Question

所以我正在解决RSPEC问题，而且我在这个网站上获得的所有帮助都很棒。尽管如此我还是有点陷入困境。

这个想法是在Ruby中创建一个字典对象，RSPEC代码看起来像

require 'dictionary'

describe Dictionary do
  before do
    @d = Dictionary.new
  end

  it 'is empty when created' do
    @d.entries.should == {}
  end

  it 'can add whole entries with keyword and definition' do
    @d.add('fish' => 'aquatic animal')
    @d.entries.should == {'fish' => 'aquatic animal'}
    @d.keywords.should == ['fish']
  end

  it 'add keywords (without definition)' do
    @d.add('fish')
    @d.entries.should == {'fish' => nil}
    @d.keywords.should == ['fish']
  end

  it 'can check whether a given keyword exists' do
    @d.include?('fish').should be_false
  end

  it "doesn't cheat when checking whether a given keyword exists" do
    @d.include?('fish').should be_false # if the method is empty, this test passes with nil returned
    @d.add('fish')
    @d.include?('fish').should be_true # confirms that it actually checks
    @d.include?('bird').should be_false # confirms not always returning true after add
  end

  it "doesn't include a prefix that wasn't added as a word in and of itself" do
    @d.add('fish')
    @d.include?('fi').should be_false
   end

  it "doesn't find a word in empty dictionary" do
    @d.find('fi').should be_empty # {}
  end

  it 'finds nothing if the prefix matches nothing' do
    @d.add('fiend')
    @d.add('great')
    @d.find('nothing').should be_empty
  end

  it "finds an entry" do
    @d.add('fish' => 'aquatic animal')
    @d.find('fish').should == {'fish' => 'aquatic animal'}
  end

  it 'finds multiple matches from a prefix and returns the entire entry (keyword + definition)' do
    @d.add('fish' => 'aquatic animal')
    @d.add('fiend' => 'wicked person')
    @d.add('great' => 'remarkable')
    @d.find('fi').should == {'fish' => 'aquatic animal', 'fiend' => 'wicked person'}
  end

  it 'lists keywords alphabetically' do
    @d.add('zebra' => 'African land animal with stripes')
    @d.add('fish' => 'aquatic animal')
    @d.add('apple' => 'fruit')
    @d.keywords.should == %w(apple fish zebra)
  end

  it 'can produce printable output like so: [keyword] "definition"' do
    @d.add('zebra' => 'African land animal with stripes')
    @d.add('fish' => 'aquatic animal')
    @d.add('apple' => 'fruit')
    @d.printable.should == %Q{[apple] "fruit"\n[fish] "aquatic animal"\n[zebra] "African land animal with stripes"}
  end
end

我有一切都在“找到一个条目”，这就是我遇到麻烦的地方。到目前为止我的代码看起来像这样。感谢本网站上帮助我构建代码的每个人。

class Dictionary
  attr_accessor :keywords, :entries

  def initialize 
    @entries = {}
  end

  def add(defs)
    defs.each do |word, definition|
      @entries[word] = definition
    end      
  end

  def keywords
    @entries.keys.sort
  end

  def include?(key)
    @entries.key?(key)
  end 

  def find(query)
    @entries.select { |word, definition| word.scan(query).join == query}  
  end    
end

我正在进行该测试的错误是：

1) Dictionary finds an entry
     Failure/Error: @d.find('fish').should == {'fish' => 'aquatic animal'}
       expected: {"fish"=>"aquatic animal"}
            got: [["fish", "aquatic animal"]] (using ==)
       Diff:
       @@ -1,2 +1,2 @@
       -"fish" => "aquatic animal"
       +[["fish", "aquatic animal"]]

所以看起来这是finder方法的输出是一个数组而不是哈希的问题。解决这个问题的最佳方法是什么？先感谢您！

Answer 1

你必须使用相当旧版本的Ruby。

在Ruby 1.8.7中：

{}.select{} # => []

使用最新的Ruby：

{}.select{} # => {}

你可以：

将Ruby升级到1.9.2 +
使用reject代替并改变您的情况
或使用我的backports gem和require 'backports/force/hash/select'在Ruby 1.8.x中获得相同的行为。

Answer 2

尝试选择并执行以下操作返回结果：

Hash[result]

或者更原始，只需制作哈希！

{result.first.first => result.first.last}

在Ruby中从数组转换为哈希

2 个答案: