Question

我有一个像这样的字符串数组：

[
  "productRankingOct12:LessPopularityEPC",
  "productRankingOct12:CtrEpcJob", 
  "productRankingOct12:DeviceSpecificLPE", 
  "guidedSearch:false", 
  "guidedSearch:true", 
  "spotfront4:true", 
  "spotfront4:false",
  "postClick4:ipsa", 
  "postClick4:false",
  "trendingModule:false"
]

我必须根据给定的键：值对构建URL。我想要元素的组合：

[
  "productRankingOct12:LessPopularityEPC",
  "productRankingOct12:CtrEpcJob", 
  "productRankingOct12:DeviceSpecificLPE",
  "guidedSearch:false", 
  "guidedSearch:true", 
  "productRankingOct12:LessPopularityEPC & guidedSearch:false"
  "productRankingOct12:CtrEpcJob & guidedSearch:false"
  "productRankingOct12:DeviceSpecificLPE & guidedSearch:false"
  "productRankingOct12:LessPopularityEPC & guidedSearch:true"
  "productRankingOct12:CtrEpcJob & guidedSearch:true"
  "productRankingOct12:DeviceSpecificLPE & guidedSearch:true"
  "productRankingOct12:LessPopularityEPC & guidedSearch:false & spotfront4:true"
  "productRankingOct12:CtrEpcJob & guidedSearch:false & "spotfront4:true""
  "productRankingOct12:DeviceSpecificLPE & guidedSearch:false & "spotfront4:true""
  "productRankingOct12:LessPopularityEPC & guidedSearch:true & "spotfront4:true""
]

我试图用这个来生成排列：

def Automate_AB_tests.permutation(arg_keyVal)
  return [arg_keyVal] if arg_keyVal.size < 2
  ch = arg_keyVal[0] 
  permutation(arg_keyVal[1..-1]).each_with_object([]) do |perm,result|
    (0..perm.size).each {|i| result << perm.dup.insert(i,ch)}
  end
end

但似乎它正在采取所有元素，并且它崩溃了。我不想置换所有元素。我想要如上所述的所需输出。请任何人都可以帮助我。

Answer 1

首先，我不能＆＃39;找出你的输出中"trendingModule:false"到处都缺失的原因。

为什么输出中缺少行"spotfront4:true"，"spotfront4:false"，"postClick4:ipsa"，"postClick4:ipsa"等？包含或排除每种组合背后的逻辑是什么？

当然，由于重复的密钥，您希望排除"key:true&key:false"之类的行。

其次，这个程序产生了所有可能的组合，但这不是你想要的，对吧？ 215个元素，比你在例子中列出的更多

opts = [
  "productRankingOct12:LessPopularityEPC",
  "productRankingOct12:CtrEpcJob", 
  "productRankingOct12:DeviceSpecificLPE", 
  "guidedSearch:false", 
  "guidedSearch:true", 
  "spotfront4:true", 
  "spotfront4:false",
  "postClick4:ipsa", 
  "postClick4:false",
  "trendingModule:false"
]

normal_opts = Hash.new
opts.each do |e|
  k, v = e.split(":")
  (normal_opts[k] ||= []) << v
end

# normal_ops looks like this
#normal_opts = {
#  "productRankingOct12" => ["LessPopularityEPC", "CtrEpcJob", "DeviceSpecificLPE"],
#  "guidedSearch" => ["false", "true"],
#  "spotfornt4" => ["true", "false"],
#  "postClick4" => ["ipsa", "false"],
#  "trendingModule" => ["false"]
#}

# nobody is ever going to mantain this
result = normal_opts.keys.length.times.map do |i|
  normal_opts.keys.combination(i + 1).map do |keys|
    combs = keys.map { |k|
      normal_opts[k].map { |e|
        "#{k}:#{e}"
      }
    }
    # good luck understanding this!
    combs.first.product(*combs[1..-1]).map{|e| e.join("&")}
  end
end

puts result.flatten

如果你更好地解释自己，我可以改进我的答案

如何生成所有可能的组合

1 个答案: