Is there an inverse 'member?' method in ruby?

问题

I often find myself checking if some value belongs to some set. As I understand, people normally use Enumerable#member? for this.

end_index = [\'.\', \',\'].member?(word[-1]) ? -3 : -2

However, this feels a little less elegant than most of things in Ruby. I\'d rather write this code as

end_index = word[-1].is_in?(\'.\', \',\') ? -3 : -2

but I fail to find such method. Does it even exist? If not, any ideas as to why?

回答1:

Not in ruby but in ActiveSupport:

characters = ["Konata", "Kagami", "Tsukasa"]
"Konata".in?(characters) # => true

回答2:

You can easily define it along this line:

class Object
  def is_in? set
    set.include? self
  end
end

and then use as

8.is_in? [0, 9, 15]   # false
8.is_in? [0, 8, 15]   # true

or define

class Object
  def is_in? *set
    set.include? self
  end
end

and use as

8.is_in?(0, 9, 15)   # false
8.is_in?(0, 8, 15)   # true

回答3:

Not the answer for your question, but perhaps a solution for your problem.

word is a String, isn't it?

You may check with a regex:

end_index = word =~ /\A[\.,]/  ? -3 : -2

or

end_index = word.match(/\A[\.,]/)  ? -3 : -2

回答4:

In your specific case there's end_with?, which takes multiple arguments.

"Hello.".end_with?(',', '.') #=> true

回答5:

Unless you are dealing with elements that have special meaning for === like modules, regexes, etc., you can do pretty much well with case.

end_index = case word[-1]; when '.', ','; -3 else -2 end

来源：https://stackoverflow.com/questions/8133397/is-there-an-inverse-member-method-in-ruby