Ruby - 如何用数组值反转哈希？

Question

寻找适用于 Ruby 1.8.7 的答案:

例如，假设我有这样的哈希:

{"Book Y"=>["author B", "author C"], "Book X"=>["author A", "author B", "author C"]}

我想得到这个:

{ 
    "author A" => ["Book X"],
    "author B" => ["Book Y", "Book X"],
    "author C" => ["Book Y", "Book X"] 
}

我为它写了一个很长的方法，但是对于大型数据集，它非常慢。

有什么优雅的解决方案吗？

Answer 1

这是一种方式:

g = {"Book Y"=>["author B", "author C"],
     "Book X"=>["author A", "author B", "author C"]}

g.each_with_object({}) do |(book,authors),h|
  authors.each { |author| (h[author] ||= []) << book }
end
  #=> {"author B"=>["Book Y", "Book X"],
  #    "author C"=>["Book Y", "Book X"],
  #    "author A"=>["Book X"]} 

步骤:

enum = g.each_with_object({})
  #=> #<Enumerator: {"Book Y"=>["author B", "author C"],
  #   "Book X"=>["author A", "author B", "author C"]}:each_with_object({})> 

我们可以看到 enum 的元素，它将通过将其转换为数组来传递到 block 中:

enum.to_a
  #=> [[["Book Y", ["author B", "author C"]], {}],
  #    [["Book X", ["author A", "author B", "author C"]], {}]]

传递给 block 并分配给 block 变量的枚举的第一个元素是:

(book,authors),h = enum.next
  #=> [["Book Y", ["author B", "author C"]], {}] 
book
  #=> "Book Y" 
authors
  #=> ["author B", "author C"] 
h
  #=> {} 

enum1 = authors.each
  #=> #<Enumerator: ["author B", "author C"]:each>
author = enum1.next
  #=> "author B"
(h[author] ||= []) << book
  #=> (h["author B"] ||= []) << "Book Y"
  #=> (h["author B"] = h["author B"] || []) << "Book Y"
  #=> (h["author B"] = nil || []) << "Book Y"
  #=> h["author B"] = ["Book Y"]
  #=> ["Book Y"]
h #=> {"author B"=>["Book Y"]} 

下一步:

author = enum1.next
  #=> "author C" 
(h[author] ||= []) << book
h #=> {"author B"=>["Book Y", "Book Y"], "author C"=>["Book Y"]} 

读完“Book X”，

(book,authors),h = enum.next
  #=> [["Book X", ["author A", "author B", "author C"]],
  #    {"author B"=>["Book Y", "Book Y"], "author C"=>["Book Y"]}]
book
  #=> "Book X" 
authors
  #=> ["author A", "author B", "author C"] 
h
  #=> {"author B"=>["Book Y", "Book Y"], "author C"=>["Book Y"]} 

我们现在重复与 “Book X” 相同的计算。唯一的区别是当我们遇到:

(h[author] ||= []) << book

相当于

(h[author] = h[author] || []) << book

在大多数情况下，等号右边的h[author]不会是nil(例如，它可能是["Book X"] ，在这种情况下，上述表达式将简化为:

h[author] << book

附录

对于战前的 Ruby 版本(例如 1.8.7)，只需先初始化散列并使用 each 而不是 each_with_object (我们在 1.9 中得到后者. 对于 1.8.7 我太年轻了，但我经常想知道没有它人们如何相处。)你只需要记住在最后返回 h ，作为 each只是返回它的接收者。

所以改成:

h = {}
g.each do |book,authors|
  authors.each { |author| (h[author] ||= []) << book }
end
h
  #=> {"author B"=>["Book Y", "Book X"],
  #    "author C"=>["Book Y", "Book X"],
  #    "author A"=>["Book X"]}