如果您有哈希哈希如何删除First Hash的值 - Ruby

Question

来自以下的最简单方法是什么：

{"Wednesday"=>{"Horse"=>{"9.0"=>1, "10.0"=>1}}, 
"Thursday"=>{"Horse"=>{"9.0"=>1, "10.0"=>1}}}

.

？

{"Wednesday"=>{"9.0"=>1, "10.0"=>1}, 
"Thursday"=>{"9.0"=>1, "10.0"=>1}}

.

我已经在我的控制台中挣扎了2个小时。

将欣赏任何答案！

p.s。这匹马是一种变化，如：法国人，Belgiansalion，Lipicanec ...

Answer 1

a functional 样式解决方案（无需修改原始哈希或使用额外变量 - 为我看起来优雅的解决方案）

hash.reduce({}) { |acc, (k, v)| acc.merge(Hash[k, *v.values]) }
# => {"Wednesday"=>{"9.0"=>1, "10.0"=>1}, "Thursday"=>{"9.0"=>1, "10.0"=>1}}

.

Answer 2

怎么样。

hash = {"Wednesday"=>{"Horse"=>{"9.0"=>1, "10.0"=>1}}, 
"Thursday"=>{"Horse"=>{"9.0"=>1, "10.0"=>1}}}
a ={}
hash.each do |k,v|
  a[k]=v.values.first
end

.

Answer 3

这样做

  a = {"Wednesday"=>{"Horse"=>{"9.0"=>1, "10.0"=>1}}, 
       "Thursday"=>{"Horse"=>{"9.0"=>1, "10.0"=>1}}}
  d = {} 
  a.each { |k,v| d[k] = v["Horse"]  }
  puts d

.

Answer 4

Hash[] 方法派上友好地构建哈希：

hash = {
  "Wednesday"=>{"Horse"=>{"9.0"=>1, "10.0"=>1}},
  "Thursday"=>{"Horse"=>{"9.0"=>1, "10.0"=>1}}
}

x = "Horse"

Hash[
  hash.collect do |k, v|
    [ k, v[x] ]
  end
]

# => {"Wednesday"=>{"9.0"=>1, "10.0"=>1}, "Thursday"=>{"9.0"=>1, "10.0"=>1}}

.

Answer 5

我会做

hash = {"Wednesday"=>{"Horse"=>{"9.0"=>1, "10.0"=>1}}, 
"Thursday"=>{"Horse"=>{"9.0"=>1, "10.0"=>1}}}

hash.each_with_object({}) { |(k,v),h| h[k] = v['Horse']}
# => {"Wednesday"=>{"9.0"=>1, "10.0"=>1}, "Thursday"=>{"9.0"=>1, "10.0"=>1}}

.

更新

hash.each_with_object({}) { |(k,v),h| h[k] = v.shift.last}
# => {"Wednesday"=>{"9.0"=>1, "10.0"=>1}, "Thursday"=>{"9.0"=>1, "10.0"=>1}}

.

Answer 6

一种方式是使用哈希＃merge 采用块来确定两个哈希中存在的键的值：

h = { "Wednesday"=>{ "Horse"=>{ "9.0"=>1, "10.0"=>1 } }, 
      "Thursday" =>{ "Horse"=>{ "9.0"=>1, "10.0"=>1 } } }

key = "Horse"

h.merge(h) { |*_,g| g[key] }
  #=> { "Wednesday"=>{ "9.0"=>1, "10.0"=>1 },
  #     "Thursday" =>{ "9.0"=>1, "10.0"=>1 } }

.