Question

如果我知道哈希中只有一个键/值对，是否可以直接检索键或值而无需获取所有键或值？< / p>

这是一个简单的例子：

hsh1 = {a: 1} 
hsh2 = {a: 2}
hsh1.keys # => [:a]
hsh2.values # => [2]
hsh1.values + hsh2.values # => [1,2]

有没有办法解决这个问题？

1 + 2 # => 3

Answer 1

hsh1 = {a: 1}
hsh2 = {b: 2}

hsh1.values.first + hsh2.values.first # => 3

Answer 2

您可以注入添加内容：

(hsh1.values + hsh2.values).inject(:+)

Answer 3

作为每个优秀的Rubyist，您应该拥有自己的私人改进库：

module MyStuff
  refine Hash do
    def key
      msg = "Method ##{__callee__} called on non-singleton hash!"
      keys.tap { |x| x.size == 1 or fail TypeError, msg }.first
    end

    def value
      msg = "Method ##{__callee__} called on non-singleton hash!"
      values.tap { |x| x.size == 1 or fail TypeError, msg }.first
    end

    def merge *args
      return super unless args.empty?
      Class.new BasicObject do
        def initialize hsh; @hsh = hsh end
        def method_missing sym, *args
          super unless args.size == 1
          @hsh.merge args.first do |_, a, b| a.send( sym, b ) end
        end
      end.new( self )
    end
  end
end

using MyStuff

hsh1 = {a: 1} 
hsh2 = {a: 2}
hsh1.key #=> :a
hsh1.value + hsh2.value #=> 3

wrong_hsh_1 = {}
wrong_hsh_2 = {a: 1, b: 2}
wrong_hsh_1.key #=> TypeError
wrong_hsh_2.value #=> TypeError

允许偶数

( hsh1.merge + hsh2 ).value
( hsh1.merge - hsh2 ).value
( hsh1.merge * hsh2 ).value
( hsh1.merge / hsh2 ).value

直接从Ruby中的单个元素哈希获取键或值

3 个答案: