ruby - 如何判断散列中是否有指定键以外的其他键？

Question

假设我有一个哈希:

a = {b: "asdfgh", c: "qwerty", d: "dvorak"}

而且我希望能够判断是否存在我在其中指定的键以外的键，如下所示:

a.has_other_keys?(:b, :c, :d)
=> false
a.has_other_keys?(:c, :d)
=> true

但如果键少于指定的键，我不希望它返回 false:

a.has_other_keys?(:b, :c, :d, :e)
=> true

在 ruby​​ 中有没有简单的方法来做到这一点？

Answer 1

Rails has an except/except! method返回删除了这些键的散列。如果你已经在使用 Rails，你可以使用它

class Hash
  # Returns a hash that includes everything but the given keys.
  #   hash = { a: true, b: false, c: nil}
  #   hash.except(:c) # => { a: true, b: false}
  #   hash # => { a: true, b: false, c: nil}
  #
  # This is useful for limiting a set of parameters to everything but a few known toggles:
  #   @person.update(params[:person].except(:admin))
  def except(*keys)
    dup.except!(*keys)
  end

  # Replaces the hash without the given keys.
  #   hash = { a: true, b: false, c: nil}
  #   hash.except!(:c) # => { a: true, b: false}
  #   hash # => { a: true, b: false }
  def except!(*keys)
    keys.each { |key| delete(key) }
    self
  end
end

使用上面你可以做:

bundle :002 >    a = {b: "asdfgh", c: "qwerty", d: "dvorak"}
 => {:b=>"asdfgh", :c=>"qwerty", :d=>"dvorak"}
bundle :006 >  a.except(:b)
 => {:c=>"qwerty", :d=>"dvorak"}
bundle :007 >  a.except(:b).length
 => 2
bundle :008 >  a.except(:b, :c, :d, :e).length
 => 0

---

在普通 Ruby 中，您可以执行以下操作:

2.2.2 :010 >  a.select{|x| ![:b, :c, :d, :e].include?(x)}
 => {}
2.2.2 :011 >  a.select{|x| ![:b, :c, :d, :e].include?(x)}.length
 => 0