紅寶石枚舉：先拿n，其中塊返回true

Question

我要採取的第一個「N」項該塊

a = 1..100_000_000 # Basically a long array 

# This iterates over the whole array -- no good 
b = a.select{|x| x.expensive_operation?}.take(n) 

我想傳遞給短路迭代一次我有n個條目，其中「昂貴「的條件是真的。

你有什麼建議？ take_while並保持n的數量？

# This is the code i have; which i think can be written better, but how? 
a = 1..100_000_000 # Basically a long array 
n = 20 
i = 0 
b = a.take_while do |x| 
    ((i < n) && (x.expensive_operation?)).tap do |r| 
    i += 1 
    end 
end 

Answer 1

紅寶石2.0工具lazy enumerables，舊版本使用的寶石enumerable-lazy：

require 'enumerable/lazy' 
(1..Float::INFINITY).lazy.select(&:even?).take(5).to_a 
#=> [2, 4, 6, 8, 10] 

Answer 2

應該用一個簡單的for循環的工作方式和break：

a = 1..100_000_000 # Basically a long array 
n = 20 
selected = [] 
for x in a 
    selected << x if x.expensive_operation? 
    break if select.length == n 
end