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

我只是一个虾纸丫 提交于 2019-11-27 15:10:24
Vasiliy Ermolovich

Not in ruby but in ActiveSupport:

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

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

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

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

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

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
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