Scala：如何按元组的第一个元素合并列表

假设我有一个清单：

[(A, a), (A, b), (A, c), (B, a), (B, d)]  

我如何将该列表变成：

[(A, [a,b,c]), (B, [a,d])]  

具有单一功能？

Thanks

The groupBy http://www.scala-lang.org/api/2.12.x/scala/collection/Seq.html#groupBy%5BK%5D(f:A=%3EK):scala.collection.immutable.Map%5BK,Repr%5D函数可以让你实现这一点：

scala> val list = List((1, 'a'), (1, 'b'), (1, 'c'), (2, 'a'), (2, 'd'))
list: List[(Int, Char)] = List((1,a), (1,b), (1,c), (2,a), (2,d))

scala> list.groupBy(_._1) // grouping by the first item in the tuple

res0: scala.collection.immutable.Map[Int,List[(Int, Char)]] = Map(2 -> List((2,a), (2,d)), 1 -> List((1,a), (1,b), (1,c)))