Ruby 中是否有一种优雅的方式来过滤散列数组的散列？答案

【问题标题】：Is there an elegant way in Ruby to filter a hash of arrays of hashses?Ruby 中是否有一种优雅的方式来过滤散列数组的散列？
【发布时间】：2020-10-23 11:57:13
【问题描述】：

假设我有这种类型的数据结构：

{
  "foo": [{state: on}, {state: off}, {state: on}],
  "bar": [{state: off}, {state: off}, {state: on}],
  "baz": [{state: on}, {state: on}, {state: on}]
}

如何以优雅的方式过滤嵌套的哈希数组，以便找回：

{
  "foo": [{state: on}, {state: on}],
  "bar": [{state: on}],
  "baz": [{state: on}, {state: on}, {state: on}]
}

【问题讨论】：

标签： ruby filter ruby-hash

【解决方案1】：

a={
    "foo": [{state: "on"}, {state: "off"}, {state: "on"}],
    "bar": [{state: "off"}, {state: "off"}, {state: "on"}],
    "baz": [{state: "on"}, {state: "on"}, {state: "on"}]
}

代码

p a.transform_values{|arr| arr.select{|h|h[:state].eql?'on'}}

结果

{:foo=>[{:state=>"on"}, {:state=>"on"}], :bar=>[{:state=>"on"}], :baz=>[{:state=>"on"}, {:state=>"on"}, {:state=>"on"}]}

【讨论】：

【解决方案2】：

给定：

data = {
  "foo": [{state: :on}, {state: :off}, {state: :on}],
  "bar": [{state: :off}, {state: :off}, {state: :on}],
  "baz": [{state: :on}, {state: :on}, {state: :on}]
}

使用#transform_values 迭代和替换哈希值，使用#select 过滤数组中的元素：

data.transform_values do |value|
  value.select { |hash| hash[:state] == :on  }
end

【讨论】：

啊哈transform_keys！我还没有看到这个。太棒了！
糟糕，应该是 transform_values