Question

我有以下数据：

List(Map("weight"->"1000","type"->"abc","match"->"first"),Ma‌p("weight"->"1000","‌type"->"abc","match"‌->"third"),Map("weig‌ht"->"182","type"->"‌bcd","match"->"secon‌d"),Map("weight"->"1‌50","type"->"bcd","m‌atch"->"fourth"),Map‌("weight"->"40","typ‌e"->"aaa","match"->"‌fifth"))

按＆＃34;分组＆＃34;我想要第一个结果＆＃34; abc＆＃34;然后＆＃34; bcd＆＃34;然后＆＃34; aaa＆＃34;

当我按＆＃34键入组时，输入＆＃34;结果映射给出第一个键为aaa，而我希望第一个键为＆＃34; abc＆＃34;和所有相应的值。

我怎样才能做到这一点？

Answer 1

如评论中所述，您需要一个有序的地图，它不是由一个简单的小组创建的。你能做的是：

val a = List(Map("weight"->"1000", "type"->"abc","match"->"first"), Map("weight"->"1000","type"->"abc","match"->"third"),Map("weig‌ht"->"182","type"->"‌bcd","match"->"secon‌d"), Map("weight"->"1‌50","type"->"bcd","m‌atch"->"fourth"), Map("weight"->"40","type"->"aaa","match"->"‌fifth"))
val sortedGrouping = SortedMap.empty[String,String] ++ a.groupBy { x => x("type") }
println(sortedGrouping)

你得到（打印）的是：

Map(aaa -> List(Map(weight -> 40, type -> aaa, match -> ‌fifth)), abc -> List(Map(weight -> 1000, type -> abc, match -> first), Map(weight -> 1000, type -> abc, match -> third)), bcd -> List(Map(weight -> 1‌50, type -> bcd, m‌atch -> fourth)), ‌bcd -> List(Map(weig‌ht -> 182, type -> ‌bcd, match -> secon‌d)))

Answer 2

我认为没有开箱即用的解决方案可以实现您的目标。 Map有两种使用方式：通过键获取值，然后迭代它。第一部分不受排序顺序的影响，因此groupBy非常有效。第二部分可以通过按所需顺序创建密钥列表，然后使用该列表以特定顺序从Map获取键值对来实现。例如：

val xs = List(Map("weight"->"1000","type"->"abc","match"->"first"),
              Map("weight"->"1000","type"->"abc","match"->"third"),
              Map("weight"->"182","type"->"bcd","match"->"second"),
              Map("weight"->"150","type"->"bcd","match"->"fourth"),
              Map("weight"->"40","type"->"aaa","match"->"‌fifth"))

import collection.mutable.{ListBuffer, Set}

// List of types in the order of appearance in your list 
val sortedKeys = xs.map(_("type")).distinct

//> sortedKeys: List[String] = List(abc, bcd, aaa)

现在，当您需要迭代时，只需执行：

val grouped = xs.groupBy(_("type"))

sortedKeys.map(k => (k, grouped(k)))
//  List[(String, List[scala.collection.immutable.Map[String,String]])] =
//  List(
//    (abc,List(Map(weight -> 1000, type -> abc, match -> first),
//              Map(weight -> 1000, type -> abc, match -> third))),
//    (bcd,List(Map(weight -> 182, type -> bcd, match -> second),
//              Map(weight -> 150, type -> bcd, match -> fourth))),
//    (aaa,List(Map(weight -> 40, type -> aaa, match -> ‌fifth)))
//  )

Scala group By First Occurrence

2 个答案: