像这样的东西怎么样？请看下面的评论：

arr1 = [{ 'nb' => '5df54g54df', 'active' => true, 'brand' => 'aisle' }, { 'nb' => '5jghfj264', 'active' => false, 'brand' => 'leg' },
        { 'nb' => '5qwercv546', 'active' => true, 'brand' => 'gem' }]
arr2 = [{ 'nb' => '5df54g54df', 'active' => true, 'brand' => 'aisle' },
        { 'nb' => '5jghfj264', 'active' => false, 'brand' => 'leg' }]

p(
  arr1.select do |h| # Select all objects from arr1, that return true in block:
    arr2.any? { |h2| h['nb'] == h2['nb'] } #Any element from arr2 that contains equal `'nb'` keys.
  end
)

#=> [{"nb"=>"5df54g54df", "active"=>true, "brand"=>"aisle"}, {"nb"=>"5jghfj264", "active"=>false, "brand"=>"leg"}]

require 'set'

arr2_nb_vals = arr2.each_with_object(Set.new) do |h,arr2_nb_vals|
  arr2_nb_vals << h["nb"]
end
  #=> #<Set: {"5df54g54df", "5jghfj264"}>

arr1.select { |h| arr2_nb_vals.include?(h["nb"]) }
  #=> [{"nb"=>"5df54g54df", "active"=>true, "brand"=>"aisle"},
  #    {"nb"=>"5jghfj264", "active"=>false, "brand"=>"leg"}]

通过首先将h["nb"]、arr2中的h的值收集到一个集合中，arr2_nb_vals.include?(h["nb"])的计算非常快，几乎与arr2_nb_vals.size无关。
这使得当用arr2.include?(h["nb"])替换arr2_nb_vals.include?(h["nb"])时，与O(arr1.size**2)相比，计算复杂度接近O(arr1.size + arr2.size)。