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)?
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.
each without a block calls to_enum under the hood.
stop traversing the input as soon as two elements are found is the key here.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)
a.lazy.select{|x| x > 3}.first(2) to bail out as soon as the first two have been found.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)
[1, 1, *1..Float::INFINITY].
partition. I've not been able to find how to break when res.size < 2, and break if was returning nil. But I'm learning it right now from your answer: { |e, res| res << e if e > 3; break res if res.size == 2 } :).partition, there is a simple way to ensure: (1..10).partition { |e| puts e; e > 3 }.first.first(2). To escape sooner you need a lazy enumerator.
arr.select { |z| z > 3 }[2]?