如何将浮点数转换为分数? [英] How to convert float to fraction?

问题描述

这是一个作业问题.我想编写一个将浮点数转换为整数对的函数:分子和分母.例如:float 0.5应该转换为(1,2).

This is a homework problem. I want to write a function to convert a float to a pair of integers: numerator and denominator. For instance: float 0.5 should be converted to (1,2).

我正在尝试做某事. (请参阅下文)，但坦率地说，它对我来说并不好.

I am trying smth. (see below) but frankly it does not look great for me.

// f is the input float
int n = 1
while(fractional_part(f) > 0)
    f *= 10;
    n++
int m = f;
gcd = gcd(m, n)
return (m/gcd, n/gcd)

如何将浮点数转换为分数?

How would you convert a float to a fraction ?

推荐答案

您可以只使用分数库.

但是，如果您想开发该算法，则建议如下:

But, if you would like to develop the algorithm, here is a suggestion:

from math import floor
from fractions import gcd

def func(v, tol=1e-4):
    """
    Don't handle negative values.
    Use binary search to find the fraction of a float.
    The algorithm is based in a very simple theorem: If a < b then a < (a+b)/2 < b.
    """
    f = v - floor(v)
    lo = (0, 1)
    hi = (1, 1)
    while True:
        # mid = (lo + hi)/2
        # if lo = a/b and hi = c/d, then mid = (ad+bc)/(2ad)
        mid = (lo[0]*hi[1] + hi[0]*lo[1], 2*lo[1]*hi[1])
        # gcd to reduce fraction
        k = gcd(mid[0], mid[1])
        mid = (mid[0]/k, mid[1]/k)

        d = 1.*mid[0]/mid[1]
        # are we close enough?
        if abs(f - d) < tol:
            break
        # if we are above our goal, get high to middle
        elif d > f:
            hi = mid
        # if we are under our goal, get lower to middle
        else:
            lo = mid
    # Add integer part
    mid = (mid[0] + int(floor(v))*mid[1], mid[1])
    # Debug comparing to Fraction library solution.
    #print v, mid, Fraction('%s' % v)
    return mid

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