Question

String = "Mod1:10022932,10828075,5946410,13321905,5491120,5030731|Mod2:22704455,22991440,22991464,21984312,21777721,21777723,21889761,21939852,23091478,22339903,23091485,22099714,21998260,22364832,21939858,21944274,21944226,22800221,22704443,21777728,21777719,21678184,21998265,21834900,21984331,22704454,21998261,21944214,21862610,21836482|Mod3:10828075,13321905,5491120,5946410,5030731,15806212,4100566,4787137,2625339,2408317,2646868,19612047,2646862,11983534,8591489,19612048,10249319,14220471,15806209,13330887,15075124,17656842,3056657,5086273|Mod4:10828075,5946410,13321905,5030731,5491120,4787137,4100566,15806212,2625339,3542205,2408317,2646862,2646868|Mod5:10022932;0.2512,10828075;0.2093,5030731;0.1465,5946410;0.1465,4787137;0.1465,2625339;0.0143,5491120;0.0143,13321905;0.0143,3542205;0.0143,15806212;0.0119,4100566;0.0119,19612047;0.0100,2408317;0.0100"

如何将其拆分出来以便我可以获得每个标题（Mod1，Mod2 ..）和属于每个标题的ID。

这是我到目前为止尝试过的，它正在删除管道之后的所有内容，这是我不想要的。

mod_name = string.split(":")[0]
mod_ids = string.split(":")[1] #This gets me the ID's but also include the |Mod*
ids = mod_mod_ids.split("|").first.strip #Only returns Id's before the first "|"

期望的输出：我需要将mod_name和mod_ids保存到各自的列

mod_name = #name ("Mod1...Mod2 etc) #string
mod_ids = #ids (All Ids after the ":" in Mod*:) #array

Answer 1

有几种方法可以做到这一点：

# This will split the string on "|" and ":" and will return:
#   %w( Mod1 id1 Mod2 id2 Mod3 id3 ... )
ids = string.split(/[|:]/)

# This will first split on "|", and for each string, split it again on ":" and returs:
#   [ %w(Mod1 id1), %w(Mod2 id2), %w(Mod3 id3), ... ]
ids = string.split("|").map { |str| str.split(":") }

Answer 2

我认为这可以满足您的需求：

ids = string.split("|").map {|part| [part.split(":")[0], part.split(":")[1].split(/,|;/)]}

Answer 3

如果您希望通过标题轻松访问哈希，那么您可以这样做：

 str.split('|').inject({}){|h,x| k,v = x.split(':'); h[k] = v.split(','); h}

  => {
      "Mod1"=>["10022932", "10828075", "5946410", "13321905", "5491120", "5030731"], 
      "Mod2"=>["22704455", "22991440", "22991464", "21984312", "21777721", "21777723", "21889761", "21939852", "23091478", "22339903", "23091485", "22099714", "21998260", "22364832", "21939858", "21944274", "21944226", "22800221", "22704443", "21777728", "21777719", "21678184", "21998265", "21834900", "21984331", "22704454", "21998261", "21944214", "21862610", "21836482"], 
      "Mod3"=>["10828075", "13321905", "5491120", "5946410", "5030731", "15806212", "4100566", "4787137", "2625339", "2408317", "2646868", "19612047", "2646862", "11983534", "8591489", "19612048", "10249319", "14220471", "15806209", "13330887", "15075124", "17656842", "3056657", "5086273"], 
      "Mod4"=>["10828075", "5946410", "13321905", "5030731", "5491120", "4787137", "4100566", "15806212", "2625339", "3542205", "2408317", "2646862", "2646868"], 
      "Mod5"=>["10022932;0.2512", "10828075;0.2093", "5030731;0.1465", "5946410;0.1465", "4787137;0.1465", "2625339;0.0143", "5491120;0.0143", "13321905;0.0143", "3542205;0.0143", "15806212;0.0119", "4100566;0.0119", "19612047;0.0100", "2408317;0.0100"]
     }

Answer 4

未测试：

all_mods = {}
string.split("|").each do |fragment|
  mod_fragments = fragment.split(":")
  all_mods[mod_fragments[0]] = mod_fragments[1].split(",")
end

Answer 5

由于@tillerjs的帮助，我最终得到了什么。

data = sting.split("|")
data.each do |mod|
   module_name = mod.split(":")[0]
   recommendations = mod.split(":")[1]
end

Ruby拆分字符串

5 个答案: