Ruby

Question

正在寻找适用于 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块中：

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

传递给块并分配给块变量的enum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"]}

下一个：

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

完成 "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。唯一不同的是当我们遇到：

(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，上面的表达式简化为：

h[author] << book

附录

对于 war 之前的 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（我们得到后者是 1.9。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。）你只需要记住最后 return h，因为 each 只是 return 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"]}

Answer 2

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

p h.inject(Hash.new([])) { |memo,(key,values)|
  values.each { |value| memo[value] += [key] }
  memo
}
# => {"author B"=>["Book Y", "Book X"], "author C"=>["Book Y", "Book X"], "author A"=>["Book X"]}

Answer 3

我会在 Ruby 1.8:

中做这样的事情

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

library = Hash.new { |h, k| h[k] = [] }

hash.each do |book, authors|
  authors.each { |author| library[author] << book }
end

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

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