Question

我想生成对应于数字的字母序列，即“A”，“DE”“GJE”等。前26个很容易，因此3返回“C”，26返回“Z”，27返回“AA”，28“AB”，依此类推。

我无法弄清楚的是如何做到这一点所以它将处理传入的任何数字。所以如果我传入4123我应该回到3个字母的组合，因为（26 * 26 * 26）允许最多+17,000种组合。

有什么建议吗？

Answer 1

class Numeric
  Alph = ("a".."z").to_a
  def alph
    s, q = "", self
    (q, r = (q - 1).divmod(26)); s.prepend(Alph[r]) until q.zero?
    s
  end
end

3.alph
# => "c"
26.alph
# => "z"
27.alph
# => "aa"
4123.alph
# => "fbo"

Answer 2

对@sawa原创的Ruby 2.0答案进行了调整，因为我无法按原样运行：

class Numeric
  Alpha26 = ("a".."z").to_a
  def to_s26
    return "" if self < 1
    s, q = "", self
    loop do
      q, r = (q - 1).divmod(26)
      s.prepend(Alpha26[r]) 
      break if q.zero?
    end
    s
  end
end

这里它从字符串反向转换为整数：

class String
  Alpha26 = ("a".."z").to_a

  def to_i26
    result = 0
    downcased = downcase
    (1..length).each do |i|
      char = downcased[-i]
      result += 26**(i-1) * (Alpha26.index(char) + 1)
    end
    result
  end

end

用法：

1234567890.to_s26 
# => "cywoqvj"

"cywoqvj".to_i26  
# => 1234567890

1234567890.to_s26.to_i26
# => 1234567890

"".to_i26
# => 0

0.to_s26
# => ""

Answer 3

字符串确实有succ方法，因此它们可用于范围。 “Z”的后继者恰好是“AA”，所以这有效：

h = {}
('A'..'ZZZ').each_with_index{|w, i| h[i+1] = w } 
p h[27] #=> "AA"

Answer 4

我喜欢这个答案：https://stackoverflow.com/a/17785576/514483

number.to_s(26).tr("0123456789abcdefghijklmnopq", "ABCDEFGHIJKLMNOPQRSTUVWXYZ")

Answer 5

使用找到here的基本转换方法。我也改变了它，因为我们在这个编号系统中缺少“0”。已经解决了最终案件。

def baseAZ(num)
  # temp variable for converting base
  temp = num

  # the base 26 (az) number
  az = ''

  while temp > 0

    # get the remainder and convert to a letter
    num26 = temp % 26
    temp /= 26

    # offset for lack of "0"
    temp -= 1 if num26 == 0

    az = (num26).to_s(26).tr('0-9a-p', 'ZA-Y') + az
  end

  return az
end

irb I / O：

>> baseAZ(1)
=> "A"
>> baseAZ(26^2 + 1)
=> "Y"
>> baseAZ(26*26 + 1)
=> "ZA"
>> baseAZ(26*26*26 + 1)
=> "YZA"
>> baseAZ(26*26*26 + 26*26 + 1)
=> "ZZA"

Answer 6

def letter_sequence(n)
    n.to_s(26).each_char.map {|i| ('A'..'Z').to_a[i.to_i(26)]}.join
end

Answer 7

根据sawa的回答，我想要一种独立工作的方法，尽管是递归的，以达到预期的效果：

def num_to_col(num)
  raise("invalid value #{num} for num") unless num > 0
  result, remainder = num.divmod(26)
  if remainder == 0
    result -= 1
    remainder = 26
  end
  final_letter = ('a'..'z').to_a[remainder-1]
  result > 0 ? previous_letters = num_to_col(result) : previous_letters = ''
  "#{previous_letters}#{final_letter}".upcase
end

Answer 8

这是一种基于性能递归的短期解决方案

class Numeric
  # 'z' is placed in the begining of the array because 26 % 26 is 0 not 26
  Alph = ['z'] + ("a".."y").to_a

  def to_alph

    # if self is 0 or negative return a blank string.
    # this is also used to end the recursive chain of to_alph calls
    # so don't replace this with raising an error or returning anything else

    return '' if self < 1

    # (otherwise) return two concatenated strings:
    # the right side is the letter for self modules 26
    # the left side is comprised of:
    #  1. minus one, because this is not a zero-based numbering system.
    #     therefore, round numbers (26, 52...) should return one digit 'z'
    #     instead of two digits 'aa' or 'ba'.
    #  2. divide by 26 and call to_alph on that.
    #     this repeats recursively for every digit of the final string,
    #     plus once more that will return '' to end the recursion chain.

    return ((self - 1) / 26).to_alph + Alph[self % 26]
  end
end

Answer 9

您可以通过从序数中减去 96 来获得字符在字母表中的数字位置，如下所示：

"a".ord - 96
=> 1

"z".ord - 96
=> 26

您还可以通过添加 96 来按字母数字位置获取字母字符，如下所示：

(1 + 96).chr
=> "a"

(26 + 96).chr
=> "z"

使用ruby生成用来表示数字的字母？

9 个答案:

这是一种基于性能递归的短期解决方案