Question

我正在阅读Ruby编程（第4版），我不太明白如何用2个括号迭代for循环。

require_relative "words_from_string.rb"
require_relative "count_frequency.rb"

raw_text = %{hey hey hey man man bro}

word_list = words_from_string(raw_text)
counts = count_frequency(word_list)
sorted = counts.sort_by {|word, count| count }
top_five = sorted.last(5)

for i in 0...5
    word = top_five[i][0]   #don't understand why there are 2 brackets. 
    count = top_five[i][1]  #how does the iterator work with 2 brackets.
    puts "#{word}: #{count}"
end

words_from_string.rb

def words_from_string(string)
    string.downcase.scan(/[\w']+/)
end

count_frequency.rb

def count_frequency(word_list)
    counts = Hash.new(0)

    for word in word_list
        counts[word] += 1
    end

    counts
end

Answer 1

如果使用sort_by进行散列，则会将散列转换为二维数组。

counts = Hash.new(0)
=> {}
counts[1] = 2
=> 2
counts[2] = 1
=> 1
counts
=> {1=>2, 2=>1}
sorted = counts.sort_by{ |k, v| v }
=> [[2, 1], [1, 2]]
sorted[0][1]
=> 1

Answer 2

top_five数组由元素组成，这些元素本身就是带有两个元素的数组 - 第一个是单词，第二个是它出现在raw_text中的次数。

循环迭代这些对并将单词（每个数组的第一个元素，索引0）提取到word，并将计数（每个数组中的第二个元素，索引1）提取到count

Answer 3

这是因为'topfive'是一个多维数组（在你的情况下是一个二维），意思是，它是一个包含数组的数组。

如果你打算玩它：如果你放了topfive [n]，它会给你一个结构为[string，int]的数组，在你的例子中是[word，count]。

代码使用两个括号，以便您访问数组中的元素。在示例中，topfive [i] [0]将获得该单词，topfive [i] [1]将获得计数或单词数

Answer 4

我玩我的控制台并根据你的答案，我终于明白了！

基本上，first []是主数组的索引，第二个[]是嵌套数组的索引。现在我懂了！辉煌！谢谢你们！

   def words_from_string(string)
       string.downcase.scan(/[\w']+/)
   end

          def count_frequency(word_list)
              counts = Hash.new(0)
              for word in word_list
              counts[word] += 1
              end
            counts
          end

raw_text = %{hey hey hey man man bro yo yo yo yo yo ye ye ya oi ui}

    word_list = words_from_string(raw_text)
    counts = count_frequency(word_list)
    sorted = counts.sort_by {|word, count| count}
    top_five = sorted.last(5)

    print word_list
    #produces
    ["hey", "hey", "hey", "man", "man", "bro", "yo", "yo", "yo", "yo", "yo", "ye", "ye", "ya", "oi", "ui"]

    print counts
    #produces
    {"hey"=>3, "man"=>2, "bro"=>1, "yo"=>5, "ye"=>2, "ya"=>1, "oi"=>1, "ui"=>1}

    print sorted
    #produces
    [["ui", 1], ["ya", 1], ["oi", 1], ["bro", 1], ["man", 2], ["ye", 2], ["hey", 3], ["yo", 5]]

    print top_five
    #produces
    [["bro", 1], ["man", 2], ["ye", 2], ["hey", 3], ["yo", 5]]

    for i in 0...5
    word = top_five[i][0]   
    count = top_five[i][1]  
    puts "#{word}: #{count}"
  end
    #produces
    bro: 1
    man: 2
    ye: 2
    hey: 3
    yo: 5


        puts top_five[4][1] 
        #produces 
        "5"       
        puts top_five[4][0] 
        #produces 
        "yo"

使用2个括号迭代“For”循环

4 个答案: