如何在 Ruby 中迭代范围数组？

发布于 2024-10-17 06:29:11 字数 410 浏览 2 评论 0原文

如果您有一系列范围，例如 [1..4, 7..11, 14..18, 21..25, 28..28]，我有哪些选项迭代元素？

我可以

ranges.each do |range|
  range.each do |date|
    puts "Do work on February #{date}"
  end
end

这样做有点冗长，或者

dates = ranges.map(&:to_a).flatten
dates.each do |date|
  puts "Do work on February #{date}"
end

如果范围很大，我可以这样做可能会使用大量内存。

还有其他选择吗？

原文

If you have an array of ranges, such as [1..4, 7..11, 14..18, 21..25, 28..28], what options do I have for iterating through the elements?

I could do

ranges.each do |range|
  range.each do |date|
    puts "Do work on February #{date}"
  end
end

which is a bit verbose, or I could do

dates = ranges.map(&:to_a).flatten
dates.each do |date|
  puts "Do work on February #{date}"
end

which could use a lot of memory if the ranges are large.

Are there any alternatives?

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君勿笑 2024-10-24 06:29:11

好吧，我不认为你的第一个答案太冗长，但如果这种模式使用得足够频繁，它可能会证明这样的情况 -

module Enumerable
  def each_node
    each do |x|
      (x.respond_to?(:each_node)) ? x.each_node{ |y| yield(y) } : yield(x)
    end
  end
end

[[[(1..5)], (1..2)],1].each_node { |x| print x }  #=> 12345121

ranges = [1..4, 7..11, 14..18, 21..25, 28..28]
ranges.each_node{ |date| puts "Do work on February #{date}" } #=>as expected

Well, I don't think your first answer is too verbose, but if that pattern is getting used often enough, it might make the case for something like this -

module Enumerable
  def each_node
    each do |x|
      (x.respond_to?(:each_node)) ? x.each_node{ |y| yield(y) } : yield(x)
    end
  end
end

[[[(1..5)], (1..2)],1].each_node { |x| print x }  #=> 12345121

ranges = [1..4, 7..11, 14..18, 21..25, 28..28]
ranges.each_node{ |date| puts "Do work on February #{date}" } #=>as expected

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