want to truncate to the second digit, but a more flexible solution would be :
number = Math.trunc(number*Math.pow(10, digits))/Math.pow(10, digits)
where digits is the amount of decimal digits to keep.
See Math.trunc specs for details and browser compatibility.
I opted to write this instead to manually remove the remainder with strings so I don't have to deal with the math issues that come with numbers:
num = num.toString(); //If it's not already a String
num = num.slice(0, (num.indexOf("."))+3); //With 3 exposing the hundredths place
Number(num); //If you need it back as a Number
This will give you "15.77" with num = 15.7784514;
Update (Jan 2021)
Depending on its range, a number in javascript may be shown in scientific notation. For example, if you type 0.0000001 in the console, you may see it as 1e-7, whereas 0.000001 appears unchanged (0.000001).
If your application works on a range of numbers for which scientific notation is not involved, you can just ignore this update and use the original answer below.
This update is about adding a function that checks if the number is in scientific format and, if so, converts it into decimal format. Here I'm proposing this one, but you can use any other function that achieves the same goal, according to your application's needs:
function toFixed(x) {
if (Math.abs(x) < 1.0) {
let e = parseInt(x.toString().split('e-')[1]);
if (e) {
x *= Math.pow(10,e-1);
x = '0.' + (new Array(e)).join('0') + x.toString().substring(2);
}
} else {
let e = parseInt(x.toString().split('+')[1]);
if (e > 20) {
e -= 20;
x /= Math.pow(10,e);
x += (new Array(e+1)).join('0');
}
}
return x;
}
Now just apply that function to the parameter (that's the only change with respect to the original answer):
function toFixedTrunc(x, n) {
x = toFixed(x)
// From here on the code is the same than the original answer
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
This updated version addresses also a case mentioned in a comment:
toFixedTrunc(0.000000199, 2) => "0.00"
Again, choose what fits your application needs at best.
Original answer (October 2017)
General solution to truncate (no rounding) a number to the n-th decimal digit and convert it to a string with exactly n decimal digits, for any n≥0.
function toFixedTrunc(x, n) {
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
where x can be either a number (which gets converted into a string) or a string.
Here are some tests for n=2 (including the one requested by OP):
0 => 0.00
0.01 => 0.01
0.5839 => 0.58
0.999 => 0.99
1.01 => 1.01
2 => 2.00
2.551 => 2.55
2.99999 => 2.99
4.27 => 4.27
15.7784514 => 15.77
123.5999 => 123.59
And for some other values of n:
15.001097 => 15.0010 (n=4)
0.000003298 => 0.0000032 (n=7)
0.000003298257899 => 0.000003298257 (n=12)
parseInt is faster then Math.floor
function floorFigure(figure, decimals){
if (!decimals) decimals = 2;
var d = Math.pow(10,decimals);
return (parseInt(figure*d)/d).toFixed(decimals);
};
floorFigure(123.5999) => "123.59"
floorFigure(123.5999, 3) => "123.599"
num = 19.66752
f = num.toFixed(3).slice(0,-1)
alert(f)
This will return 19.66
Simple do this
number = parseInt(number * 100)/100;
Just truncate the digits:
function truncDigits(inputNumber, digits) {
const fact = 10 ** digits;
return Math.floor(inputNumber * fact) / fact;
}
This is not a safe alternative, as many others commented examples with numbers that turn into exponential notation, that scenery is not covered by this function
// typescript
// function formatLimitDecimals(value: number, decimals: number): number {
function formatLimitDecimals(value, decimals) {
const stringValue = value.toString();
if(stringValue.includes('e')) {
// TODO: remove exponential notation
throw 'invald number';
} else {
const [integerPart, decimalPart] = stringValue.split('.');
if(decimalPart) {
return +[integerPart, decimalPart.slice(0, decimals)].join('.')
} else {
return integerPart;
}
}
}
console.log(formatLimitDecimals(4.156, 2)); // 4.15
console.log(formatLimitDecimals(4.156, 8)); // 4.156
console.log(formatLimitDecimals(4.156, 0)); // 4
console.log(formatLimitDecimals(0, 4)); // 0
// not covered
console.log(formatLimitDecimals(0.000000199, 2)); // 0.00
These solutions do work, but to me seem unnecessarily complicated. I personally like to use the modulus operator to obtain the remainder of a division operation, and remove that. Assuming that num = 15.7784514:
num-=num%.01;
This is equivalent to saying num = num - (num % .01).
I fixed using following simple way-
var num = 15.7784514;
Math.floor(num*100)/100;
Results will be 15.77
My version for positive numbers:
function toFixed_norounding(n,p)
{
var result = n.toFixed(p);
return result <= n ? result: (result - Math.pow(0.1,p)).toFixed(p);
}
Fast, pretty, obvious. (version for positive numbers)
The answers here didn't help me, it kept rounding up or giving me the wrong decimal.
my solution converts your decimal to a string, extracts the characters and then returns the whole thing as a number.
function Dec2(num) {
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length == 1) {
return Number(num);
}
else {
return Number(numarr[0]+"."+numarr[1].charAt(0)+numarr[1].charAt(1));
}
}
else {
return Number(num);
}
}
Dec2(99); // 99
Dec2(99.9999999); // 99.99
Dec2(99.35154); // 99.35
Dec2(99.8); // 99.8
Dec2(10265.985475); // 10265.98
The following code works very good for me:
num.toString().match(/.\*\\..{0,2}|.\*/)[0];
This worked well for me. I hope it will fix your issues too.
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
// 5.56789 ----> $5.56
// 0.342 ----> $0.34
// -10.3484534 ----> $-10.34
// 600 ----> $600.00
function convertNumber(){
var result = toFixedNumber(document.getElementById("valueText").value);
document.getElementById("resultText").value = result;
}
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
<div>
<input type="text" id="valueText" placeholder="Input value here..">
<br>
<button onclick="convertNumber()" >Convert</button>
<br><hr>
<input type="text" id="resultText" placeholder="result" readonly="true">
</div>
An Easy way to do it is the next but is necessary ensure that the amount parameter is given as a string.
function truncate(amountAsString, decimals = 2){
var dotIndex = amountAsString.indexOf('.');
var toTruncate = dotIndex !== -1 && ( amountAsString.length > dotIndex + decimals + 1);
var approach = Math.pow(10, decimals);
var amountToTruncate = toTruncate ? amountAsString.slice(0, dotIndex + decimals +1) : amountAsString;
return toTruncate
? Math.floor(parseFloat(amountToTruncate) * approach ) / approach
: parseFloat(amountAsString);
}
console.log(truncate("7.99999")); //OUTPUT ==> 7.99
console.log(truncate("7.99999", 3)); //OUTPUT ==> 7.999
console.log(truncate("12.799999999999999")); //OUTPUT ==> 7.99
Here you are. An answer that shows yet another way to solve the problem:
// For the sake of simplicity, here is a complete function:
function truncate(numToBeTruncated, numOfDecimals) {
var theNumber = numToBeTruncated.toString();
var pointIndex = theNumber.indexOf('.');
return +(theNumber.slice(0, pointIndex > -1 ? ++numOfDecimals + pointIndex : undefined));
}
Note the use of + before the final expression. That is to convert our truncated, sliced string back to number type.
Hope it helps!
truncate without zeroes
function toTrunc(value,n){
return Math.floor(value*Math.pow(10,n))/(Math.pow(10,n));
}
or
function toTrunc(value,n){
x=(value.toString()+".0").split(".");
return parseFloat(x[0]+"."+x[1].substr(0,n));
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.4
truncate with zeroes
function toTruncFixed(value,n){
return toTrunc(value,n).toFixed(n);
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.40
If you exactly wanted to truncate to 2 digits of precision, you can go with a simple logic:
function myFunction(number) {
var roundedNumber = number.toFixed(2);
if (roundedNumber > number)
{
roundedNumber = roundedNumber - 0.01;
}
return roundedNumber;
}
I used (num-0.05).toFixed(1) to get the second decimal floored.
It's more reliable to get two floating points without rounding.
Reference Answer
var number = 10.5859;
var fixed2FloatPoints = parseInt(number * 100) / 100;
console.log(fixed2FloatPoints);
Thank You !
My solution in typescript (can easily be ported to JS):
/**
* Returns the price with correct precision as a string
*
* #param price The price in decimal to be formatted.
* #param decimalPlaces The number of decimal places to use
* #return string The price in Decimal formatting.
*/
type toDecimal = (price: number, decimalPlaces?: number) => string;
const toDecimalOdds: toDecimal = (
price: number,
decimalPlaces: number = 2,
): string => {
const priceString: string = price.toString();
const pointIndex: number = priceString.indexOf('.');
// Return the integer part if decimalPlaces is 0
if (decimalPlaces === 0) {
return priceString.substr(0, pointIndex);
}
// Return value with 0s appended after decimal if the price is an integer
if (pointIndex === -1) {
const padZeroString: string = '0'.repeat(decimalPlaces);
return `${priceString}.${padZeroString}`;
}
// If numbers after decimal are less than decimalPlaces, append with 0s
const padZeroLen: number = priceString.length - pointIndex - 1;
if (padZeroLen > 0 && padZeroLen < decimalPlaces) {
const padZeroString: string = '0'.repeat(padZeroLen);
return `${priceString}${padZeroString}`;
}
return priceString.substr(0, pointIndex + decimalPlaces + 1);
};
Test cases:
expect(filters.toDecimalOdds(3.14159)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 2)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 0)).toBe('3');
expect(filters.toDecimalOdds(3.14159, 10)).toBe('3.1415900000');
expect(filters.toDecimalOdds(8.2)).toBe('8.20');
Any improvements?
Another solution, that truncates and round:
function round (number, decimals, truncate) {
if (truncate) {
number = number.toFixed(decimals + 1);
return parseFloat(number.slice(0, -1));
}
var n = Math.pow(10.0, decimals);
return Math.round(number * n) / n;
};
function limitDecimalsWithoutRounding(val, decimals){
let parts = val.toString().split(".");
return parseFloat(parts[0] + "." + parts[1].substring(0, decimals));
}
var num = parseFloat(15.7784514);
var new_num = limitDecimalsWithoutRounding(num, 2);
Roll your own toFixed function: for positive values Math.floor works fine.
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}
For negative values Math.floor is round of the values. So you can use Math.ceil instead.
Example,
Math.ceil(-15.778665 * 10000) / 10000 = -15.7786
Math.floor(-15.778665 * 10000) / 10000 = -15.7787 // wrong.
Gumbo's second solution, with the regular expression, does work but is slow because of the regular expression. Gumbo's first solution fails in certain situations due to imprecision in floating points numbers. See the JSFiddle for a demonstration and a benchmark. The second solution takes about 1636 nanoseconds per call on my current system, Intel Core i5-2500 CPU at 3.30 GHz.
The solution I've written involves adding a small compensation to take care of floating point imprecision. It is basically instantaneous, i.e. on the order of nanoseconds. I clocked 2 nanoseconds per call but the JavaScript timers are not very precise or granular. Here is the JS Fiddle and the code.
function toFixedWithoutRounding (value, precision)
{
var factorError = Math.pow(10, 14);
var factorTruncate = Math.pow(10, 14 - precision);
var factorDecimal = Math.pow(10, precision);
return Math.floor(Math.floor(value * factorError + 1) / factorTruncate) / factorDecimal;
}
var values = [1.1299999999, 1.13, 1.139999999, 1.14, 1.14000000001, 1.13 * 100];
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 2));
}
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 4));
}
console.log("type of result is " + typeof toFixedWithoutRounding(1.13 * 100 / 100, 2));
// Benchmark
var value = 1.13 * 100;
var startTime = new Date();
var numRun = 1000000;
var nanosecondsPerMilliseconds = 1000000;
for (var run = 0; run < numRun; run++)
toFixedWithoutRounding(value, 2);
var endTime = new Date();
var timeDiffNs = nanosecondsPerMilliseconds * (endTime - startTime);
var timePerCallNs = timeDiffNs / numRun;
console.log("Time per call (nanoseconds): " + timePerCallNs);
Building on David D's answer:
function NumberFormat(num,n) {
var num = (arguments[0] != null) ? arguments[0] : 0;
var n = (arguments[1] != null) ? arguments[1] : 2;
if(num > 0){
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length > 1) {
if(n > 0){
var temp = numarr[0] + ".";
for(var i = 0; i < n; i++){
if(i < numarr[1].length){
temp += numarr[1].charAt(i);
}
}
num = Number(temp);
}
}
}
}
return Number(num);
}
console.log('NumberFormat(123.85,2)',NumberFormat(123.85,2));
console.log('NumberFormat(123.851,2)',NumberFormat(123.851,2));
console.log('NumberFormat(0.85,2)',NumberFormat(0.85,2));
console.log('NumberFormat(0.851,2)',NumberFormat(0.851,2));
console.log('NumberFormat(0.85156,2)',NumberFormat(0.85156,2));
console.log('NumberFormat(0.85156,4)',NumberFormat(0.85156,4));
console.log('NumberFormat(0.85156,8)',NumberFormat(0.85156,8));
console.log('NumberFormat(".85156",2)',NumberFormat(".85156",2));
console.log('NumberFormat("0.85156",2)',NumberFormat("0.85156",2));
console.log('NumberFormat("1005.85156",2)',NumberFormat("1005.85156",2));
console.log('NumberFormat("0",2)',NumberFormat("0",2));
console.log('NumberFormat("",2)',NumberFormat("",2));
console.log('NumberFormat(85156,8)',NumberFormat(85156,8));
console.log('NumberFormat("85156",2)',NumberFormat("85156",2));
console.log('NumberFormat("85156.",2)',NumberFormat("85156.",2));
// NumberFormat(123.85,2) 123.85
// NumberFormat(123.851,2) 123.85
// NumberFormat(0.85,2) 0.85
// NumberFormat(0.851,2) 0.85
// NumberFormat(0.85156,2) 0.85
// NumberFormat(0.85156,4) 0.8515
// NumberFormat(0.85156,8) 0.85156
// NumberFormat(".85156",2) 0.85
// NumberFormat("0.85156",2) 0.85
// NumberFormat("1005.85156",2) 1005.85
// NumberFormat("0",2) 0
// NumberFormat("",2) 0
// NumberFormat(85156,8) 85156
// NumberFormat("85156",2) 85156
// NumberFormat("85156.",2) 85156
Already there are some suitable answer with regular expression and arithmetic calculation, you can also try this
function myFunction() {
var str = 12.234556;
str = str.toString().split('.');
var res = str[1].slice(0, 2);
document.getElementById("demo").innerHTML = str[0]+'.'+res;
}
// output: 12.23
Here is what is did it with string
export function withoutRange(number) {
const str = String(number);
const dotPosition = str.indexOf('.');
if (dotPosition > 0) {
const length = str.substring().length;
const end = length > 3 ? 3 : length;
return str.substring(0, dotPosition + end);
}
return str;
}