迭代紅寶石中的回圈？

我是紅寶石新手。我正在嘗試創建一個字母計數器。我的預期輸出應該是 [2,"I"] 但我一直得到 [3,"D]。任何有助于理解我出錯的地方都會非常有幫助，謝謝。

class LetterCounter
      def initialize(text)
        @text = text
      end
    
      def calculate_most_common()
        counter = Hash.new(1)
        most_common = nil
        most_common_count = 1
        @text.chars.each do |char|
          next unless is_letter?(char)
          counter[char] = (counter[char] || 1)   1
          if counter[char] > most_common_count
            most_common = char
            most_common_count  = counter[char]
          end
        end
        return [most_common_count, most_common]
      end
    
      private
    
      def is_letter?(letter)
        return letter =~ /[a-z]/i
      end
    end
    
    counter = LetterCounter.new("Digital Punk")
    p counter.calculate_most_common

# Intended output:
# [2, "i"]

uj5u.com熱心網友回復：

嘗試：

class LetterCounter
    def initialize(text)
        @text = text
    end
    
    def calculate_most_common()
        arr=@text.scan(/[a-z]/i)
        arr.each_with_object(Hash.new(0)) { |n,h| h[n]  = 1 }.max_by(&:last)
    end
    
end
    
counter = LetterCounter.new("Digital Punk")
p counter.calculate_most_common

印刷：

["i", 2]

如果您想修復您的問題，請嘗試：

class LetterCounter
    def initialize(text)
        @text = text
    end
    
    def calculate_most_common()
        counter = Hash.new(0)
        most_common = nil
        most_common_count = 0
        @text.chars.each do |char|
            next unless is_letter?(char)
            counter[char]  = 1
            if counter[char]>most_common_count
                most_common=char 
                most_common_count=counter[char]
            end 
        end
        return [most_common_count, most_common]
    end
    
    private
    
    def is_letter?(letter)
        return letter =~ /[a-z]/i
    end
end

counter = LetterCounter.new("Digital Punk")
p counter.calculate_most_common

uj5u.com熱心網友回復：

class LetterCounter
  def initialize(text)
    @text = text
  end

  def calculate_most_common()
    counter = Hash.new(0)  # wrong initialization in your code
    most_common = nil
    most_common_count = 0
    @text.chars.each do |char|
      next unless is_letter?(char)
      counter[char] = (counter[char] || 1)   1
      puts "#{char} #{counter[char]}"
      if counter[char] > most_common_count
        puts "most common: #{most_common_count} #{char} #{counter[char]}"
        most_common = char
        most_common_count = counter[char] # error in your code
      end
    end
    return [most_common_count, most_common]
  end

  private

  def is_letter?(letter)
    return letter =~ /[a-z]/i
  end
end

counter = LetterCounter.new("Digital Punk")
p counter.calculate_most_common

uj5u.com熱心網友回復：

初始計數器為 1。這些應該為 0，因為無論如何您都在回圈內部遞增。

在第一次迭代中 - 哈希 counter[char] 被初始化為 1（ From Hash.new(1) ），然后 counter[char] 在第一個回圈中增加到 1 并且 most_common_count 也增加了 1。這導致 D有值 3。

因為 most_common_count 已經是 3 - 回圈將不再進入 if 條件 - 因為其他字符只會達到 2（ 1 來自 Hash.new 和 1 來自 counter[char] 1 ）

if 條件是 > most_common_count 而不是 >= - 因此即使 i 達到 3 - 條件也不會執行。

因此輸出 [3, 'D']

試試這個：

類 LetterCounter def 初始化（文本）@text = 文本結束

  def calculate_most_common()
    counter = Hash.new(0)
    most_common = nil
    most_common_count = 0
    @text.chars.each do |char|
      next unless is_letter?(char)

      puts char   "  ==> "   counter[char].to_s # To know the value on each iteration
      counter[char]  = 1
      if counter[char] > most_common_count
        most_common = char
        most_common_count = counter[char]
        puts [most_common_count, most_common] # To know when the if condition is executed
      end
    end
    return [most_common_count, most_common]
  end

  private

  def is_letter?(letter)
    return letter =~ /[a-z]/i
  end
end

counter = LetterCounter.new("Digital Punk")
p counter.calculate_most_common

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