array.select最多n个元素

Question

Consider a Ruby array with a lot of elements.

I want to reverse iterate over the array and select the first n results for which a given block returns true.

Eg

seniors = persons.reverse.select do |person|
  person.age > 60
end

This is okay, however it collects all seniors, instead of n seniors at most. What's the nicest way to prematurely terminate the iteration when n results has been collected?

Answer 1

seniors = []
persons.reverse.each{ |p| 
  seniors << p if p.age > 60
  break if seniors.count >= n
}

Answer 2

Lazy evaluation is great because it makes possible to write this kind of code without changing the abstractions you would normally use in strict evaluation (Haskell has proved how powerful this is). Happily Ruby 2.0 will ship with some infrastructure on this regard, when that time comes you'll be able to write this:

# Ruby >= 2.0
seniors = persons.reverse.lazy.select { |person| person.age > 60 }.take(n)

For Ruby < 2.0: https://github.com/yhara/enumerable-lazy

Answer 3

You can chain take_while and each_with_object :

seniors = []
persons.take_while.each_with_object(seniors) do |e, o| 
  o << e if e.age > 60 && o.count < 1000
  o.count < 1000
end

require 'ostruct'
require 'benchmark'

persons = 1000000.times.map {|i| OpenStruct.new age: Random.rand(50..85) }

Benchmark.bm do |b|
  b.report do
    seniors = []
    persons.take_while.each_with_object(seniors) do |e, o| 
      o << e if e.age > 60 && o.count < 1000
      o.count < 1000
    end
  end
end

#=> finishes in 1-3ms on my machine