如何将浮点数转换为人类可读的分数? [英] How to convert floats to human-readable fractions?

问题描述

假设我们有0.33，我们需要输出1/3.
如果我们有0.4，我们需要输出2/5.

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

这个想法是为了让用户能够理解y 中的 x 部分"，从而更好地理解数据.

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.

我知道百分比是一个很好的替代品，但我想知道是否有一种简单的方法可以做到这一点?

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

推荐答案

我找到了 David Eppstein 的 找到给定实数的有理近似 C 代码正是您所要求的.它基于连分数理论，速度非常快且相当紧凑.

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.

我已经使用了针对特定分子和分母限制定制的版本.

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
",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
", 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
", m[0][0], m[1][0],
           startx - ((double) m[0][0] / (double) m[1][0]));
}

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