Question

我有以下列表

val listA = List(("tcp",22,"All","sshd"), ("tcp6",22,"All","sshd"), ("tcp6",443,"All","docker-proxy"), ("tcp6",8000,"All","docker-proxy"), ("tcp6",4100,"All","docker-proxy"), ("tcp6",5000,"All","docker-proxy"), ("tcp",5000,"All","docker-proxy"),("tcp6",4200,"All","docker-proxy"), ("tcp6",80,"All","docker-proxy"))

listA的类型为List[(String, Int, String, String)]，我的预期输出为

val output = List(("tcp",22,"All","sshd"), ("tcp6",443,"All","docker-proxy"), ("tcp6",8000,"All","docker-proxy"), ("tcp6",4100,"All","docker-proxy"), ("tcp6",5000,"All","docker-proxy"), ("tcp6",4200,"All","docker-proxy"), ("tcp6",80,"All","docker-proxy"))

这里想要匹配listA.map(_._2) distinct我试过

val output = listA.groupBy(_._2).map {
  case (key, value) =>
  if (value.map(_._2).contains(key)) {
    value
  }
}

但上面没有给出我预期的结果（它与listA相同）。

如何获得任何人都知道的预期输出？

Answer 1

如果我理解你想要达到的目标，那就是你需要的

listA.groupBy(_._2).map{ case (_, value) => value.head }

它从每个端口组中选择第一个元素。

Scala列表中的元组匹配

1 个答案: