Question

我意识到我的沟通技巧并不那么出色（我甚至在尝试查找时遇到问题），所以我让代码发言：

easy_as_123 = ("a".."c").to_a
10.times do |j|
  if j >= easy_as_123.length
    puts "j is #{j}, letter is #{easy_as_123[j % easy_as_123.length]}"
  else 
    puts "j is #{j}, letter is #{easy_as_123[j]}"
  end
end

是否有一个更优雅，更简洁的解决方案，可以不断迭代我的[a，b，c]数组？

编辑，我发布的代码是作品，这正是我正在寻找的结果，但它不简洁也不好看，是否有方法能够以更优雅的方式实现同样的结果？

Answer 1

Enumerable#cycle是你的朋友。无限循环：

(?a..?c).cycle.with_index do |letter, j|
  puts "j is #{j}, letter is #{letter}"
end

在10次迭代后打破一个循环：

(?a..?c).cycle.with_index do |letter, j|
  break if j >= 10
  puts "j is #{j}, letter is #{letter}"
end

使用#take（@Stefan的信用）：

(?a..?c).cycle.take(10).each.with_index do |letter, j|
  puts "j is #{j}, letter is #{letter}"
end

反过来说：

enum = (?a..?c).cycle
10.times do |j|
  puts "j is #{j}, letter is #{enum.next}"
end

没有#cycle：

enum = (?a..?c).to_a
10.times do |j|
  puts "j is #{j}, letter is #{enum[j % enum.length]}"
end

评论：

if是多余的;
to_a是多余的，可能会枚举范围。

Answer 2

在Enumerator的新实例中定义无限循环：

enum = Enumerator.new do |y|
  j = 0
  arr = ['a','b','c'].cycle
  loop do
    y << "j is #{j}, letter is #{arr.next}"
    j += 1
  end
end

enum.next #=> "j is 0, letter is a"
enum.next #=> "j is 1, letter is b"
enum.next #=> "j is 2, letter is c"
enum.next #=> "j is 3, letter is a"

想回到开头？使用Enumerator#rewind：

enum.rewind
enum.next #=> "j is 0, letter is a"

使用take：

输出前十次迭代

puts enum.take 10 #=> 
# j is 0, letter is a
# j is 1, letter is b
# j is 2, letter is c
# j is 3, letter is a
# j is 4, letter is b
# j is 5, letter is c
# j is 6, letter is a
# j is 7, letter is b
# j is 8, letter is c
# j is 9, letter is a

或使用times和next：

10.times { p enum.next } #=>
# "j is 0, letter is a"
# "j is 1, letter is b"
# "j is 2, letter is c"
# "j is 3, letter is a"
# "j is 4, letter is b"
# "j is 5, letter is c"
# "j is 6, letter is a"
# "j is 7, letter is b"
# "j is 8, letter is c"
# "j is 9, letter is a"

继续遍历数组

2 个答案: