Arrays 在ruby中基于引用数组对结构进行排序

我有一个结构定义为

Player = Struct.new(:name, :level)

和一个样本数组

levels = [ 'level-1','level-2','beta','alpha','level4','level7']

我想对结构进行排序，使其成员按所述级别的递增顺序出现 乙二醇

---

你几乎成功了。sort_by block的结果应该是您想要比较的结果。您希望使用级别索引，而不是包含级别索引的数组

arr.sort_by { |e| levels.index(e.level) }

---

确保你的水平线是正确的。您可能希望使用常量来避免输入错误。

您已经差不多做到了。sort_by block的结果应该是您想要比较的结果。您希望使用级别索引，而不是包含级别索引的数组

arr.sort_by { |e| levels.index(e.level) }

---

确保你的水平线是正确的。您可能希望使用常量来避免输入错误。

这里有一个解决方案，它的计算复杂度接近于开，因为哈希查找的时间不是非常恒定的

我们得到：

Player = Struct.new(:name, :level)

arr = [["Tom", "level4"], ["Ryan", "beta"], ["Chris", "alpha"],
       ["Drew", "level4"]].map { |a| Player.new(*a) }
  #=> [#<struct Player name="Tom",   level="level4">,
  #    #<struct Player name="Ryan",  level="beta">,
  #    #<struct Player name="Chris", level="alpha">,
  #    #<struct Player name="Drew",  level="level4">]

然后

步骤如下：

h = arr.group_by(&:level)
  #=> {"level4"=>[#<struct Player name="Tom",   level="level4">,
  #               #<struct Player name="Drew",  level="level4">],
  #    "beta"  =>[#<struct Player name="Ryan",  level="beta">],
  #    "alpha" =>[#<struct Player name="Chris", level="alpha">]}

---

这是一个计算复杂度接近于On的解决方案，因为哈希查找不是固定时间

我们得到：

Player = Struct.new(:name, :level)

arr = [["Tom", "level4"], ["Ryan", "beta"], ["Chris", "alpha"],
       ["Drew", "level4"]].map { |a| Player.new(*a) }
  #=> [#<struct Player name="Tom",   level="level4">,
  #    #<struct Player name="Ryan",  level="beta">,
  #    #<struct Player name="Chris", level="alpha">,
  #    #<struct Player name="Drew",  level="level4">]

然后

步骤如下：

h = arr.group_by(&:level)
  #=> {"level4"=>[#<struct Player name="Tom",   level="level4">,
  #               #<struct Player name="Drew",  level="level4">],
  #    "beta"  =>[#<struct Player name="Ryan",  level="beta">],
  #    "alpha" =>[#<struct Player name="Chris", level="alpha">]}

---

我只想指出，您可以通过对levels数组进行预索引来优化时间复杂度，例如，将其转换为{=>}散列，这样您就不必调用。index repeatedly为@maxpleaner的建议添加一些内容：对于arr=[[Tom，level4]，[Ryan，beta]，[Chris，alpha]]。map{a | Player.new*a}=>[，]，构建level_to_idx=levels.each_与_index.to_a.to_h=>{level-1=>0，level-2=>1，beta=>2，alpha=>3，level-4=>4，level-7=>5}，…@CarySwoveland我以前在采访中提出过这个问题，我总是说它将从^2开始，但我突然想问这是真的吗？sort_by的时间复杂度为ON？…然后arr.sort_by{i | level_to_idx[i.level]}=>[，]。这将计算复杂度从**2降低到了Onlogn。@maxpleaner，你在说什么？我说的是Onlogn。坦白：我在一篇被删除的评论中说到。我只想指出，通过对levels数组进行预索引，可以优化这里的时间复杂度，例如，将其转换为{=>}散列，这样您就不必调用。重复索引可以让@maxpleaner的建议更生动：对于arr=[[Tom，level4]，[Ryan，beta]，[Chris，alpha]]。map{| a | Player.new*a}=>[，]，构造级别u to_idx=levels.each_与_index.to_a.to_h=>{level-1=>0，level-2=>1，beta=>2，alpha=>3，level4=>4，level7=>5}，…@CarySwoveland我以前在采访中提出过这个问题，我总是说这是从^2到^2，但我突然想问，这是否真的是真的？排序的时间复杂度是吗？…然后arr.sort_by{I | level|u to_idx[I.level]=>[，]。这将计算复杂度从**2降低到Onlogn。@maxpleaner，你在说什么？我说的是Onlogn。坦白：我在删除的评论中说了On。

    arr.group_by(&:level).values_at(*levels).flatten.compact
  #=>[#<struct Player name="Ryan",  level="beta">,
  #   #<struct Player name="Chris", level="alpha">,
  #   #<struct Player name="Tom",   level="level4">,
  #   #<struct Player name="Drew",  level="level4">]

h = arr.group_by(&:level)
  #=> {"level4"=>[#<struct Player name="Tom",   level="level4">,
  #               #<struct Player name="Drew",  level="level4">],
  #    "beta"  =>[#<struct Player name="Ryan",  level="beta">],
  #    "alpha" =>[#<struct Player name="Chris", level="alpha">]}

a = h.values_at(*levels)
  #=> [nil,
  #    nil,
  #    [#<struct Player name="Ryan",  level="beta">],
  #    [#<struct Player name="Chris", level="alpha">],
  #    [#<struct Player name="Tom",   level="level4">,
  #     #<struct Player name="Drew",  level="level4">],
  #    nil]

b = a.compact
  #=>  [[#<struct Player name="Ryan",  level="beta">],
  #     [#<struct Player name="Chris", level="alpha">],
  #     [#<struct Player name="Tom",   level="level4">,
  #      #<struct Player name="Drew",  level="level4">]]

c = b.flatten
  #=>  [#<struct Player name="Ryan",  level="beta">,
  #     #<struct Player name="Chris", level="alpha">,
  #     #<struct Player name="Tom",   level="level4">,
  #     #<struct Player name="Drew",  level="level4">]