Scala：在Seq of Seq中查找唯一的Seq

我甚至很难用语言表达我想问的问题，所以让我举个例子

我有一个类似这样的

Seq

，其中

事件

是一个案例类：

Seq(
  Seq(Occurrence1, Occurrence2, Occurrence3), 
  Seq(Occurrence2, Occurrence3, Occurrence1),
  Seq(Occurrence4),
  Seq(Occurrence5, Occurrence6),
  Seq(Occurrence6, Occurrence5)
)

我想把它浓缩成这样：

Seq(
  Seq(Occurrence1, Occurrence2, Occurrence3),
  Seq(Occurence4),
  Seq(Occurrence5, Occurrence6)
)

是否有允许我这样做的集合API？我想对该序列中的项目进行排序

您可以对列表中的元素进行排序

并将列表转换为

Set

，因为Set只包含不同的元素

例如

final case class Occurance(times: Int)

val Occurrence1 = Occurance(times = 1)
val Occurrence2 = Occurance(times = 2)
val Occurrence3 = Occurance(times = 3)
val Occurrence4 = Occurance(times = 4)
val Occurrence5 = Occurance(times = 5)
val Occurrence6 = Occurance(times = 6)

val data = Seq(
        Seq(Occurrence1, Occurrence2, Occurrence3), 
        Seq(Occurrence2, Occurrence3, Occurrence1),
        Seq(Occurrence4),
        Seq(Occurrence5, Occurrence6),
        Seq(Occurrence6, Occurrence5)
      )

data.map(_.sortBy {_.times}).toSet.toSeq

//res0: Seq[Seq[Occurance]] = Vector(List(Occurance(1), Occurance(2), Occurance(3)), List(Occurance(4)), List(Occurance(5), Occurance(6)))

data.map(_.sortBy{_.times}).distinct
//res1: Seq[Seq[Occurance]] = List(List(Occurance(1), Occurance(2), Occurance(3)), List(Occurance(4)), List(Occurance(5), Occurance(6)))

• 您也可以

  不重复
  ，而不是
  .toSet.toSeq
  ，

例如

final case class Occurance(times: Int)

val Occurrence1 = Occurance(times = 1)
val Occurrence2 = Occurance(times = 2)
val Occurrence3 = Occurance(times = 3)
val Occurrence4 = Occurance(times = 4)
val Occurrence5 = Occurance(times = 5)
val Occurrence6 = Occurance(times = 6)

val data = Seq(
        Seq(Occurrence1, Occurrence2, Occurrence3), 
        Seq(Occurrence2, Occurrence3, Occurrence1),
        Seq(Occurrence4),
        Seq(Occurrence5, Occurrence6),
        Seq(Occurrence6, Occurrence5)
      )

data.map(_.sortBy {_.times}).toSet.toSeq

//res0: Seq[Seq[Occurance]] = Vector(List(Occurance(1), Occurance(2), Occurance(3)), List(Occurance(4)), List(Occurance(5), Occurance(6)))

data.map(_.sortBy{_.times}).distinct
//res1: Seq[Seq[Occurance]] = List(List(Occurance(1), Occurance(2), Occurance(3)), List(Occurance(4)), List(Occurance(5), Occurance(6)))

相关问题-

• 您可以对列表中的元素进行排序
• 并将列表转换为

  Set

  ，因为Set只包含不同的元素

例如

final case class Occurance(times: Int)

val Occurrence1 = Occurance(times = 1)
val Occurrence2 = Occurance(times = 2)
val Occurrence3 = Occurance(times = 3)
val Occurrence4 = Occurance(times = 4)
val Occurrence5 = Occurance(times = 5)
val Occurrence6 = Occurance(times = 6)

val data = Seq(
        Seq(Occurrence1, Occurrence2, Occurrence3), 
        Seq(Occurrence2, Occurrence3, Occurrence1),
        Seq(Occurrence4),
        Seq(Occurrence5, Occurrence6),
        Seq(Occurrence6, Occurrence5)
      )

data.map(_.sortBy {_.times}).toSet.toSeq

//res0: Seq[Seq[Occurance]] = Vector(List(Occurance(1), Occurance(2), Occurance(3)), List(Occurance(4)), List(Occurance(5), Occurance(6)))

data.map(_.sortBy{_.times}).distinct
//res1: Seq[Seq[Occurance]] = List(List(Occurance(1), Occurance(2), Occurance(3)), List(Occurance(4)), List(Occurance(5), Occurance(6)))

• 您也可以

  不重复
  ，而不是
  .toSet.toSeq
  ，

例如

final case class Occurance(times: Int)

val Occurrence1 = Occurance(times = 1)
val Occurrence2 = Occurance(times = 2)
val Occurrence3 = Occurance(times = 3)
val Occurrence4 = Occurance(times = 4)
val Occurrence5 = Occurance(times = 5)
val Occurrence6 = Occurance(times = 6)

val data = Seq(
        Seq(Occurrence1, Occurrence2, Occurrence3), 
        Seq(Occurrence2, Occurrence3, Occurrence1),
        Seq(Occurrence4),
        Seq(Occurrence5, Occurrence6),
        Seq(Occurrence6, Occurrence5)
      )

data.map(_.sortBy {_.times}).toSet.toSeq

//res0: Seq[Seq[Occurance]] = Vector(List(Occurance(1), Occurance(2), Occurance(3)), List(Occurance(4)), List(Occurance(5), Occurance(6)))

data.map(_.sortBy{_.times}).distinct
//res1: Seq[Seq[Occurance]] = List(List(Occurance(1), Occurance(2), Occurance(3)), List(Occurance(4)), List(Occurance(5), Occurance(6)))

---

不知何故相关的问题-

很高兴能提供帮助，不过您可能需要找到排序功能。欢迎使用scala

data.map（u.toSet）.toSet

应该足够了，我想我很乐意为您提供帮助，不过您可能需要找到排序函数。欢迎使用scala

data.map（u.toSet）.toSet

应该足够了，我认为scala 2.13的可能副本添加了

distinctBy

，这在这里非常有用。2.12及更早版本没有。有人系统地浏览了我的个人资料，并对我在Stackoverflow上提供的所有问题和答案进行了否决。管理员们似乎也很满足于让它发生。因此，我想请这篇文章的任何后续访问者真诚地投票，并留下评论，解释你为什么认为这篇文章有用或不有用。我并不是想沉溺于一些可悲的针锋相对，我只是想帮助下一个人。非常感谢。Scala 2.13的可能副本通过添加了

distinctBy，这在这里非常有用。2.12及更早版本没有。有人系统地浏览了我的个人资料，并对我在Stackoverflow上提供的所有问题和答案进行了否决。管理员们似乎也很满足于让它发生。因此，我想请这篇文章的任何后续访问者真诚地投票，并留下评论，解释你为什么认为这篇文章有用或不有用。我并不是想沉溺于一些可悲的针锋相对，我只是想帮助下一个人。非常感谢。