Ruby（Hard Way，Ex48）匹配数组到hash然后为struct赋值？

Question

我查看了关于struct，map，array＆amp;的Ruby文档中的所有内容。散列;但无论出于什么原因还没有抓住如何做到这一点。我有两个问题。

1. 查看数组和散列是否有共同值的好方法是什么？
2. 从哈希中，我怎样才能找到一个值并“自动”获取它的密钥（或反之亦然）并将它们包含在其他东西的新实例中（例如Pair.new）？

我看到了这个SO回复，但没有多大帮助...... Learn Ruby Hard Way ex. 48

也看了这个，但它没有用＆amp; Zed说使用map func ... Using Ruby, what is the most efficient way to check if any key in a hash matches any values within an Array

可在此处找到练习说明。 Ruby The Hard Way EX.48 Instructions

我的代码（尝试＃2080）

class Lexicon

Pair = Struct.new(:token, :key)

def scan(stuff)
    @words = stuff.split(" ")
    return analyze
end

def analyze

    hash = { :direction => "north", :direction => "south", 
            :direction => "east", :direction => "west", :verb => "go",  
                                 :verb => "stop"}

    @words.map do |word|
        if word == hash[:direction]
        #i need something here that says if word matches a value in hash... 
        #assign matching key/value to a new instance of Pair
            Pair.new(word)
        else
                            *#puts to see if anything is sticking* 
            puts "Oh god its not working #{word}"
            puts Pair[]
            puts hash[:direction]
           end
       end
  end
end

a = Lexicon.new()
a.scan("north mama jeffrey homie")

TERMINAL

$ ruby lexicon.rb
Oh god its not working north
#<struct Lexicon::Pair token=nil, key=nil>
west

Oh god its not working mama
#<struct Lexicon::Pair token=nil, key=nil>
west

Oh god its not working Jeffrey
#<struct Lexicon::Pair token=nil, key=nil>
west

Oh god its not working homie
#<struct Lexicon::Pair token=nil, key=nil>
west

MY CODE＃2081与上述相同，但

   hash.each do |k, v|
      if v == @words
      #i need something here that says if word matches a value in hash... 
      #assign matching key/value to a new instance of Pair
        Pair.new(k,v)
      else
        puts "Oh god its not working"
        puts Pair[]
        puts hash[:direction]
        puts @words
      end
   end

TERMINAL

Oh god its not working
#<struct Lexicon::Pair token=nil, key=nil>
...
...

Answer 1

这里有几个问题。最重要的是，你的哈希不起作用;哈希是使用唯一键建立在键值对上的，因此您的哈希实际上与以下内容相同：

hash = { :direction => "west", :verb => "stop"}

您可能最好更换哈希中的键值对，如下所示：

hash = { "north" => :direction, "south" => :direction, 
        "east" => :direction, "west" => :direction, "go" => :verb,  
                             "stop" => :verb }

@words.map do |word|
  hash.keys.include?(word) ? Pair.new(hash[word], word) : Pair.new(:error, word)
end

Answer 2

def have_common_value?(array, hash)
  hash.values.any? { |v| array.include? v }
end

def get_pair_from_value(hash, value)
  hash.find { |k,v| v == value }
end

Answer 3

哈希每个值只能有一个唯一键，因此，部分问题是上面生成的哈希只会返回一个:direction和一个:location。

这应该有助于让您更接近您所寻找的目标：

class Lexicon

  Pair = Struct.new(:token, :key)

  def scan(stuff)
    @words = stuff.split(" ")
    return analyze
  end

  def analyze
    hash = { :directions => { :north => "north", :south => "south", :east => "east", :west => "west" },
             :actions => { :verb => "go", :verb => "stop" } }

    @words.map do |word|
      if hash[:directions].values.include?(word)
        Pair.new(word)
      else
        puts "Oh god its not working #{word}"
        puts Pair[]
        puts hash[:directions]
      end
    end
  end
end

a = Lexicon.new()
a.scan("north mama jeffrey homie")