How to convert floats to human-readable fractions?

Question

Let's say we have 0.33, we need to output "1/3".
If we have "0.4", we need to output "2/5".

The idea is to make it human-readable to make the user understand "x parts out of y" as a better way of understanding data.

I know that percentages is a good substitute but I was wondering if there was a simple way to do this?

Answer 1

I think the best way to do this is to first convert your float value to an ascii representation. In C++ you could use ostringstream or in C, you could use sprintf. Here's how it would look in C++:

ostringstream oss;
float num;
cin >> num;
oss << num;
string numStr = oss.str();
int i = numStr.length(), pow_ten = 0;
while (i > 0) {
    if (numStr[i] == '.')
        break;
    pow_ten++;
    i--;
}
for (int j = 1; j < pow_ten; j++) {
    num *= 10.0;
}
cout << static_cast<int>(num) << "/" << pow(10, pow_ten - 1) << endl;

A similar approach could be taken in straight C.

Afterwards you would need to check that the fraction is in lowest terms. This algorithm will give a precise answer, i.e. 0.33 would output "33/100", not "1/3." However, 0.4 would give "4/10," which when reduced to lowest terms would be "2/5." This may not be as powerful as EppStein's solution, but I believe this is more straightforward.

Answer 2

From Python 2.6 on there is the fractions module.

(Quoting from the docs.)

>>> from fractions import Fraction
>>> Fraction('3.1415926535897932').limit_denominator(1000)
Fraction(355, 113)

>>> from math import pi, cos
>>> Fraction.from_float(cos(pi/3))
Fraction(4503599627370497, 9007199254740992)
>>> Fraction.from_float(cos(pi/3)).limit_denominator()
Fraction(1, 2)

Answer 3

Here is implementation for ruby http://github.com/valodzka/frac

Math.frac(0.2, 100)  # => (1/5)
Math.frac(0.33, 10)  # => (1/3)
Math.frac(0.33, 100) # => (33/100)

Answer 4

I have found David Eppstein's find rational approximation to given real number C code to be exactly what you are asking for. Its based on the theory of continued fractions and very fast and fairly compact.

I have used versions of this customized for specific numerator and denominator limits.

/*
** find rational approximation to given real number
** David Eppstein / UC Irvine / 8 Aug 1993
**
** With corrections from Arno Formella, May 2008
**
** usage: a.out r d
**   r is real number to approx
**   d is the maximum denominator allowed
**
** based on the theory of continued fractions
** if x = a1 + 1/(a2 + 1/(a3 + 1/(a4 + ...)))
** then best approximation is found by truncating this series
** (with some adjustments in the last term).
**
** Note the fraction can be recovered as the first column of the matrix
**  ( a1 1 ) ( a2 1 ) ( a3 1 ) ...
**  ( 1  0 ) ( 1  0 ) ( 1  0 )
** Instead of keeping the sequence of continued fraction terms,
** we just keep the last partial product of these matrices.
*/

#include <stdio.h>

main(ac, av)
int ac;
char ** av;
{
    double atof();
    int atoi();
    void exit();

    long m[2][2];
    double x, startx;
    long maxden;
    long ai;

    /* read command line arguments */
    if (ac != 3) {
        fprintf(stderr, "usage: %s r d\n",av[0]);  // AF: argument missing
        exit(1);
    }
    startx = x = atof(av[1]);
    maxden = atoi(av[2]);

    /* initialize matrix */
    m[0][0] = m[1][1] = 1;
    m[0][1] = m[1][0] = 0;

    /* loop finding terms until denom gets too big */
    while (m[1][0] *  ( ai = (long)x ) + m[1][1] <= maxden) {
        long t;
        t = m[0][0] * ai + m[0][1];
        m[0][1] = m[0][0];
        m[0][0] = t;
        t = m[1][0] * ai + m[1][1];
        m[1][1] = m[1][0];
        m[1][0] = t;
        if(x==(double)ai) break;     // AF: division by zero
        x = 1/(x - (double) ai);
        if(x>(double)0x7FFFFFFF) break;  // AF: representation failure
    } 

    /* now remaining x is between 0 and 1/ai */
    /* approx as either 0 or 1/m where m is max that will fit in maxden */
    /* first try zero */
    printf("%ld/%ld, error = %e\n", m[0][0], m[1][0],
           startx - ((double) m[0][0] / (double) m[1][0]));

    /* now try other possibility */
    ai = (maxden - m[1][1]) / m[1][0];
    m[0][0] = m[0][0] * ai + m[0][1];
    m[1][0] = m[1][0] * ai + m[1][1];
    printf("%ld/%ld, error = %e\n", m[0][0], m[1][0],
           startx - ((double) m[0][0] / (double) m[1][0]));
}

Answer 5

You can do this in any programming language using the following steps:

1. Multiply and Divide by 10^x where x is the power of 10 required to make sure that the number has no decimal places remaining. Example: Multiply 0.33 by 10^2 = 100 to make it 33 and divide it by the same to get 33/100
2. Reduce the numerator and the denominator of the resulting fraction by factorization, till you can no longer obtain integers from the result.
3. The resulting reduced fraction should be your answer.

Example: 0.2 =0.2 x 10^1/10^1 =2/10 =1/5

So, that can be read as '1 part out of 5'

Answer 6

As many people have stated you really can't convert a floating point back to a fraction (unless its extremely exact like .25). Of course you could create some type of look up for a large array of fractions and use some sort of fuzzy logic to produce the result you are looking for. Again this wouldn't be exact though and you would need to define a lower bounds of how large your want the denominator to go.

.32 < x < .34 = 1/3 or something like that.