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

Question

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?

You also can use alias for member? method: include?. This may look better in your context: ['.', ','].include?(word[-1]). — Aliaksei Kliuchnikau, Nov 15 '11 at 08:35

score 17 · Accepted Answer · edited Feb 06 '13 at 16:18

17

Not in ruby but in ActiveSupport:

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

edited Feb 06 '13 at 16:18

Joachim Sauer

answered Nov 15 '11 at 08:31

undur_gongor · Answer 2 · 2011-11-15T08:52:23.343

10

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

edited Nov 15 '11 at 08:52

answered Nov 15 '11 at 08:43

undur_gongor

Yeah, I knew I could. But I'd rather not drag this code to every new project. :-) – Sergio Tulentsev Nov 15 '11 at 09:02
@Sergei Tulentsev: The choice is yours. – undur_gongor Nov 15 '11 at 11:17
Just put it in a gem or something, then. – Marnen Laibow-Koser Nov 15 '11 at 15:12

score 1 · Answer 3 · answered Nov 15 '11 at 09:28

1

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

answered Nov 15 '11 at 09:28

knut

Yeah, that would solve this exact problem, but there are many others, like [:admin, :moderator].include?(current_user.role) – Sergio Tulentsev Nov 16 '11 at 01:42

score 1 · Answer 4 · answered Nov 15 '11 at 12:34

1

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

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

answered Nov 15 '11 at 12:34

steenslag

The question was about general case, I just pasted the latest occurence in code. But thanks for the tip, I didn't know about this method. – Sergio Tulentsev Nov 16 '11 at 01:45

score 0 · Answer 5 · answered Feb 06 '13 at 16:25

0

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

answered Feb 06 '13 at 16:25

sawa

5 Answers5