سؤال
In JavaScript, is there any way to convert a decimal number (such as 0.0002) to a fraction represented as a string (such as "2/10000")?
If a function called decimalToFraction had been written for this purpose, then decimalToFraction(0.0002) would return the string "2/10000".
المحلول
You can use Erik Garrison's fraction.js library to do that and more fractional operations.
https://stackoverflow.com/questions/14783869
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Russianvar f = new Fraction(2, 10000);
console.log(f.numerator + '/' + f.denominator);
To to do .003 you can just do
var f = new Fraction(.003);
console.log(f.numerator + '/' + f.denominator);
نصائح أخرى
A little googling with the term "decimal to fraction js" the first yielded this:
http://wildreason.com/wildreason-blog/2010/javascript-convert-a-decimal-into-a-simplified-fraction/
It seems to work:
function HCF(u, v) {
var U = u, V = v
while (true) {
if (!(U%=V)) return V
if (!(V%=U)) return U
}
}
//convert a decimal into a fraction
function fraction(decimal){
if(!decimal){
decimal=this;
}
whole = String(decimal).split('.')[0];
decimal = parseFloat("."+String(decimal).split('.')[1]);
num = "1";
for(z=0; z<String(decimal).length-2; z++){
num += "0";
}
decimal = decimal*num;
num = parseInt(num);
for(z=2; z<decimal+1; z++){
if(decimal%z==0 && num%z==0){
decimal = decimal/z;
num = num/z;
z=2;
}
}
//if format of fraction is xx/xxx
if (decimal.toString().length == 2 &&
num.toString().length == 3) {
//reduce by removing trailing 0's
decimal = Math.round(Math.round(decimal)/10);
num = Math.round(Math.round(num)/10);
}
//if format of fraction is xx/xx
else if (decimal.toString().length == 2 &&
num.toString().length == 2) {
decimal = Math.round(decimal/10);
num = Math.round(num/10);
}
//get highest common factor to simplify
var t = HCF(decimal, num);
//return the fraction after simplifying it
return ((whole==0)?"" : whole+" ")+decimal/t+"/"+num/t;
}
// Test it
alert(fraction(0.0002)); // "1/5000"
I used this site http://mathforum.org/library/drmath/view/51886.html to build a function but as the article mentions you will get an unreasonable large number for radicals or pi.
Hope it helps though.
function Fraction(){}
Fraction.prototype.convert = function(x, improper)
{
improper = improper || false;
var abs = Math.abs(x);
this.sign = x/abs;
x = abs;
var stack = 0;
this.whole = !improper ? Math.floor(x) : 0;
var fractional = !improper ? x-this.whole : abs;
/*recursive function that transforms the fraction*/
function recurs(x){
stack++;
var intgr = Math.floor(x); //get the integer part of the number
var dec = (x - intgr); //get the decimal part of the number
if(dec < 0.0019 || stack > 20) return [intgr,1]; //return the last integer you divided by
var num = recurs(1/dec); //call the function again with the inverted decimal part
return[intgr*num[0]+num[1],num[0]]
}
var t = recurs(fractional);
this.numerator = t[0];
this.denominator = t[1];
}
Fraction.prototype.toString = function()
{
var l = this.sign.toString().length;
var sign = l === 2 ? '-' : '';
var whole = this.whole !== 0 ? this.sign*this.whole+' ': sign;
return whole+this.numerator+'/'+this.denominator;
}
//var frac = new Fraction()
//frac.convert(2.56, false)
//console.log(frac.toString())
//use frac.convert(2.56,true) to get it as an improper fraction
If you just want a self-contained function that only returns the numerator and denominator then use the function below.
var toFraction = function (dec) {
var is_neg = dec < 0;
dec = Math.abs(dec);
var done = false;
//you can adjust the epsilon to a larger number if you don't need very high precision
var n1 = 0, d1 = 1, n2 = 1, d2 = 0, n = 0, q = dec, epsilon = 1e-13;
while (!done) {
n++;
if (n > 10000) {
done = true;
}
var a = parseInt(q);
var num = n1 + a * n2;
var den = d1 + a * d2;
var e = (q - a);
if (e < epsilon) {
done = true;
}
q = 1 / e;
n1 = n2;
d1 = d2;
n2 = num;
d2 = den;
if (Math.abs(num / den - dec) < epsilon || n > 30) {
done = true;
}
}
return [is_neg ? -num : num, den];
};
//Usage:
//var frac = toFraction(0.5);
//console.log(frac)
//Output: [ 1, 2 ]
Very old question but maybe someone can find this useful. It's iterative, not recursive and doesn't require factorization
function getClosestFraction(value, tol) {
var original_value = value;
var iteration = 0;
var denominator=1, last_d = 0, numerator;
while (iteration < 20) {
value = 1 / (value - Math.floor(value))
var _d = denominator;
denominator = Math.floor(denominator * value + last_d);
last_d = _d;
numerator = Math.ceil(original_value * denominator)
if (Math.abs(numerator/denominator - original_value) < tol)
break;
iteration++;
}
return {numerator: numerator, denominator: denominator};
};
There is a very simple solution using string representation of numbers
string = function(f){ // returns string representation of an object or number
return f+"";
}
fPart = function(f){ // returns the fraction part (the part after the '.') of a number
str = string(f);
return str.indexOf(".")<0?"0":str.substring(str.indexOf(".") + 1);
}
wPart = function(f){ // returns the integer part (the part before the '.') of a number
str = string(f);
return str.indexOf(".")<0?str:str.substring(0, str.indexOf(".")); // possibility 1
//return string(f - parseInt(fPart(f))); // just substract the fPart
}
power = function(base, exp){
var tmp = base;
while(exp>1){
base*=tmp;
--exp;
}
return base;
}
getFraction = function(f){ // the function
var denominator = power(10, fPart(f).length), numerator = parseInt(fPart(f)) + parseInt(wPart(f))*denominator;
return "[ " + numerator + ", " + denominator + "]";
}
console.log(getFraction(987.23));
which will just check how many numbers are in the fraction and then expands the fraction of f/1 until f is an integer. This can lead to huge fractions, so you can reduce it by dividing both numerator and denominator by the greatest common divisor of both, e.g.
// greatest common divisor brute force
gcd = function(x,y){
for(var i = Math.min(x, y);i>0;i--) if(!(x%i||y%i)) return i;
return 1;
}
The good news is that it's possible, but you'll have to convert it to code.
Let's go with 2.56 for no reason at all.
Use the decimal portion of the number .56
There are 2 digits in .56, write .56 as 56/100.
So we have 2 + 56/100 and need to reduce this fraction to lowest terms by dividing both the numerator and denominator by the greatest common divisor, which is 4 in this case.
So, this fraction reduced to lowest terms is 2 + 14/25.
To add those whole 2, we multiply by the divisor and add to the 14
(2*25 + 14)/25 = 64/25
I did what popnoodles suggested and here it is
function FractionFormatter(value) {
if (value == undefined || value == null || isNaN(value))
return "";
function _FractionFormatterHighestCommonFactor(u, v) {
var U = u, V = v
while (true) {
if (!(U %= V)) return V
if (!(V %= U)) return U
}
}
var parts = value.toString().split('.');
if (parts.length == 1)
return parts;
else if (parts.length == 2) {
var wholeNum = parts[0];
var decimal = parts[1];
var denom = Math.pow(10, decimal.length);
var factor = _FractionFormatterHighestCommonFactor(decimal, denom)
return (wholeNum == '0' ? '' : (wholeNum + " ")) + (decimal / factor) + '/' + (denom / factor);
} else {
return "";
}
}
This may be a little old but the code that was posted fails on 0 values I have fixed that error and will post the updated code below
//function to get highest common factor of two numbers (a fraction)
function HCF(u, v) {
var U = u, V = v
while (true) {
if (!(U%=V)) return V
if (!(V%=U)) return U
}
}
//convert a decimal into a fraction
function fraction(decimal){
if(!decimal){
decimal=this;
}
whole = String(decimal).split('.')[0];
decimal = parseFloat("."+String(decimal).split('.')[1]);
num = "1";
for(z=0; z<String(decimal).length-2; z++){
num += "0";
}
decimal = decimal*num;
num = parseInt(num);
for(z=2; z<decimal+1; z++){
if(decimal%z==0 && num%z==0){
decimal = decimal/z;
num = num/z;
z=2;
}
}
//if format of fraction is xx/xxx
if (decimal.toString().length == 2 &&
num.toString().length == 3) {
//reduce by removing trailing 0's
// '
decimal = Math.round(Math.round(decimal)/10);
num = Math.round(Math.round(num)/10);
}
//if format of fraction is xx/xx
else if (decimal.toString().length == 2 &&
num.toString().length == 2) {
decimal = Math.round(decimal/10);
num = Math.round(num/10);
}
//get highest common factor to simplify
var t = HCF(decimal, num);
//return the fraction after simplifying it
if(isNaN(whole) === true)
{
whole = "0";
}
if(isNaN(decimal) === true)
{
return ((whole==0)?"0" : whole);
}
else
{
return ((whole==0)?"0 " : whole+" ")+decimal/t+"/"+num/t;
}
}
I just want a leave one alternative that I found to convert decimal numbers into fractions and reducing fractions, it's a JS library.
The library calls fraction.js, it was really helpful for me and save me a lot time and work. Hope can be useful to somebody else!
I know this is an old question, but I have created a function that has been greatly simplified.
Math.fraction=function(x){
return x?+x?x.toString().includes(".")?x.toString().replace(".","")/(function(a,b){return b?arguments.callee(b,a%b):a;})(x.toString().replace(".",""),"1"+"0".repeat(x.toString().split(".")[1].length))+"/"+("1"+"0".repeat(x.toString().split(".")[1].length))/(function(a,b){return b?arguments.callee(b,a%b):a;})(x.toString().replace(".",""),"1"+"0".repeat(x.toString().split(".")[1].length)):x+"/1":NaN:void 0;
}
Call it with Math.fraction(2.56)
It will:
string (use Math.fraction(2.56).split("/") for an array containing the numerator and denominator)Please note that this uses the deprecated arguments.callee, and thus may be incompatible in some browsers.
Test it here
Have tried something like this?
var cnum = 3.5,
deno = 10000,
neww;
neww = cnum * deno;
while (!(neww % 2 > 0) && !(deno % 2 > 0)) {
neww = neww / 2;
deno = deno / 2;
}
while (!(neww % 3 > 0) && !(deno % 3 > 0)) {
neww = neww / 3;
deno = deno / 3;
}
while (!(neww % 5 > 0) && !(deno % 5 > 0)) {
neww = neww / 5;
deno = deno / 5;
}
while (!(neww % 7 > 0) && !(deno % 7 > 0)) {
neww = neww / 7;
deno = deno / 7;
}
while (!(neww % 11 > 0) && !(deno % 11 > 0)) {
neww = neww / 11;
deno = deno / 11;
}
while (!(neww % 13 > 0) && !(deno % 13 > 0)) {
neww = neww / 13;
deno = deno / 13;
}
while (!(neww % 17 > 0) && !(deno % 17 > 0)) {
neww = neww / 17;
deno = deno / 17;
}
while (!(neww % 19 > 0) && !(deno % 19 > 0)) {
neww = neww / 19;
deno = deno / 19;
}
console.log(neww + "/" + deno);Although old, I see this question has had some edits recently and its still the first thing that came up on search engines for me when I was looking into this.
For that reason I wanted to share: if anyone comes to this looking for a simple (naive) solution that doesn't require pulling in a library, this is what I ended up using for my purpose.
Its a simple brute-force that first looks if 1/⌊1/x⌉ is a good approximation, if not it checks 2/⌊2/x⌉, 3/⌊3/x⌉ and so on until the first result within the given error bound.
function getFrac(x, maxErr){
let s = x<0?-1:1
x = Math.abs(x),
i = Math.floor(x),
d = x-i,
maxErr = maxErr ? maxErr : Math.pow(10,-6);
if(d<maxErr) return [i,1];
let n = 1,
nmax = Math.ceil(d*Math.min(
Math.pow(10,Math.ceil(Math.abs(Math.log10(maxErr)))),
Number.MAX_SAFE_INTEGER
)),
min = Infinity,
best = [0,0];
while(n <= nmax){
let err = Math.abs(d - n/Math.round(n/d));
if(err < maxErr) return [s*(n+i*Math.round(n/d)), Math.round(n/d)];
else if(err < min){
min = err;
best = [s*(n+i*Math.round(n/d)), Math.round(n/d)];
}
n++;
}
return best[1] == 0 ? false : best;
}
Example output:
getFrac(0) // [0, 1]
getFrac(0.28) // [7, 25]
getFrac(0.28282828) // [28, 99]
getFrac(2.56) // [64, 25]
getFrac(-1.33333333) // [-4, 3]
getFrac(Math.E) // [2721, 1001]
getFrac(Math.PI) // [355, 113]
getFrac(Math.PI, 0.01) // [22, 7]
getFrac(Math.PI, 0.001) // [201, 64]
getFrac(Math.PI, 10**-4) // [333, 106]
getFrac(Math.PI, 10**-12) // [4272943, 1360120]
Although this is the naive approach, it runs surprisingly well, fast enough for most needs. Both JSbench and tests in my console using performance.now() indicate getFrac takes ~4 microseconds to return on average when fed random input using the default error level of 6 decimal places. The linked JSBench was giving me a result of ~200 ops/sec on my machine when calling getFrac 1000 times each op. I include below the timer script I used in the console.
let N = 10000,
nums = new Array(N).fill(0).map(v=>Math.random()*(Math.random()<2/3?1:100)),
t = 0;
for(let n of nums){
const t0 = performance.now();
getFrac(n);
const t1 = performance.now();
t += t1-t0;
}
console.log((t/N)*1000,'micros'); // 4.039999999850989 micros
