Question

我是ruby的新手，我很难搞清楚如何将数组数组转换为数组哈希的哈希值。

例如，说我有：

[ [38, "s", "hum"], 
  [38, "t", "foo"], 
  [38, "t", "bar"], 
  [45, "s", "hum"], 
  [45, "t", "ram"], 
  [52, "s", "hum"], 
  [52, "t", "cat"], 
  [52, "t", "dog"]
]

我最终想要：

{38 => {"s" => ["hum"],
        "t" => ["foo", "bar"]
       },
 45 => {"s" => ["hum"],
        "t" => ["ram"]
       },
 52 => {"s" => ["hum"],
        "t" => ["cat", "dog"]
       }
 }

我已尝试过group_by和Hash，但两人都没有给我我想要的东西。

Answer 1

也许有一种更简洁的方法可以做到这一点，但我决定采用直截了当的方式：

input = [ [38, "s", "hum"],
  [38, "t", "foo"],
  [38, "t", "bar"],
  [45, "s", "hum"],
  [45, "t", "ram"],
  [52, "s", "hum"],
  [52, "t", "cat"],
  [52, "t", "dog"]
]

output = {}

# I'll talk through the first iteration in the comments.

input.each do |outer_key, inner_key, value|
  # Set output[38] to a new hash, since output[38] isn't set yet.
  # If it were already set, this line would do nothing, so
  # output[38] would keep its previous data.
  output[outer_key] ||= {}

  # Set output[38]["s"] to a new array, since output[38]["s"] isn't set yet.
  # If it were already set, this line would do nothing, so
  # output[38]["s"] would keep its previous data.
  output[outer_key][inner_key] ||= []

  # Add "hum" to the array at output[38]["s"].
  output[outer_key][inner_key] << value
end

所以，你实际使用的部分，全部整理了：

output = {}

input.each do |outer_key, inner_key, value|
  output[outer_key] ||= {}
  output[outer_key][inner_key] ||= []
  output[outer_key][inner_key] << value
end

Answer 2

在这种情况下，inject（1.9中的a.k.a. reduce）是一个很棒的工具：

input.inject({}) do |acc, (a, b, c)|
  acc[a] ||= {}
  acc[a][b] ||= []
  acc[a][b] << c
  acc
end

它将为input中的每个项目调用块一次，通过累加器和项目。它第一次将参数作为累加器传递，后续调用将最后一次调用的返回值作为累加器。

Answer 3

根据你的敏感度，这可能被视为可怕或优雅：

input.inject(Hash.new {|h1,k1| h1[k1] = Hash.new {|h2,k2| h2[k2] = Array.new}}) {|hash,elem| hash[elem[0]][elem[1]].push(elem[2]); hash}
=> {38=>{"s"=>["hum"], "t"=>["foo", "bar"]}, 45=>{"s"=>["hum"], "t"=>["ram"]}, 52=>{"s"=>["hum"], "t"=>["cat", "dog"]}}

理想情况下，更易读的版本是：

input.inject(Hash.new(Hash.new(Array.new))) {|hash,elem| hash[elem[0]][elem[1]].push(elem[2]); hash}

也就是说，从空哈希开始，默认值等于空哈希值，默认值等于空数组。然后迭代输入，将元素存储在适当的位置。

后一种语法的问题是Hash.new（Hash.new（Array.new））将导致所有散列和数组在内存中具有相同的位置，因此值将被覆盖。前一种语法每次都会创建一个新对象，从而得到所需的结果。

Answer 4

问题中给出的示例对于每个元素数组的长度为3，但下面的方法使用递归，并且可以用于任意长度。

a = [ [38, "s", "hum", 1], 
    [38, "t", "foo", 2],
    [38, "t", "bar", 3], 
    [45, "s", "hum", 1], 
    [45, "t", "ram", 1], 
    [52, "s", "hum", 3], 
    [52, "t", "cat", 3], 
    [52, "t", "dog", 2]
]

class Array
  def rep
    group_by{|k, _| k}.
    each_value{|v| v.map!{|_, *args| args}}.
    tap{|h| h.each{|k, v| h[k] = (v.first.length > 1 ? v.rep : v.flatten(1))}}
  end
end

p a.rep

具有重复值的哈希数组的哈希数组

4 个答案: