Question

我想获取通过块

的第一个“n”条目

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)

我希望在有'n'条件成真的条目后，将迭代短路。

你有什么建议？ take_while并记住n？

# 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

Ruby 2.0实现lazy enumerables，旧版本使用gem enumerable-lazy：

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

Answer 2

它应该使用简单的for循环和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

Ruby Enumeration：首先取n，其中block返回true

2 个答案: