ruby 将数组转换为哈希保留重复键

Question

我需要将 git ls-remote 的结果下拉到一个数组中，然后将该数组转换为这样的哈希：{commit_hash => reference}。有时，两个提交哈希是相同的（但可能有不同的引用）。所以我得到了这种东西：

["19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/auto",
 "8f6f47c6e8023540b022586e368c68e1e814ce6d","refs/heads/callout_hooks",  
 "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8","refs/heads/elab",
 "d38a9a26ef887c08b306bdab210b39882f58e587","refs/heads/elab_6.1",
 "19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/master",
 "906dfe6eebff832baf0f92683d751432fcc98ab7","refs/heads/regression"]

我想转换成：

{"19d97e408ee3f993745b053e281ac9dc69519e06" => "refs/heads/auto"...}

但是 master 和 auto 具有相同的哈希值，因此其中一个会在转换中被丢弃。

如何 1.) 获取转换中删除的值的列表，或 2.) 通过向键添加特殊字符（如 *）使键唯一？

Answer 1

你为你想要做的事情提供了两个选项：

• 获取转换中删除的值的列表
• 通过向键添加特殊字符使键唯一

我认为第二种方法是一个坏主意，原因有两个：a）您必须有一种修改密钥的方法，以允许它们有多个重复的可能性； b) 在原件和复制件之间建立联系会很尴尬。此外，它会很丑陋。

我看到其他人提出了第三种可能性：更改结果哈希的形式，以便为字符串数组赋值。这可能对你有好处，但这不是你所要求的，所以我选择建立一个被删除的值的列表；即，除了第一个。

代码

def create_hash_and_save_extras(arr)
  arr.each_slice(2).with_object([{},[]]) { |(k,v),(h,ex)|
    h.update({k=>v}) { |k, ov, nv| ex << {k=>nv}; ov } }
end

示例

create_hash_and_save_extras(arr)
  #=> [{"19d97e408ee3f993745b053e281ac9dc69519e06"=>"refs/heads/auto",
  #     "8f6f47c6e8023540b022586e368c68e1e814ce6d"=>"refs/heads/callout_hooks",
  #     "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8"=>"refs/heads/elab",
  #     "d38a9a26ef887c08b306bdab210b39882f58e587"=>"refs/heads/elab_6.1",
  #     "906dfe6eebff832baf0f92683d751432fcc98ab7"=>"refs/heads/regression"},
  #   [{"19d97e408ee3f993745b053e281ac9dc69519e06"=>"refs/heads/master"}]]

说明

Enumerable#each_slice 发送到arr 返回一个枚举器：

enum1 = arr.each_slice(2)
  #=> #<Enumerator: [
  #      "19d97e408ee3f993745b053e281ac9dc69519e06", "refs/heads/auto",
  #      "8f6f47c6e8023540b022586e368c68e1e814ce6d", "refs/heads/callout_hooks",
  #      ...
  #      "906dfe6eebff832baf0f92683d751432fcc98ab7", "refs/heads/regression"
  #   ]:each_slice(2)>

Enumerator#with_object 创建一个由初始空散列（由块变量 h 表示）和一个由块变量 ex 表示的初始空数组（用于“附加”）组成的数组，其中然后发送到enum1 以创建另一个枚举器（您可以将其视为“复合枚举器”——注意下面对each_slice(2)>:with_object({}) 的引用）。

enum2 = enum1.with_object([{},[]])
  #=> #<Enumerator: #<Enumerator: [
  #      "19d97e408ee3f993745b053e281ac9dc69519e06", "refs/heads/auto",
  #      "8f6f47c6e8023540b022586e368c68e1e814ce6d", "refs/heads/callout_hooks",
  #      ...
  #      "906dfe6eebff832baf0f92683d751432fcc98ab7", "refs/heads/regression"
  #   ]:each_slice(2)>:with_object([{},[])>

我们可以将enum2 转换为一个数组，看看它将传递给它的块：

enum2.to_a
#=> [[["19d97e408ee3f993745b053e281ac9dc69519e06", "refs/heads/auto"],
#       [{}, []]],
#    [["8f6f47c6e8023540b022586e368c68e1e814ce6d", "refs/heads/callout_hooks"],
#       [{}, []]],
#    [["3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8", "refs/heads/elab"],
#       [{}, []]],
#    [["d38a9a26ef887c08b306bdab210b39882f58e587", "refs/heads/elab_6.1"],
#       [{}, []]],
#    [["19d97e408ee3f993745b053e281ac9dc69519e06", "refs/heads/master"],
#       [{}, []]],
#    [["906dfe6eebff832baf0f92683d751432fcc98ab7", "refs/heads/regression"],
#       [{}, []]],

enum2 传入其块的第一个元素是

[["19d97e408ee3f993745b053e281ac9dc69519e06", "refs/heads/auto"], [{}, []]]]]

因此，块变量分配如下：

k => "19d97e408ee3f993745b053e281ac9dc69519e06"
v => "refs/heads/auto"
h => {}
ex = []

我们现在使用Hash#update（又名Hash#merge!）将{k,v}合并到h（h最初为空。）因此

h.update({k=>v}) { |k, ov, nv| extras << {k=>nv}; ov }

变成

h.update({"19d97e408ee3f993745b053e281ac9dc69519e06"=>"refs/heads/auto"})

紧随其后

{ |k, ov, nv| ex << {k=>nv}; ov }

但该块仅适用于哈希合并哈希 (h) 和正在合并的哈希 (update 的参数) 共享相同的密钥 k，在这种情况下 ov 和 nv 是分别与 h 的这些键关联的值和要合并的哈希值。键 k 的合并值将是块返回的值。是的，这将在我们遇到重复项时适用。

那么现在

h #=> {"19d97e408ee3f993745b053e281ac9dc69519e06"=>"refs/heads/auto"}

我们继续以这种方式处理enum2 的其他每个元素。当我们遇到

k = "19d97e408ee3f993745b053e281ac9dc69519e06"
v = "refs/heads/master"
h = {"19d97e408ee3f993745b053e281ac9dc69519e06"=>"refs/heads/auto",
      "8f6f47c6e8023540b022586e368c68e1e814ce6d"=>"refs/heads/callout_hooks",
      "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8"=>"refs/heads/elab",
      "d38a9a26ef887c08b306bdab210b39882f58e587"=>"refs/heads/elab_6.1"}

我们发现k已经在合并哈希h中，因此对该块进行评估以确定合并哈希h中k的值。我们希望保留当前值h[k]，即ov，这就是块返回的值。然而，首先，我们在（仍然为空的）数组ex 中附加重复值，以哈希表示。

ex << {"19d97e408ee3f993745b053e281ac9dc69519e06" => "refs/heads/master"}

Answer 2

我希望你会喜欢这个：

ary = [
       "19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/auto",
       "8f6f47c6e8023540b022586e368c68e1e814ce6d","refs/heads/callout_hooks",  
       "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8","refs/heads/elab",
       "d38a9a26ef887c08b306bdab210b39882f58e587","refs/heads/elab_6.1",
       "19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/master",
       "906dfe6eebff832baf0f92683d751432fcc98ab7","refs/heads/regression"
     ]

array_hash = ary.each_slice(2).with_object(Hash.new { |h,k| h[k] = []}) do |(k,v),hash|
  hash[k] << v 
end

# the main advantage is here you wouldn't loose any data, all are with you. You can
# use it as per your need. I think it is a better approach to deal with your situation.
array_hash
# => {"19d97e408ee3f993745b053e281ac9dc69519e06"=>
#      ["refs/heads/auto", "refs/heads/master"],
#     "8f6f47c6e8023540b022586e368c68e1e814ce6d"=>["refs/heads/callout_hooks"],
#     "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8"=>["refs/heads/elab"],
#     "d38a9a26ef887c08b306bdab210b39882f58e587"=>["refs/heads/elab_6.1"],
#     "906dfe6eebff832baf0f92683d751432fcc98ab7"=>["refs/heads/regression"]}

Answer 3

如果你对 hash_value => refs 数组进行散列，你将保留所有内容：

array = ["19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/auto",
 "8f6f47c6e8023540b022586e368c68e1e814ce6d","refs/heads/callout_hooks",  
 "3cbdb4b2fcb85bc7f0ed08b62e2bf2445a7659e8","refs/heads/elab",
 "d38a9a26ef887c08b306bdab210b39882f58e587","refs/heads/elab_6.1",
 "19d97e408ee3f993745b053e281ac9dc69519e06","refs/heads/master",
 "906dfe6eebff832baf0f92683d751432fcc98ab7","refs/heads/regression"
]

array.each_slice(2).reduce({}) { |h, (k, v)| (h[k] ||= []) << v; h }

看起来奥雅纳和我的想法是一样的......