使用Ruby将数字转换为单词？

Question

使用Ruby将数字转换为单词？

ruby-on-railsrubynumbers

15

如何在Ruby中将数字转换为单词？

我知道有一个gem可以实现。但我想尝试在没有gem的情况下实现它。我只需要将整数转换成英文单词。找到了这个，但是代码很混乱。如果你有任何想法可以实现更简洁易读的方法，请分享。

http://raveendran.wordpress.com/2009/05/29/ruby-convert-number-to-english-word/

这是我一直在努力的工作内容，但是在实现比例尺时遇到了一些问题。代码还很混乱，希望在它正常运行时能够使其更易读。

   class Numberswords
    def in_words(n)

    words_hash = {0=>"zero",1=>"one",2=>"two",3=>"three",4=>"four",5=>"five",6=>"six",7=>"seven",8=>"eight",9=>"nine",
                    10=>"ten",11=>"eleven",12=>"twelve",13=>"thirteen",14=>"fourteen",15=>"fifteen",16=>"sixteen",
                     17=>"seventeen", 18=>"eighteen",19=>"nineteen",
                    20=>"twenty",30=>"thirty",40=>"forty",50=>"fifty",60=>"sixty",70=>"seventy",80=>"eighty",90=>"ninety"}

     scale = [000=>"",1000=>"thousand",1000000=>" million",1000000000=>" billion",1000000000000=>" trillion", 1000000000000000=>" quadrillion"]


    if words_hash.has_key?(n) 
      words_hash[n]

      #still working on this middle part. Anything above 999 will not work
     elsif n>= 1000  
     print  n.to_s.scan(/.{1,3}/) do |number|
            print number
      end



      #print value = n.to_s.reverse.scan(/.{1,3}/).inject([]) { |first_part,second_part| first_part << (second_part == "000" ? "" : second_part.reverse.to_i.in_words) }
      #(value.each_with_index.map { |first_part,second_part| first_part == "" ? "" : first_part + scale[second_part] }-[""]).reverse.join(" ")

    elsif n <= 99
       return [words_hash[n - n%10],words_hash[n%10]].join(" ")
    else
      words_hash.merge!({ 100=>"hundred" })
      ([(n%100 < 20 ? n%100 : n.to_s[2].to_i), n.to_s[1].to_i*10, 100, n.to_s[0].to_i]-[0]-[10])
        .reverse.map { |num| words_hash[num] }.join(" ")
    end
  end
end

#test code
test = Numberswords.new
 print test.in_words(200)

- hken27

8个回答

15

你考虑过humanize吗？

https://github.com/radar/humanize

- rhernando

9

简单的答案是使用humanize宝石，您将获得所需的输出。

直接安装它。

gem install humanize

或将其添加到您的Gemfile中

gem 'humanize'

你可以使用它

require 'humanize'

1.humanize       #=> 'one'
345.humanize     #=> 'three hundred and forty-five'
1723323.humanize #=> 'one million, seven hundred and twenty-three thousand, three hundred and twenty-three'

如果您正在使用Rails，您可以直接使用此功能。

注意：如下评论中sren所提到的。由ActiveSupport提供的humanize方法与humanize宝石不同。

- Deepak Mahakale

4

不要混淆这个独立的 gem 和 ActiveSupport 的 humanize 方法。https://github.com/radar/humanize - sren

6

你可以使用to_words宝石。

这个宝石将整数转换为单词。

例如：

1.to_words # one ,

100.to_words # one hundred ,

101.to_words # one hundred and one

它还可以转换负数。

- taimur akhtar

1

这个问题已经问了很久了。现在Rails有内置的解决方案。 https://api.rubyonrails.org/classes/ActionView/Helpers/NumberHelper.html

number_to_human(1234567)                                      # => "1.23 Million"
number_to_human(1234567890)                                   # => "1.23 Billion"
number_to_human(1234567890123)                                # => "1.23 Trillion"
number_to_human(1234567890123456)                             # => "1.23 Quadrillion"
number_to_human(1234567890123456789)                          # => "1230 Quadrillion"

- Nishutosh Sharma

1

我可以看到你正在寻找的内容，你可能想查看这篇StackOverflow帖子: Number to English Word Conversion Rails

这里是摘要:

不，你必须自己编写一个函数。最接近你想要的是number_to_human，但它不能将1转换为One。

以下是一些有用的网址:

http://codesnippets.joyent.com/posts/show/447 http://raveendran.wordpress.com/2009/05/29/ruby-convert-number-to-english-word/ http://deveiate.org/projects/Linguistics/

- Richard Peck

0

def subhundred number

  ones = %w{zero one two three four five six seven eight nine
            ten eleven twelve thirteen fourteen fifteen
            sixteen seventeen eighteen nineteen}

  tens = %w{zero ten twenty thirty **forty** fifty sixty seventy eighty ninety}

  subhundred = number % 100

  return [ones[subhundred]] if subhundred < 20

  return [tens[subhundred / 10]] if subhundred % 10 == 0

  return [tens[subhundred / 10], ones[subhundred % 10]]

end




def subthousand number

  hundreds = (number % 1000) / 100

  tens = number % 100

  s = []

  s = subhundred(hundreds) + ["hundred"] unless hundreds == 0

  s = s + ["and"] unless hundreds == 0 or tens == 0

  s = s + [subhundred(tens)] unless tens == 0


end




def decimals number

  return [] unless number.to_s['.']

  digits = number.to_s.split('.')[1].split('').reverse

  digits = digits.drop_while {|d| d.to_i == 0} . reverse

  digits = digits.map {|d| subhundred d.to_i} . flatten

  digits.empty? ? [] : ["and cents"] + digits

end





def words_from_numbers number

  steps = [""] + %w{thousand million billion trillion quadrillion quintillion sextillion}

  result = []

  n = number.to_i

  steps.each do |step|

    x = n % 1000

    unit = (step == "") ? [] : [step]

    result = subthousand(x) + unit  + result unless x == 0

    n = n / 1000

  end



  result = ["zero"] if result.empty?

  result = result + decimals(number)




  result.join(' ').strip

end






 def words_from_numbers(number)

    ApplicationHelper.words_from_numbers(number)

  end

- dsn raghavendra rao

0

我不太确定这对你是否有效。方法可以这样调用。

n2w(33123) {|i| puts i unless i.to_s.empty?}

这里是一个方法（我还没有完全测试过它，但我认为它可以处理百万级别的数据。代码很丑陋，有很多重构的空间。）

def n2w(n)
  words_hash = {0=>"zero",1=>"one",2=>"two",3=>"three",4=>"four",5=>"five",6=>"six",7=>"seven",8=>"eight",9=>"nine",
                    10=>"ten",11=>"eleven",12=>"twelve",13=>"thirteen",14=>"fourteen",15=>"fifteen",16=>"sixteen",
                     17=>"seventeen", 18=>"eighteen",19=>"nineteen",
                    20=>"twenty",30=>"thirty",40=>"forty",50=>"fifty",60=>"sixty",70=>"seventy",80=>"eighty",90=>"ninety"}
  scale = {3=>"hundred",4 =>"thousand",6=>"million",9=>"billion"}

  if words_hash.has_key?n
    yield words_hash[n] 
  else
    ns = n.to_s.split(//)
      while ns.size > 0      
        if ns.size == 2
            yield("and")
            yield words_hash[(ns.join.to_i) - (ns.join.to_i)%10]            
            ns.shift
        end
        if ns.size > 4
          yield(words_hash[(ns[0,2].join.to_i) - (ns[0,2].join.to_i) % 10])
        else
          yield(words_hash[ns[0].to_i]) 
        end
        yield(scale[ns.size])
        ns.shift
      end
    end
end

- Bala

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- NoNonsense · Accepted Answer

我的看法

def in_words(int)
  numbers_to_name = {
      1000000 => "million",
      1000 => "thousand",
      100 => "hundred",
      90 => "ninety",
      80 => "eighty",
      70 => "seventy",
      60 => "sixty",
      50 => "fifty",
      40 => "forty",
      30 => "thirty",
      20 => "twenty",
      19=>"nineteen",
      18=>"eighteen",
      17=>"seventeen", 
      16=>"sixteen",
      15=>"fifteen",
      14=>"fourteen",
      13=>"thirteen",              
      12=>"twelve",
      11 => "eleven",
      10 => "ten",
      9 => "nine",
      8 => "eight",
      7 => "seven",
      6 => "six",
      5 => "five",
      4 => "four",
      3 => "three",
      2 => "two",
      1 => "one"
    }
  str = ""
  numbers_to_name.each do |num, name|
    if int == 0
      return str
    elsif int.to_s.length == 1 && int/num > 0
      return str + "#{name}"      
    elsif int < 100 && int/num > 0
      return str + "#{name}" if int%num == 0
      return str + "#{name} " + in_words(int%num)
    elsif int/num > 0
      return str + in_words(int/num) + " #{name} " + in_words(int%num)
    end
  end
end



puts in_words(4) == "four"
puts in_words(27) == "twenty seven"
puts in_words(102) == "one hundred two"
puts in_words(38_079) == "thirty eight thousand seventy nine"
puts in_words(82102713) == "eighty two million one hundred two thousand seven hundred thirteen"