Ruby 按键值分组哈希

Question

我有一个数组，它是由 MongoDB 执行的 map/reduce 方法输出的，它看起来像这样：

[{"minute"=>30.0, "hour"=>15.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>0.0, "count"=>299.0}, 
{"minute"=>30.0, "hour"=>15.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>10.0, "count"=>244.0}, 
{"minute"=>30.0, "hour"=>15.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>1.0, "count"=>204.0}, 
{"minute"=>45.0, "hour"=>15.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>0.0, "count"=>510.0}, 
{"minute"=>45.0, "hour"=>15.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>10.0, "count"=>437.0}, 
{"minute"=>0.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>0.0, "count"=>469.0}, 
{"minute"=>0.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>10.0, "count"=>477.0}, 
{"minute"=>15.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>0.0, "count"=>481.0}, 
{"minute"=>15.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>10.0, "count"=>401.0}, 
{"minute"=>30.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>0.0, "count"=>468.0}, 
{"minute"=>30.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>10.0, "count"=>448.0}, 
{"minute"=>45.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>0.0, "count"=>485.0}, 
{"minute"=>45.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "type"=>10.0, "count"=>518.0}] 

您会注意到type 有三个不同的值，在本例中为0、1 和2，现在要做的是将这个哈希数组按其@987654327 的值分组@key，所以例如这个数组最终看起来像：

{
  :type_0 => [
    {"minute"=>30.0, "hour"=>15.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>299.0}, 
    {"minute"=>45.0, "hour"=>15.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>510.0}, 
    {"minute"=>0.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>469.0}, 
    {"minute"=>15.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>481.0}, 
    {"minute"=>30.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>468.0}, 
    {"minute"=>45.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>485.0}
  ],

  :type_1 => [
    {"minute"=>30.0, "hour"=>15.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>204.0}
  ],

  :type_10 => [
    {"minute"=>30.0, "hour"=>15.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>244.0}, 
    {"minute"=>45.0, "hour"=>15.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>437.0},
    {"minute"=>0.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>477.0}, 
    {"minute"=>15.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>401.0}, 
    {"minute"=>30.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>448.0}, 
    {"minute"=>45.0, "hour"=>16.0, "date"=>5.0, "month"=>9.0, "year"=>2011.0, "count"=>518.0}
  ]
} 

所以我知道这些示例数组确实很大，但我认为这可能是一个比我想象的更简单的问题

所以基本上每个哈希数组都将按其type 键的值分组，然后作为带有每种类型数组的哈希返回，任何帮助都会非常有帮助，甚至只是一些有用的提示将不胜感激。

Answer 1

array.group_by {|x| x['type']}

或者如果你想要符号键，你甚至可以

array.group_by {|x| "type_#{x['type']}".to_sym}

我认为这最能表达“所以基本上每个哈希数组都会按其类型键的值分组，然后作为哈希返回每种类型的数组”，即使它在输出哈希中单独留下:type 键。

Answer 2

大概是这样的吧？

mangled = a.group_by { |h| h['type'].to_i }.each_with_object({ }) do |(k,v), memo|
    tk = ('type_' + k.to_s).to_sym
    memo[tk] = v.map { |h| h = h.dup; h.delete('type'); h }
end

或者如果您不关心保留原始数据：

mangled = a.group_by { |h| h['type'].to_i }.each_with_object({ }) do |(k,v), memo|
    tk = ('type_' + k.to_s).to_sym
    memo[tk] = v.map { |h| h.delete('type'); h } # Drop the h.dup in here
end

Answer 3

by_type = {}

a.each do |h|
   type = h.delete("type").to_s
   # type = ("type_" + type ).to_sym

   by_type[ type ] ||= []
   by_type[ type ] << h      # note: h is modified, without "type" key

end

注意：这里的哈希键略有不同，我直接使用类型值作为键

如果您必须拥有示例中的哈希键，则可以添加被注释掉的行。

---

P.S.：我刚刚看到了 Tapio 的解决方案——它非常好而且很短！请注意，它仅适用于 Ruby >= 1.9

Answer 4

group_by 收集一个可枚举的集合，按块的结果分组。您不限于简单地获取此块中的键值，因此如果您想在这些集合中省略 'type'，您可以这样做，例如：

array.group_by {|x| "type_#{x.delete('type').to_i}".to_sym}

这将完全符合您的要求。

高级：这有点超出了问题的范围，但是如果您想保留原始数组，则必须复制其中的每个对象。这样就可以了：

array.map(&:dup).group_by {|x| "type_#{x.delete('type').to_i}".to_sym}