如何根据键将数据值排列到数组中 - Ruby答案

【问题标题】：How to arrange data values into arrays based on their key - Ruby如何根据键将数据值排列到数组中 - Ruby
【发布时间】：2016-02-13 21:32:25
【问题描述】：

我从我的模型中获得了一些数据（在一个名为 results 的变量中），我想根据它们的键将它组织成数组。即，每个genreName 值都应该被推入genres[] 数组，每个actorID 应该被推入cast[] 数组。

results：

#<Genre genreName: "Fantasy">
#<Genre genreName: "Comedy">
#<Genre genreName: "Children">
#<Genre genreName: "Animation">
#<Genre genreName: "Adventure">
#<Actor actorID: "tom_hanks", actorName: "Tom Hanks">
#<Actor actorID: "tim_allen", actorName: "Tim Allen">
#<Actor actorID: "don_rickles", actorName: "Don Rickles">
#<Actor actorID: "jim_varney", actorName: "Jim Varney">
#<Actor actorID: "wallace_shawn", actorName: "Wallace Shawn">
#<Director directorID: "john_lasseter", directorName: "John Lasseter">
#<Country countryName: "USA">
#<Location locationName: "N/A">

你有什么建议？我正在尝试在 Ruby 中完成此操作。

编辑

也许将所有对象值添加到哈希中会更有意义，但我担心唯一的键名......

【问题讨论】：

您应该将results 显示为有效的Ruby 对象。假设它是一个字符串数组：results = ["#<Genre genreName: "Fantasy">, ..."#<Location locationName: "N/A">"].
...如果results 是一个类实例数组，那需要说明。

标签： arrays ruby sorting

【解决方案1】：

genre_names = results.map { |object| object.genreName if object.class.name == "Genre" }

这将遍历结果中的每个对象，并通过仅包含“流派”类的对象返回一个包含每个流派名称字段的数组。然后，您可以对其他字段执行类似操作。

【讨论】：

【解决方案2】：

（编辑：我可能误解了这个问题。我将数据视为字符串数组，但它们似乎更有可能是类实例。）

results = <<_.lines
#<Genre genreName: "Fantasy">
#<Genre genreName: "Comedy">
#<Genre genreName: "Children">
#<Genre genreName: "Animation">
#<Genre genreName: "Adventure">
#<Actor actorID: "tom_hanks", actorName: "Tom Hanks">
#<Actor actorID: "tim_allen", actorName: "Tim Allen">
#<Actor actorID: "don_rickles", actorName: "Don Rickles">
#<Actor actorID: "jim_varney", actorName: "Jim Varney">
#<Actor actorID: "wallace_shawn", actorName: "Wallace Shawn">
#<Director directorID: "john_lasseter", directorName: "John Lasseter">
#<Country countryName: "USA">
#<Location locationName: "N/A">
_

你可以这样做：

R = /
    \b        # Match a word break
    (genreName|actorID|cast) # Match one of three strings in capture group 1
    \b        # Match a word break 
    (?=       # Begin a positive lookahead 
      :\s+\"  # Match : >= 1 whitespace double quote
      (\w+)   # Match >= 1 word characters in capture group 2
      \"      # Match double quote
    )         # End postive lookahead
    /x        # Extended/free-spacing regex definition mode


h = results.each_with_object({ genreName: [], actorID: [], cast: [] }) { |s,h|
  s.scan(R) { h[$1.to_sym] << $2 } }
  #=> {:genreName=>["Fantasy", "Comedy", "Children", "Animation", "Adventure"],
  #    :actorID=>["tom_hanks", "tim_allen", "don_rickles", "jim_varney", "wallace_shawn"],
  #    :cast=>[]}

然后

actorID = h[:actorID]

等等。

【讨论】：