list - Scala根据另一个列表中的值对一个列表进行排序

Question

我有两个IndexedSeq

works[Work]
order[Int]

每个对象Work都有一个带有整数值的id字段:Work.id
在订单列表中有ID，这些ID符合我们整理工作的顺序。就像，在位置0有第一个ID，因此我们需要在Works数组中找到具有与该ID对应的ID的作品，并将其放在0处，依此类推。
有没有办法通过scala做到这一点而无需经历两个循环？
一些伪数据，例如:

order = 33, 22, 11, 55

works = (33, "some text"), (55, "eeeee"), (22, "fdsfs"), (11, "fdsffds")

排序后:

order = 33, 22, 11, 55

works = (33, "some text"),(22, "fdsfs"),  (11, "fdsffds"), (55, "eeeee"),

Answer 1

您可以为此使用字典，其复杂度为O(N)(与两个嵌套循环的N * N相比，它更好):

scala> val order = List(33, 22, 11, 55)
order: List[Int] = List(33, 22, 11, 55)

scala> val works = List((33, "some text"), (55, "eeeee"), (22, "fdsfs"), (11, "fdsffds"))
works: List[(Int, String)] = List((33,some text), (55,eeeee), (22,fdsfs), (11,fdsffds))

scala> val worksMap = works.toMap
worksMap: scala.collection.immutable.Map[Int,String] = Map(33 -> some text, 55 -> eeeee, 22 -> fdsfs, 11 -> fdsffds)

scala> val newWorks = order zip order.map(worksMap)
newWorks: List[(Int, String)] = List((33,some text), (22,fdsfs), (11,fdsffds), (55,eeeee))

如果您的实体不仅仅是元组:

scala> val worksMap = (works map (_._1) zip works).toMap //any other key extraction, like `_.myKey` may be applied instead of `_._1`
worksMap: scala.collection.immutable.Map[Int,(Int, String)] = Map(33 -> (33,some text), 55 -> (55,eeeee), 22 -> (22,fdsfs), 11 -> (11,fdsffds))

scala> order.map(worksMap)
res13: List[(Int, String)] = List((33,some text), (22,fdsfs), (11,fdsffds), (55,eeeee))

如果您不想为Map花费内存-只需使用find而不是Map.apply(但它会是O(N * N)，因此速度较慢):

scala> val newWorks = order.map(x => works.find(_._1 == x).get)
newWorks: List[(Int, String)] = List((33,some text), (22,fdsfs), (11,fdsffds), (55,eeeee))

如果您不想在order不包含 key 的情况下出现异常，则可以使用flatMap:

scala> val newWorks = order.flatMap(x => works.find(_._1 == x))
newWorks: List[(Int, String)] = List((33,some text), (22,fdsfs), (11,fdsffds), (55,eeeee))

scala> order.flatMap(worksMap.get)
res15: List[(Int, String)] = List((33,some text), (22,fdsfs), (11,fdsffds), (55,eeeee))