Ruby，带有元素数量的Array.select

Question

I have this array

[1, 2, 3, 4, 5, 6]

I would like to get the first 2 elements that are bigger than 3.

I can do:

elements = []

[1, 2, 3, 4, 5, 6].each do |element|
  elements << element if element > 3
  break if elements.size == 2
end

puts elements

Is there a more elegant way to do this?

Is there something in the Ruby core like Array.select(num_elements, &block) ?

Answer 1

a = [1, 2, 3, 4, 5, 6]

p a.filter {|x| x > 3}.first(2)

Or

p a.select{|x| x > 3}.first(2)

output

[4, 5]

As Cary suggest, the given below code wouldn't be a performance hit if array is bigger, it would stop executing further if 2 elements are found

a.lazy.select{|x| x > 3}.first(2)

Answer 2

You were nearly there. Just use break with a parameter:

[1, 2, 3, 4, 5, 6].each_with_object([]) do |element, acc|
  acc << element if element > 3
  break acc if acc.size >= 2
end

Another way to accomplish it, would be to use Enumerator::Lazy with array.lazy.select , or an explicit Enumerator instance with Enumerable#take (here it's a definite overkill, posting mostly for educational purposes.)

enum =
  Enumerator.new do |y|
    i = [1, 2, 3, 4, 5, 6].each
    loop { i.next.tap { |e| y << e if e > 3 } }
  end
enum.take(2)
#⇒ [4, 5]

Sidenote: both examples above would stop traversing the input as soon as two elements are found.

Answer 3

Just for having a couple of options more..

ary.each_with_object([]) { |e, res| res << e if e > 3 && res.size < 2 }

or

ary.partition { |e| e > 3 }.first.first(2)