Question

我有以下列表：

hash_list = 
{ "a"=>{"unit_id"=>"43", "dep_id"=>"153","_destroy"=>"false"},
  "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"},
  "c"=>{"unit_id"=>"43", "dep_id"=>"154", "_destroy"=>"false"},
  "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"}
}

我需要结果为：

{
 "a"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false"},
 "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"},
 "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"}
}

基本上，unit_id不应该重复。但是，所有_destroy=="1"条目都可以显示在列表中。

请帮忙。

Answer 1

试试这个：

> hash_list.to_a.uniq{|_,v|v["unit_id"] unless v["_destroy"] == "1"}.to_h

#=> {
      "a"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false"},
      "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"},
      "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"}
    }

这将检查unit_id为uniq，并且还会显示"_destory" == "1"个条目。

Answer 2

此代码：

keepers = hash_list.select { |_k, v| v["_destroy"] == "1" }.to_h
new_hash_list = hash_list.to_a.uniq { |_k, v| v["unit_id"] }.to_h
new_hash_list.merge!(keepers)

针对此数据运行时：

hash_list = { 
  "a"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false"},
  "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"},
  "c"=>{"unit_id"=>"43", "dep_id"=>"154", "_destroy"=>"false"},
  "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"},
  "e"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"1"},
}

产生这个结果：

{
  "a"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false"},
  "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"},
  "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"},
  "e"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"1"},
}

Answer 3

另一种方式：

new_hash_list =
  { "a"=>{ "unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false" },
    "b"=>{ "unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"     },
    "c"=>{ "unit_id"=>"43", "dep_id"=>"154", "_destroy"=>"false" },
    "d"=>{ "unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false" },
    "e"=>{ "unit_id"=>"43", "dep_id"=>"cat", "_destroy"=>"1"     }}

require 'set'

s = Set.new
new_hash_list.each_with_object({}) do |(k,v),h|
  (h[k] = v) if v["_destroy"] == "1" || s.add?(v["unit_id"])
end
  #=> {"a"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false"},
  #    "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"    },
  #    "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"},
  #    "e"=>{"unit_id"=>"43", "dep_id"=>"cat", "_destroy"=>"1"    }}

如何根据多个值删除哈希值

3 个答案: