Scala 的几种group汇合操作
scala的汇合中有如下几种group操作
groupBy按特定条件对汇合元素进行分类grouped将汇合拆分成指定长度的子集合groupMap应用办法按特定条件对汇合的元素进行分类并解决每个元素groupMapReduce应用办法按特定条件对汇合中的元素进行分类，别离进行解决，最初将它们reduce
这些办法在scala.collection.IterableOps中定义
groupBy
def groupBy[K](f: (A) => K): immutable.Map[K, C]返回immutable Map，每个Map由一个键和一个原始类型的值的汇合组成
为了在生成的Map中解决这个值的汇合，Scala提供了mapValues办法
def mapValues[W](f: (V) => W): Map[K, W]这个groupBy / mapValues组合对于解决从分组生成的Map的值十分不便
然而从Scala 2.13开始，办法mapValues不再可用
// Example: groupBycase class Pet(species: String, name: String, age: Int) val pets = List( Pet("cat", "sassy", 2), Pet("cat", "bella", 3), Pet("dog", "poppy", 3), Pet("dog", "bodie", 4), Pet("dog", "poppy", 2), Pet("bird", "coco", 2), Pet("bird", "kiwi", 1))pets.groupBy(_.species)// res5: scala.collection.immutable.Map[String,List[Pet]] = HashMap(// bird -> List(Pet(bird,coco,2), Pet(bird,kiwi,1)), // dog -> List(Pet(dog,poppy,3), Pet(dog,bodie,4), Pet(dog,poppy,2)), // cat -> List(Pet(cat,sassy,2), Pet(cat,bella,3)))groupBy / mapValues组合
// Example: groupBy and mapValuespets.groupBy(_.species).mapValues(_.map(_.name)).toMap// warning: method mapValues in trait MapOps is deprecated (since 2.13.0): Use .view.mapValues(f). A future // // version will include a strict version of this method (for now, .view.mapValues(f).toMap).// val res7: scala.collection.immutable.Map[String,List[String]] = Map(// bird -> List(coco, kiwi), // dog -> List(poppy, bodie, poppy), // cat -> List(sassy, bella))groupMap
呈现了一种新办法groupMap，按特定条件对汇合的元素进行分类并解决每个元素
def groupMap[K, B](key: (A) => K)(f: (A) => B): immutable.Map[K, CC[B]]比照下面的mapValues办法
// Example: groupMappets.groupMap(_.species)(_.name)// val res8: scala.collection.immutable.Map[String,List[String]] = Map(// cat -> List(sassy, bella), // bird -> List(coco, kiwi), // dog -> List(poppy, bodie, poppy))groupMapReduce
有时，咱们须要在对汇合进行分组后对Map值进行reduce
def groupMapReduce[K, B](key: (A) => K)(f: (A) => B)(reduce: (B, B) => B): immutable.Map[K, B]// Example groupMapReducepets.groupMapReduce(_.species)(_ => 1)(_ + _)//val res9: scala.collection.immutable.Map[String,Int] = Map(// cat -> 2, // bird -> 2, // dog -> 3)grouped
汇合拆分成指定长度的子集合
def grouped(size: Int): Iterator[C]尝试将一个长度为9的汇合进行切分
val list = List(1, 2, 3, 4, 5, 6, 7, 8, 9)list.grouped(4).toList// val res11: List[List[Int]] = List(List(1, 2, 3, 4), List(5, 6, 7, 8), List(9))