Ruby：尊重最高版本的唯一哈希陣列

我正在嘗試創建一個具有唯一值的新哈希陣列，并尊重重復哈希的最高版本。哈希如下所示：

old_hash = [
{"dependency"=>"websocket", "version"=>"2.8.0", "repo"=>"repo1"},
{"dependency"=>"rails", "version"=>"6.2.0", "repo"=>"repo2"},
{"dependency"=>"httparty", "version"=>"6.0.3.5", "repo"=>"repo2"},
{"dependency"=>"httparty", "version"=>"6.1.0.2", "repo"=>"repo2"},
{"dependency"=>"httparty", "version"=>"6.1.3.2", "repo"=>"repo2"},
{"dependency"=>"rails", "version"=>"6.1.0", "repo"=>"repo3"},
{"dependency"=>"metasploit", "version"=>"2.8.0", "repo"=>"repo3"}
]

如您所見，第三個、第四個和第五個散列具有相同的鍵值dependencywhich ishttparty和repowhich is repo2，但第五個散列具有這三個中的最高版本。因此，我想創建一個具有第一個、第二個、第五個、第六個和第七個哈希值的唯一哈希值。所以我想要的結果應該是這樣的：

unique_hash = [
{"dependency"=>"websocket", "version"=>"2.8.0", "repo"=>"repo1"},
{"dependency"=>"rails", "version"=>"6.2.0", "repo"=>"repo2"},
{"dependency"=>"httparty", "version"=>"6.1.3.2", "repo"=>"repo2"},
{"dependency"=>"rails", "version"=>"6.1.0", "repo"=>"repo3"},
{"dependency"=>"metasploit", "version"=>"2.8.0", "repo"=>"repo3"}
]

關于版本比較，我正在考慮使用這種方法來比較它們：

def version_greater? (version1, version2)
  Gem::Version.new(version1) > Gem::Version.new(version2)
end

true在 version1 大于 version2 的情況下回傳。

我將不勝感激任何有助于解決此問題的建議。

uj5u.com熱心網友回復：

一種方法是使用Hash#update（又名merge!）的形式，它采用一個塊（此處{ |_,o,n| n["version"] > o["version"] ? n : o }）來確定存在于被合并的兩個散列中的鍵的值。

old_hash = [
  {"dependency"=>"websocket",  "version"=>"2.8.0",   "repo"=>"repo1"},
  {"dependency"=>"rails",      "version"=>"6.2.0",   "repo"=>"repo2"},
  {"dependency"=>"httparty",   "version"=>"6.0.3.5", "repo"=>"repo2"},
  {"dependency"=>"httparty",   "version"=>"6.1.0.2", "repo"=>"repo2"},
  {"dependency"=>"httparty",   "version"=>"6.1.3.2", "repo"=>"repo2"},
  {"dependency"=>"rails",      "version"=>"6.1.0",   "repo"=>"repo3"},
  {"dependency"=>"metasploit", "version"=>"2.8.0",   "repo"=>"repo3"},
  {"dependency"=>"rails",      "version"=>"6.1.9",   "repo"=>"repo2"}
]

請注意，我在問題中添加了一個哈希值old_hash。（順便說一句，“old_hash”可能不是陣列的最佳名稱。）

old_hash.each_with_object({}) do |g,h|
  h.update([g["dependency"],g["repo"]]=>g) do |_,o,n|
    n["version"] > o["version"] ? n : o
  end
end.values
  #=> [{"dependency"=>"websocket",  "version"=>"2.8.0",   "repo"=>"repo1"},
  #    {"dependency"=>"rails",      "version"=>"6.2.0",   "repo"=>"repo2"},
  #    {"dependency"=>"httparty",   "version"=>"6.1.3.2", "repo"=>"repo2"},
  #    {"dependency"=>"rails",      "version"=>"6.1.0",   "repo"=>"repo3"},  
  #    {"dependency"=>"metasploit", "version"=>"2.8.0",   "repo"=>"repo3"}]

的接收者values可以看出如下。

  {["websocket", "repo1"] =>{"dependency"=>"websocket",  "version"=>  "2.8.0", "repo"=>"repo1"},
   ["rails", "repo2"]     =>{"dependency"=>"rails",      "version"=>  "6.2.0", "repo"=>"repo2"},
   ["httparty", "repo2"]  =>{"dependency"=>"httparty",   "version"=>"6.1.3.2", "repo"=>"repo2"},
   ["rails", "repo3"]     =>{"dependency"=>"rails",      "version"=>  "6.1.0", "repo"=>"repo3"},
   ["metasploit", "repo3"]=>{"dependency"=>"metasploit", "version"=>  "2.8.0", "repo"=>"repo3"}}

有關三個塊變數的描述，請參閱檔案：（_公共鍵，這里是一個下劃線，表示它未用于塊計算），o正在構造的散列中的公共鍵的值（想想“舊”） , 和n，正在合并的散列中的公共鍵的值（想想“新”）。

uj5u.com熱心網友回復：

問題是通過使用解決的：

old_hash.group_by {|h| h.values_at("dependency","repo")}.map {|_,v| v.max_by {|h| Gem::Version.new(h["version"])}}

感謝@engineersmnky。

標籤：红宝石 散列

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