红宝石数字到人类可读的字符串转换

Question

我需要一个列表，其中每个列表项的 id 为 #one、#two 等。 这是最有效的方法还是我在这里缺少内置的 ruby​​ 函数？

-num_array = ["one", "two", "three", "four", "five", "six", "seven"]
-navigation[:primary_level].each_with_index do |primary_item, idx|
   %li{ :id => "#{num_array[idx]}"}

Answer 1

humanize gem 将数字转换为单词。

Answer 2

在使用humanize gem 之外，使用哈希比使用数组更容易：

lookup = {"one" => 1, "two" => 2, "three" => 3, etc...}
text = "two"
num = lookup[text]

Answer 3

我确信这远远超出了您的需要，但在 Rosetta Code 有代码可以做到这一点

Answer 4

这是我对 Ruby 解决方案的尝试。它可能不是最理想的，并且尚未检查其正确性。

<<documentation

Converting Numbers to Human Readable Pretty Print Strings

General Description
===================

- Divide the number into groups of three
    - e.g. turn 87012940 -> 87,012,940
- Parse each individual group
    - e.g. 940 -> "nine hundred forty"
    - Only parse the rightmost two numbers
       - 0 -> 12: special cases; use hardcoded switch statements
            - e.g. "one, two, three ... ten, eleven, twelve"
       - 13 -> 19: same hardcoded switch statement + a "-teen" prefix
            - e.g. "thirteen, fourteen, fifteen ... nineteen"
       - 20 -> 99:
            - Parse left digit and return according to the following rule:
                - "twenty, thirty, forty, fifty, sixty ... ninety"
            - Return the simple name of the right digit:
               - "one, two, nine"
                - special case: zero -> ""
      - This is because the hundredth's place follows a simple prefix rule
          - e.g. one-hundred, two-hundred, three-hundred ... nine-hundred
          - special case: zero -> " "
- Add place modifiers according to each group's placement
      - e.g. the middle '012' -> "twelve thousand"
- Concatenate all and return as solution


Algorithm (slightly modified)
=============================

Modifications
-------------

- No need to divide number into groups of three; simply parse right-to-left one at a time
  - When finished processing one group, insert the result leftmost into our final solution string

documentation


def convert(num)
  return 'zero' if (num == 0)

  answer = ''

  places = ['',
               'thousand ',
               'million ',
               'billion ',
               'trillion ',
               'quadrillion ',
               'quintillion ']
  place = 0

  loop do
    break if num == 0

    # Get the rightmost group of three
    first_three_digits = num % 1000

    # Truncate the original number by those three digits
    num /= 1000

    answer.insert(0, convert_group_of_three(first_three_digits) + places[place])
    place += 1
  end

  answer.strip!
end

def convert_group_of_three(num)
  str = ''

  # Zero returns an empty string
  special_cases = ['', 'one ', 'two ', 'three ', 'four ', 'five ', 'six ', 'seven ', 'eight ', 'nine ', 'ten ',
                   'eleven ', 'twelve ', 'thirteen ', 'fourteen ', 'fifteen ', 'sixteen ', 'seventeen ', 'eighteen ', 'nineteen ']
  return special_cases[num % 100] if (0 .. special_cases.length - 1).include? (num % 100)

  # If not in special cases, num must be at least a two digit number
  # Pull the first digit
  first_digit = num % 10
  num /= 10
  str.insert(0, special_cases[first_digit])

  # Pull the second digit
  second_digit = num % 10
  num /= 10

  second_digit_str = ''
  case second_digit
    when 2
      second_digit_str = 'twenty '
    when 3
      second_digit_str = 'thirty '
    when 4
      second_digit_str = 'forty '
    when 5
      second_digit_str = 'fifty '
    when 6
      second_digit_str = 'sixty '
    when 7
      second_digit_str = 'seventy '
    when 8
      second_digit_str = 'eighty '
    when 9
      second_digit_str = 'ninety '
  end
  str.insert(0, second_digit_str)

  # If there is a third digit
  if num > 0
    third_digit = num % 10
    str.insert(0, special_cases[third_digit] + 'hundred ')
  end

  str
end

p convert(2389475623984756)

Output:

"two quadrillion three hundred eighty nine trillion four hundred seventy five billion six hundred twenty three million nine hundred eighty four thousand seven hundred fifty six"