How to format a Rational number as a decimal?

问题:

Given an arbitrary large (or small) Rational number that has a finite decimal representation, e.g.:

r = Rational(1, 2**15)
#=> (1/32768)

How can I get its full decimal value as a string?

The expected output for the above number is:

"0.000030517578125"

to_f apparently doesn't work:

r.to_f
#=> 3.0517578125e-05

And sprintf requires me to specify the number of digits:

sprintf('%.30f', r)
#=> "0.000030517578125000000000000000"

回答1:

Most ten-year-olds know how to do that: use long division!¹

Code

def finite_long_division(n,d)
  return nil if d.zero?
  sign = n*d >= 0 ? '' : '-'
  n, d = n.abs, d.abs
  pwr =
  case n <=> d
  when 1 then power(n,d)
  when 0 then 0
  else        -power(d,n)-1
  end            
  n *= 10**(-pwr) if pwr < 0
  d *= 10**(pwr)  if pwr >= 0
  s = ld(n,d)
  t = s.size == 1 ? '0' : s[1..-1]
  "%s%s.%s x 10^%d" % [sign, s[0], t, pwr]
end

def power(n, d)
  # n > d
  ns = n.to_s
  ds = d.to_s
  pwr = ns.size - ds.size - 1
  pwr += 1 if ns[0, ds.size].to_i >= ds.to_i
  pwr
end

def ld(n,d)
  s = ''
  loop do # .with_object('') do |s|
    m,n = n.divmod(d)
    s << m.to_s
    return s if n.zero?
    n *= 10
  end
end

Examples²

finite_long_division(1, 2**15)
  #=> "3.0517578125 x 10^-5"
finite_long_division(-1, 2**15)
  #=> "-3.0517578125 x 10^-5"
finite_long_division(-1, -2**15)
  #=> "3.0517578125 x 10^-5"

finite_long_division(143, 16777216)
  #=> "8.523464202880859375 x 10^-6"
143/16777216.0
  #=> 8.52346420288086e-06 

finite_long_division(8671,
  803469022129495137770981046170581301261101496891396417650688)
  #=> "1.079195309486679194852923588206549145803161531099624\
  #      804222395643336829571798416196370119711226461255452\
  #      67714596064934085006825625896453857421875 x 10^-56"      

Recall that every rational number has either a decimal representation or contains an infinitely-repeating sequence of digits (e.g., 1/3 #=> 0.33333..., 3227/555 #=> 5.8144144144... and 1/9967 #=> 0.00010033109260559848...³). This method would therefore never terminate if the rational was of the repeating sequence variety. Since one generally doesn't know in advance which type a rational number is, it might be useful to modify the method to first determine if the rational number has a finite decimal representation. It is known that a rational number n/d which cannot be reduced (by removing common factors) has this property if and only if d is divisible by 2 or by 5 and is not divisible by any other prime number.⁴ We could easily construct a method to determine if an already-reduced rational number has that property.

require 'prime'

def decimal_representation?(n, d)
  primes = Prime.prime_division(d).map(&:first)
  (primes & [2,5]).any? && (primes - [2, 5]).empty?
end

^{1 At least that was true when I was a kid.}

^{2 See here for a partial list of rational numbers that have finite decimal representations}.

^{3 This rational number's repeating sequence contains 9,966 digits.}

^{4 Reference}.

回答2:

a = sprintf('%.30f', r)
a.gsub(/0*\z/,'')

That's all :) (or should :P) It's not the best way, if the value have more than 30 decimals, you need to add more than 30 zeroes in sprintf. I think there is a better way to do it, but this way it works

Edited

require 'bigdecimal'
require 'bigdecimal/util'
b = BigDecimal.new(r, (r.denominator * r.numerator))
b.to_digits

Note about this solution. (r.denominator * r.numerator) It's the precision, the precision will never be bigger than denominator * numerator (I think, but a mathematician can tell you this)

Edit 2

r = BigDecimal("1") / (BigDecimal("2") ** BigDecimal("99"))
r.to_digits
# Example
r = BigDecimal("1") / (BigDecimal("2")**BigDecimal("99"))
r.to_digits
# "0.000000000000000000000000000001577721810442023610823457130565572459346412870218046009540557861328125"

But reeeeelly big numbers, like:

r = BigDecimal("1") / (BigDecimal("2")**BigDecimal("999999999999"))
# RangeError: integer 999999999999 too big to convert to `int'

If you need something better, I think you need to got with your own implementation of of "string divisions".

回答3:

Bigdecimal to_s has an "F" option. It takes some converting to kick this rational into shape however.

require "bigdecimal"
r = Rational(1, 2**15)
p   BigDecimal.new(r.to_f.to_s).to_s("F") # => "0.000030517578125"