find first element in array for which block returns true and return the block's return value

Question

I need to iterate over an array and apply a supplied block to each element, and return the first true value returned by the block, which implies that I need to stop immediately as soon as I get a true value.

below is my code. I am a ruby newbie, and I am not sure if this code is reinventing the wheel. Maybe there is a library method or methods that can do that already? or may be this code can be simplified?

RS = {
  :x => %w(\d+ a\d+ bb\d+ ccc\d+).map{|x| /^#{x}$/},
  :y => %w(\w+ 1\w+ 22\w+ 333\w+).map{|x| /^#{x}$/}
}.freeze

def find s, t
  r = RS[s]
  if r
    r.each do |p|
      m = p.match t
      return m if m
    end
    nil
  end
end

p find :x, 'bb12345'

Answer 1

If you want the result of the block you could do it this way. This will iterate over the whole array, but wont evaluate any matches after the first one.

def find(s,t)
  RS[s].inject(nil) {|m, p| m || p.match(t)}
end

You can break out early doing something like this

RS[s].inject(nil) {|m, p| (m && (break m)) || p.match(t)}

Answer 2

This is duplicated with: Ruby - Array.find, but return the value the block

You want a lazy map:

[nil, 1, 2, 3].lazy.map{|i| i && i.to_s}.find{|i| i}    
# => "1"