根据Scala中“List of List”的条件创建地图 [英] Create map based on condition from Future of List in Scala

问题描述

我有一个param类型的方法 Future [List [MyRes]] 。 MyRes 有两个选项字段 id 和 name 。现在，我想要创建 id 和名称的地图，如果两者都存在。我可以使用默认值创建地图，但是我不希望有默认值，只要跳过空值为0的条目。

I have method with param type Future[List[MyRes]]. MyRes has two option fields id and name. Now I want to create map of id and name if both present. I am able to create map with default value as follow but I don't want to have default value just skip the entry with null value on either.

def myMethod(myRes: Future[List[MyRes]]): Future[Map[Long, String]] = {
  myRes.map (
    _.map(
      o =>
      (o.id match {
        case Some(id) => id.toLong
        case _ => 0L 
      }) ->
      (o.name match {
        case Some(name) => name
        case _ => ""
      })
  ).toMap)

任何建议？

推荐答案

p>您正在寻找收集：）

myRes.map {  
   _.iterator
    .map { r => r.id -> r.name }
    .collect { case(Some(id), Some(name) => id -> name }
    .toMap
}

如果您的 MyRes thingy是一个案例类，那么您不需要第一个 .map ：

If your MyRes thingy is a case class, then you don't need the first .map:

myRes.map { 
  _.collect { case MyRes(Some(id), Some(name)) => id -> name }
  .toMap
}

收集就像 .map ，但它需要一个PartialFunction，并跳过未定义的元素，它有点像你的匹配语句，但没有默认值。

collect is like .map, but it takes a PartialFunction, and skips over elements on which it is not defined. It is kinda like your match statement but without the defaults.

更新：
如果我正在正确阅读您的评论，并且您想要在任一字段为无，收集将不会帮助，但您可以执行 flatMap ：

Update: If I am reading your comment correctly, and you want to log a message when either field is a None, collect won't help with that, but you can do flatMap:

 myRes.map {
   _.flatMap {
     case MyRes(Some(id), Some(name)) => Some(id -> name)
     case x => loger.warn(s"Missing fields in $x."); None
   }
   .toMap
 }

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