通过嵌套数组对计算公共元素的最简洁方法

Question

我有两个嵌套数组：

one = [
  ["Hiking", "fishing", "photography"], 
  ["The Avengers", "The Dark Knight", "Lord of the Rings"], 
  ["Firefly", "Battlestar Galactica", "The Expanse"], 
  ["The Hobbit", "1984", "Dune", "Ender's Game"]
]
two = [
  ["Hiking", "photography"], 
  ["Whiplash", "Pulp Ficiton", "The Avengers"], 
  ["Firefly", "Battlestar Galactica", "The Expanse"], 
  ["The Hobbit", "1984", "Dune", "Ender's Game"]
]

我想我可以逐个迭代和比较，但是有没有更好的方法呢？

Answer 1

可能的解决方案(假设您希望通过数组对查找公共元素)：

one.zip(two).flat_map { |f, s| f & s }.count
#=> 10

Answer 2

我会这样做：

(one.flatten & two.flatten).size
#=> 10

Answer 3

您可以考虑以下几点，以避免创建临时数组one.zip(two)。

one.size.times.reduce(0) { |t,i| t + (one[i] & two[i]).size }
  #=> 10