Ruby OpenStruct

Question

我正在使用一组OpenStruct对象，如下所示：

a=[<OpenStruct name="test1", x="6", id="1">,<OpenStruct name="test2", x="5", id="2"><OpenStruct name="test1", x="8", id="3">...]

我想将具有相同名称的OpenStruct对象分组，如下所示：

a=[<OpenStruct name="test1",x=["6","8"], id=["1","3"]>,<OpenStruct name="test2", x="5", id="2">]

我该怎么做？

Answer 1

您可以使用group_by和map方法。我认为代码是自我解释的。

a = [
  OpenStruct.new(name: "test1", x: "6", id: "1"),
  OpenStruct.new(name: "test2", x: "5", id: "2"),
  OpenStruct.new(name: "test1", x: "8", id: "3")
]

a.group_by(&:name).map do |name, as|
  OpenStruct.new(
    name: name,
    x: as.map(&:x),
    id: as.map(&:id)
  )
end
# => [#<OpenStruct name="test1", x=["6", "8"], id=["1", "3"]>, #<OpenStruct name="test2", x=["5"], id=["2"]>]

Answer 2

这样的事情应该有效：

a = [
  OpenStruct.new(name: "test1", x: "6", id: "1"),
  OpenStruct.new(name: "test2", x: "5", id: "2"),
  OpenStruct.new(name: "test1", x: "8", id: "3")
]

a.each_with_object({}) { |o, h|
  h[o.name] ||= OpenStruct.new(name: o.name, x: [], id: [])
  h[o.name][:x] << o.x
  h[o.name][:id] << o.id
}.values
#=> [#<OpenStruct name="test1", x=["6", "8"], id=["1", "3"]>, #<OpenStruct name="test2", x=["5"], id=["2"]>]

请注意x的{​​{1}}和id属性也会转换为数组。这通常是首选。

重命名属性以指示数组可能是个好主意，即test2而不是ids。