scala 如何从序列创建不可变的映射/集？
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How to create an immutable map/set from a seq?
提问by Michael Barker
I am try to construct immutable Sets/Maps from a Seq. I am currently doing the following:
我试图从 Seq 构造不可变的 Sets/Maps。我目前正在做以下事情：
val input: Seq[(String, Object)] = //.....
Map[String, Object]() ++ input
and for sets
和套
val input: Seq[String] = //.....
Set[String]() ++ input
Which seems a little convoluted, is there a better way?
这似乎有点令人费解，有没有更好的方法？
回答by Eastsun
In Scala 2.8:
在 Scala 2.8 中：
Welcome to Scala version 2.8.0.r20327-b20091230020149 (Java HotSpot(TM) Client VM, Java 1.6.
Type in expressions to have them evaluated.
Type :help for more information.
scala> val seq: Seq[(String,Object)] = ("a","A")::("b","B")::Nil
seq: Seq[(String, java.lang.Object)] = List((a,A), (b,B))
scala> val map = Map(seq: _*)
map: scala.collection.immutable.Map[String,java.lang.Object] = Map(a -> A, b -> B)
scala> val set = Set(seq: _*)
set: scala.collection.immutable.Set[(String, java.lang.Object)] = Set((a,A), (b,B))
scala>
Edit 2010.1.12
编辑 2010.1.12
I find that there is a more simple way to create set.
我发现有一种更简单的方法来创建集合。
scala> val seq: Seq[(String,Object)] = ("a","A")::("b","B")::Nil
seq: Seq[(String, java.lang.Object)] = List((a,A), (b,B))
scala> val set = seq.toSet
set: scala.collection.immutable.Set[(String, java.lang.Object)] = Set((a,A), (b,B))
回答by Chris Stivers
To convert a Seqto a Map, simply call toMapon the Seq. Note that the elements of the Seqmust be Tuple2ie. (X,Y)or (X->Y)
要转换Seq到Map，简单地调用toMap上Seq。请注意， 的元素Seq必须是Tuple2ie。(X,Y)或者(X->Y)
scala> val seq: Seq[(String,String)] = ("A","a")::("B","b")::("C","c")::Nil
seq: Seq[(java.lang.String, java.lang.String)] = List((A,a), (B,b), (C,c))
scala> seq.toMap
res0: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map((A,a), (B,b), (C,c))
To convert a Seqto a Set, simply call toSeton the Seq.
要转换Seq到Set，简单地调用toSet上Seq。
scala> val seq: Seq[String] = "a"::"b"::"c"::Nil
seq: Seq[java.lang.String] = List(a, b, c)
scala> seq.toSet
res1: scala.collection.immutable.Set[java.lang.String] = Set(a, b, c)
